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71. Let for every non-negative random variable $\xi$ the equalities hold
$$
\begin{aligned}
& L_{*} \xi=\sup \sum_{i} \mathrm{P}\left(A_{i}\right) \inf _{\omega \in A_{i}} \xi(\omega), \\
& L^{*} \xi=\inf \sum_{i} \mathrm{P}\left(A_{i}\right) \sup _{\omega \in A_{i}} \xi(\omega),
\end{aligned}
$$
where sup and inf ar... | Solution. The inequality $L_{*} \xi \leqslant L^{*} \xi$ follows from
$$
\sum_{i} \mathrm{P}\left(A_{i}\right) \inf _{\omega \in A_{i}} \xi(\omega) \leqslant \sum_{i} \mathrm{P}\left(A_{i}\right) \sup _{\omega \in A_{i}} \xi(\omega)
$$
(see the definitions of $L_{*} \xi$ and $L^{*} \xi$).
Now let the quantity $\xi$ ... | L_{*}\xi=L^{*}\xi=\mathrm{E}\xi | Algebra | proof | Yes | Yes | olympiads | false | 33,799 |
73. (On the connection between Lebesgue and Riemann integration.) Let the Borel function $f=f(x), x \in \mathbb{R}$, be integrable with respect to the Lebesgue measure $\lambda$ $\left(\int_{\mathbb{R}}|f(x)| d \lambda < \infty\right)$. For any $\varepsilon > 0$, there exist
(a) a step function $f_{\varepsilon}(x)=\su... | Solution. (a) Without loss of generality, we will assume that $f$ takes non-negative values (in the general case, one should consider the decomposition $f=f^{+}-f^{-}$). Fix any $\varepsilon>0$ and take $n=n_{\varepsilon} \geqslant 1$ such that
$$
\int_{f(x) \geqslant n \varepsilon} f(x) d \lambda+\int_{|x|>n} f(x) d ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,801 |
7. (See [12].) The Lebesgue integral $\mathrm{E} \xi$ is not defined when $\mathrm{E} \xi^{-} = \mathrm{E} \xi^{+} = \infty$ (see B1.II.6, definition 2). Suppose that for all $a, b > 0$ the limit
$$
\overline{\mathrm{E}} \xi = \lim _{n}\left(\mathrm{E}\left[\xi^{+}\right]_{a n} - \mathrm{E}\left[\xi^{-}\right]_{b n}\r... | Solution. (a) Suppose that $\tilde{\mathrm{E}} \xi$ exists and $n \mathrm{P}(|\xi|>n) \rightarrow 0$. For given $a, b>0$, choose (any) integer $c>a \vee b$. Then
$$
\begin{aligned}
\left|\mathrm{E} \xi I(|\xi|a n)+b n \mathrm{P}(|\xi|>b n)+ \\
+\mathrm{E}|\xi| I(a n\min \{a n, b n\}) \rightarrow 0
\end{aligned}
$$
In... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,803 |
76. Show that the function defined on $[0, \infty)$
$$
f(x)= \begin{cases}1, & x=0 \\ \frac{\sin x}{x}, & x>0\end{cases}
$$
is Riemann integrable and the integral of it is the Dirichlet integral
$$
(\mathrm{R}) \int_{0}^{\infty} f(x) d x=\frac{\pi}{2}
$$
However, establish that the function $f(x)$ is not Lebesgue i... | Solution. Integrating by parts, we find that for $0<a<b$ the equality holds
$$
\text { (R) } \int_{a}^{b} \frac{\sin x}{x} d x=\frac{1-\cos b}{b}-\frac{1-\cos a}{a}+(\mathrm{R}) \int_{a}^{b} \frac{1-\cos x}{x^{2}} d x
$$
Therefore, for such $a$ and $b$, we have
$$
\left|(\mathrm{R}) \int_{a}^{b} \frac{\sin x}{x} d x... | \frac{\pi}{2} | Calculus | proof | Yes | Yes | olympiads | false | 33,804 |
77. Provide an example of a function $f=f(x)$ on $[0,1]$, which is bounded and Lebesgue integrable, but for which it is impossible to find a function $g=g(x)$, Riemann integrable and coinciding with $f=f(x)$ almost everywhere according to the Lebesgue measure on $[0,1]$. | Solution. As $f$, one can take the indicator of a nowhere dense set $\mathscr{C}$ with positive Lebesgue measure. Then the set of discontinuity points of the function $f$ is precisely $\mathscr{C}$, and by the criterion for Riemann integrability (see problem II.6.72), $f$ is not Riemann integrable. On the other hand, t... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,805 |
78. Newton used the concept of the integral as an operation inverse to differentiation. In other words, a function $f=f(x)$ on $[a, b]$, $a<b$, was considered integrable if there existed an antiderivative $F=F(x)$ such that $F^{\prime}(x)=f(x)$ on $[a, b]$ (one-sided derivatives are taken at the endpoints $a$ and $b$).... | Solution. As $f(x)$, one should take $F^{\prime}(x)$, where
$$
F(x)=x^{2} \sin \frac{1}{x^{2}}
$$
Remark. In this context, we will present the construction of the generalized Riemann integral, which includes, as can be shown, the Lebesgue, Riemann, and Newton integrals (see [44]). Specifically, the number $I=(\mathrm... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,806 |
79. When determining the Riemann integral of a function \( f = f(x) \), it is not assumed that this function is Borel measurable. Provide an example of a function \( f(x) \) that is not Borel measurable but is Riemann integrable on the interval \([0,1]\). | Solution. The set of Borel subsets of the interval $[0,1]$ has the power of the continuum $\mathfrak{c}$, while the power of Lebesgue subsets is $2^{\mathfrak{c}}$ (see problem II.2.28). From this, it follows that if $\mathscr{C}$ is a Cantor set on $[0,1]$, then there exists a subset $C \subseteq \mathscr{C}$ which is... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,807 |
80. Provide an example of a bounded Borel function $f=f(x, y)$ on $\mathbb{R}^{2}$ such that (for $y \in \mathbb{R}$ and $x \in \mathbb{R}$ respectively)
$$
\int_{\mathbb{R}} f(x, y) d x=0, \quad \int_{\mathbb{R}} f(x, y) d y=0
$$
however, this function is not Lebesgue integrable on ( $\mathbb{R}^{2}, \mathscr{B}\lef... | Solution. Take any bounded odd Borel function $g=g(x), x \in \mathbb{R}$,
$$
0<\int_{\mathbb{R}}|g(x)| d x<\infty
$$
and define $f(x, y)=g(x-y)$. Then
$$
\int_{\mathbb{R}} f(x, y) d x=\int_{\mathbb{R}} g(x-y) d x=\int_{\mathbb{R}} g(x) d x=0
$$
Similarly, it can be established that $\int_{\mathbb{R}} f(x, y) d y=0$... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,808 |
81. (On Fubini's Theorem.) Show that Fubini's theorem remains valid if the finiteness of the measures \(\mu_{1}\) and \(\mu_{2}\) involved in its formulation is replaced by their \(\sigma\)-finiteness.
The following example shows that without the assumption of \(\sigma\)-finiteness, the statement of Fubini's theorem m... | Solution. The required statement can be derived by a limiting transition from Fubini's theorem for finite measures (which become probability measures after appropriate normalization). | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,809 |
82. Let it be known that for a random variable $\xi$ its expected value $\mathrm{E} \xi$ is negative and $\mathrm{E} e^{\theta \xi}=1$ for some $\theta \neq 0$. Show that then $\theta>0$. | Solution. By Jensen's inequality $e^{\theta \mathrm{E} \xi} \leqslant \mathrm{E} e^{\theta \xi}=1$, which for $\mathrm{E} \xi0$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,810 |
83. Let $h=h(t, x)$ be a function defined on the set $[a, b] \times \mathbb{R}$, where $a, b \in \mathbb{R}$ and $a < b$.
(a) Suppose that
1) for each $x \in \mathbb{R}$, the function $h(t, x), t \in [a, b]$, is continuous;
2) for each $t \in [a, b]$, the function $h(t, x), x \in \mathbb{R}$, is $\mathscr{B}(\mathbb{... | Solution. (a) The functions
$$
\begin{aligned}
h_{n}(t, x) & = \\
= & \sum_{i=0}^{n-1} h\left(a+(b-a) \frac{i}{n}, x\right) I_{\left[a+(b-a) \frac{i}{n}, a+(b-a) \frac{i+1}{n}\right]}(t)+h(b, x) I_{\{b\}}(t)
\end{aligned}
$$
are $\mathscr{B}([a, b]) \times \mathscr{B}(\mathbb{R})$-measurable and converge to $h(t, x)$... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,811 |
84. (a) Consider the equation
$$
Z_{t}=B_{t}+\int_{0}^{t} Z_{s-} d A_{s}
$$
where $A_{t}$ and $B_{t}$ are right-continuous (for $t \geqslant 0$) and have left limits (for $t>0$) functions of (locally) bounded variation.
Show that in the class of locally bounded functions, this equation has a unique solution $\mathsc... | Solution. (a) Let $C_{t}=A_{t}-\sum_{00$, and $G_{0}=1$ by the formula for integration by parts we get
$$
\begin{aligned}
\mathscr{E}_{t}(A)=F_{t} G_{t} & =1+\int_{0}^{t} F_{s} d G_{s}+\int_{0}^{t} G_{s-} d F_{s}= \\
& =1+\sum_{00$ for all $0 \leqslant s \leqslant t$ and, secondly,
$$
\begin{aligned}
Y_{t} \leqslant ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,812 |
85. Let $f=f(x)$ be a convex function defined on $\mathbb{R}$.
(a) The Legendre transformation of the function $f$ is defined as
$$
g(y)=\sup _{x}[x y-f(x)]
$$
Establish that $g=g(y)$ is a convex function on $\mathbb{R}$, and the Legendre transformation $\sup _{y}[x y-g(y)]$ of $g$ is nothing but $f=f(x)$.
Remark. ... | Solution. (a) The convexity of the function $g$ is established by direct verification.
It is easy to deduce from the convexity of the function $f$ that at any point $x \in \mathbb{R}$, $f$ has right $h_{+}(x)$ and left $h_{-}(x)$ derivatives, and $h_{-}(x)$ and $h_{+}(x)$ are non-decreasing in $x$ and $h_{-}(x) \leqsl... | (y)=\int_{0}^{y}F^{-1}(z)=F^{-1}(y)y-\mathrm{E}(F^{-1}(y)-X)^{+} | Algebra | proof | Yes | Yes | olympiads | false | 33,813 |
86. Let $f(x)=\sup _{\alpha}\left(a_{\alpha}^{\top} x+b_{\alpha}\right)$ be a finite function in a convex domain $x \in C \subset \mathbb{R}^{d}$ for some $a_{\alpha} \in \mathbb{R}^{d}, b_{\alpha} \in \mathbb{R}$, where $\alpha$ runs through an arbitrary set of indices.
(a) Prove that for any $x, y \in C$, for all $\... | Solution. (a) The statement is verified by elementary checking.
(b) We will show that for any convex function $f=f(x)$ and any point $x_{0} \in C$, there exists a hyperplane of the form $y=a^{\top} x+b$, passing through $x_{0}$, such that
$$
f(x) \geqslant a^{\top} x+b, \quad x \in C
$$
This, of course, will ensure ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,814 |
87. Show that for $a, b \in \mathbb{R}$ and $r \geqslant 0$ the so-called $c_{r}$-inequalities hold
$$
|a+b|^{r} \leqslant c_{r}\left(|a|^{r}+|b|^{r}\right)
$$
where $c_{r}=1$ for $r<1$ and $c_{r}=2^{r-1}$ for $r \geqslant 1$. | Solution. For $r \geqslant 1$, the given inequality follows from the convexity of the function $f(x)=|x|^{r}, x \in \mathbb{R}$:
$$
\left|\frac{a+b}{2}\right|^{r}=f\left(\frac{a+b}{2}\right) \leqslant \frac{f(a)+f(b)}{2}=\frac{|a|^{r}+|b|^{r}}{2}
$$
For $0 \leqslant r<1$, the function $g(x)=x^{r}, x \geqslant 0$, is ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,815 |
88. (Weyl's Uncertainty Principle.) Let the random variable $\xi$ have a smooth density $f=f(x)$, and
$$
x f(x) \rightarrow 0, \quad|x| \rightarrow \infty
$$
Prove that
$$
\mathrm{E} \xi^{2} \cdot \mathrm{E}\left|\frac{f^{\prime}(\xi)}{f(\xi)}\right|^{2} \geqslant 1
$$ | Solution. The statement follows from the Cauchy-Bunyakovsky inequality:
$$
1=\int_{\mathbb{R}} f(x) d x=-\int_{\mathbb{R}} x f^{\prime}(x) d x=-\mathrm{E} \xi \cdot \frac{f^{\prime}(\xi)}{f(\xi)} \leqslant \sqrt{\mathrm{E} \xi^{2}} \cdot \sqrt{\mathrm{E}\left|\frac{f^{\prime}(\xi)}{f(\xi)}\right|^{2}}
$$ | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,816 |
90. Let $\xi$ and $\zeta$ be non-negative random variables such that for any $x>0$ the inequality
$$
\mathrm{P}(\xi \geqslant x) \leqslant x^{-1} \mathrm{E} \zeta I(\xi \geqslant x)
$$
holds. Show that then for any $p>1$ the inequality
$$
\mathrm{E} \xi^{p} \leqslant\left(\frac{p}{p-1}\right)^{p} \mathrm{E} \zeta^{p... | Solution. By Fubini's theorem
$$
\begin{aligned}
\mathrm{E} \xi^{p}=\int_{\mathbb{R}_{+}} p x^{p-1} \mathrm{P}(\xi>x) d x \leqslant \int_{\mathbb{R}_{+}} p x^{p-2} \mathrm{E} \zeta I(\xi \geqslant x) d x= \\
=\mathrm{E} \zeta \int_{0}^{\xi} p x^{p-2} d x=\frac{p \cdot \mathrm{E} \zeta \xi^{p-1}}{p-1} .
\end{aligned}
$... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,818 |
91. Let $X$ and $Y$ be non-negative random variables for which there exist $\alpha > 1$ and $\beta > 0$ such that for all $x > 0$ the inequality
$$
\mathrm{P}(X > \alpha x, Y \leqslant x) \leqslant \beta \mathrm{P}(X > x)
$$
holds. Let also $f$ be a non-negative increasing function, $f(0) = 0$, and
$$
L_{f}(\alpha) ... | Solution. Since $\mathrm{E} f(X)x) d f(x) \geqslant \int_{\mathbb{R}_{+}} \mathbf{P}(X>\alpha x, Y \leqslant x) d f(x)= \\
= & \mathrm{E} \int_{\mathbb{R}_{+}} I(X>\alpha x, Y \leqslant x) d f(x) \geqslant \mathbf{E} \int_{\mathbb{R}_{+}} I(X>\alpha x) d f(x)- \\
& -\mathbb{E} \int_{\mathbb{R}_{+}} I(Y>y) d f(y)=\mathb... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,819 |
92. Let $X$ be a random variable with distribution function $F=F(x)$, satisfying the Lipschitz condition:
$$
|F(x)-F(y)| \leqslant L|x-y|, \quad x, y \in \mathbb{R},
$$
for some constant $L>0$. Show that $X$ has a density $f=f(x)$ and for almost all $x \in \mathbb{R}$ (with respect to the Lebesgue measure) the inequa... | Solution. Let $\mathrm{P}_{X}$ denote the distribution of the random variable $X$, which is a probability measure on $(\mathbb{R}, \mathscr{B}(\mathbb{R}))$. We will show that this measure is absolutely continuous with respect to the Lebesgue measure $\lambda$ defined on Borel sets.
By the regularity of the Lebesgue m... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,820 |
93. (See [100].) Let the random variable $\xi$ have a unimodal distribution density with a maximum at the point $m=0$, which is non-decreasing to the left and non-increasing to the right of $m$. Such a point of maximum is called the mode or peak of the distribution.
(a) Let $g$ be an even function that is increasing o... | Solution. (a) The distribution function $F$ of the quantity $|\xi|$ is concave (since the density $f$ of this quantity decreases on $\mathbb{R}_{+}$), therefore $F(x) \leqslant F(\varepsilon)+f(\varepsilon)(x-\varepsilon)$ for all $x \geqslant 0$. The latter means that $|\xi| \succcurlyeq \zeta \eta$, i.e.
$$
\mathrm{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,821 |
94. Let $\xi_{1}, \ldots, \xi_{n}$ be i.i.d. random variables, $\mathrm{P}\left(\xi_{1}>0\right)=1$ and $\mathrm{D} \ln \xi_{1}=\sigma^{2}$. Show that for $\varepsilon>0$ the inequality
$$
\mathrm{P}\left(\xi_{1} \cdots \xi_{n} \leqslant\left(\mathrm{E} \xi_{1}\right)^{n} e^{n \varepsilon}\right) \geqslant 1-\frac{\si... | Solution. The statement follows from Chebyshev's inequality
$$
\mathrm{P}\left(\left|S_{n}-\mathrm{E} S_{n}\right| \leqslant n \varepsilon\right) \geqslant 1-\frac{\mathrm{D} S_{n}}{n^{2} \varepsilon^{2}}
$$
applied to $S_{n}=\sum_{i=1}^{n} \ln \xi_{i}$, as well as Jensen's inequality $\ln \mathrm{E} \xi_{1} \geqslan... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,822 |
95. Let $\mathrm{P}$ and $\mathrm{Q}$ be two probability measures on ( $\Omega, \mathscr{F}$ ). Establish the equivalence of the following properties:
(1) the measure $Q$ is absolutely continuous with respect to $P$, and for some constant $L \geqslant 1$ the inequality
$$
\frac{d \mathbf{Q}}{d \mathrm{P}} \leqslant L... | Solution. In the case when $L=1$ or $\alpha=1$, measures $\mathbf{P}$ and $\mathbf{Q}$ coincide, and thus the equivalence of properties (1) and (2) is obvious. Let us now consider $L>1$, and $\alpha \in(0,1)$. To establish the implication (1) $\Rightarrow$ (2), it is sufficient to take $\alpha=1 / L$ and set
$$
\mathb... | proof | Other | proof | Yes | Yes | olympiads | false | 33,823 |
96. Let $\xi$ and $\zeta$ be independent random variables, and $\mathrm{E} \xi=0$. Show that
$$
\mathrm{E}|\zeta| \leqslant \mathrm{E}|\xi+\zeta|
$$
Generalize this property by establishing that
$$
\mathrm{E} \max _{1 \leqslant i \leqslant n}\left|\zeta_{i}\right| \leqslant \mathrm{E} \max _{1 \leqslant i \leqslant ... | Solution. If $\mathrm{E}|\zeta|=\infty$, then $\mathrm{E}|\xi+\zeta| \geqslant \mathrm{E}|\zeta|-\mathrm{E}|\xi|=\infty$ (the finiteness of $\mathrm{E}|\xi|$ is assumed by the condition). Now suppose that $\mathrm{E}|\zeta|<\infty$. Then $\mathrm{E}|\xi+\zeta|$ is also finite. From the independence of $\xi$ and $\eta$,... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,824 |
97. Let $X_{1}, X_{2}, \ldots-$ be i.i.d. bounded random variables, $\mathrm{E} X_{1}=0, \mathrm{a}$
$$
S_{n}=X_{1}+\ldots+X_{n}, \quad n \geqslant 1
$$
Prove that for any $p>0$ the following equality holds
$$
\mathrm{E}\left|S_{n}\right|^{p}=O\left(n^{p / 2}\right)
$$ | Solution. Since
$$
\sqrt[p]{\mathrm{E}\left|S_{n}\right|^{p}} \leqslant \sqrt[2 m]{\mathrm{ES} S_{n}^{2 m}}
$$
for any integer $m \geqslant p$, it is sufficient to prove that
$$
E S_{n}^{2 m}=O\left(n^{m}\right)
$$
Assume first that $X_{1} \stackrel{d}{=}-X_{1}$. Then $\mathrm{E} X_{i}^{l}=0$ for odd (integer) numb... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,825 |
98. Let $X_{1}, X_{2}, \ldots$ be i.i.d. random variables. Define $S_{0}=0$, $S_{n}=X_{1}+\ldots+X_{n}$. Define recursively the following ladder indices (also called ladder times):
$$
\tau_{0}=0, \quad \tau_{k}=\inf \left\{n>\tau_{k-1}: S_{n}-S_{\tau_{k-1}}>0\right\}, \quad k \geqslant 1,
$$
with the usual convention... | Solution. The statement follows from problem II.12.9.
Remark. Using ladder indices, one can define the quantity
$$
M=\sup \left\{0, S_{1}, S_{2}, \ldots\right\}
$$
Specifically, $M=\sup \left\{S_{\tau_{i}}: \tau_{i}0$ we have
$$
\mathrm{E} M^{p}=\mathrm{P}\left(\tau_{1}=\infty\right) \sum_{n \geqslant 1} \mathrm{E}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,826 |
99. Let (in the notation and conditions of the previous problem)
$$
A=\sum_{n \geqslant 1} \frac{\mathrm{P}\left(S_{n} \leqslant 0\right)}{n}, \quad B=\sum_{n \geqslant 1} \frac{\mathrm{P}\left(S_{n}>0\right)}{n} .
$$
Show that
$$
\mathrm{P}\left(\tau_{1}<\infty\right)= \begin{cases}1, & \text { if } B=\infty \\ 1-e... | $$
\begin{aligned}
& \mathrm{P}\left(\tau_{1} \leqslant n\right)=\mathrm{P}\left(\max _{1 \leqslant k \leqslant n} S_{k}>0\right)=\mathrm{P}\left(\max _{1 \leqslant k \leqslant n} S_{k} \geqslant 1\right)=\mathrm{P}\left(\min _{1 \leqslant k \leqslant n} S_{k} \leqslant 0\right) \\
& \mathrm{E} s^{\tau_{1}}=\sum_{n=1}^... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,827 |
102. Let $\xi_{1}, \xi_{2}, \ldots$ be a sequence of independent random variables uniformly distributed on $[0,1]$. Define
$$
\nu=\inf \left\{n \geqslant 2: \xi_{n}>\xi_{n-1}\right\}, \quad \mu(x)=\inf \left\{n \geqslant 1: \xi_{1}+\ldots+\xi_{n}>x\right\},
$$
where $0<n)=x^{n} / n!, n \geqslant 1$
(b) $\nu \stackre... | Solution. (a) For $x \in(0,1]$, the probability $\mathrm{P}(\mu(x)>n)=\mathrm{P}\left(\xi_{1}+\ldots\right.$ $\left.\ldots+\xi_{n} \leqslant x\right)$ is equal to the volume of the $n$-dimensional simplex $\left\{\left(x_{1}, \ldots, x_{n}\right) \in \mathbb{R}_{+}^{n}: x_{1}+\ldots\right.$ $\left.\ldots+x_{n}\xi_{2}>\... | e | Other | proof | Yes | Yes | olympiads | false | 33,830 |
103. Show that for any $\lambda>0$ the Boolean transformation
$$
x \rightarrow x-\frac{\lambda}{x} \quad(x \neq 0)
$$
preserves the Lebesgue measure on $\mathbb{R}$, i.e., for any integrable function $f$ on $\mathbb{R}$, the following equality holds:
$$
\int_{\mathbb{R}} f(x) d x=\int_{\mathbb{R}} f\left(x-\frac{\la... | Solution. Let $\lambda=1$ initially. Then the desired relation follows from the following equalities, valid for any function $f$ integrable on $\mathbb{R}:$
$$
\begin{gathered}
\int_{0}^{\infty} f\left(x-\frac{1}{x}\right) d x=\int_{-\infty}^{0} f\left(y-\frac{1}{y}\right) \frac{d y}{y^{2}} \\
\int_{\mathbb{R}} f\left... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,831 |
104. Let $f=f(x)$ be a convex non-decreasing function, $f(-\infty)=0$. Using the fact that any convex function $f$ has a non-decreasing derivative $f^{\prime}$ almost everywhere, show that
$$
f(x)=\int_{\mathbb{R}}(x-u)^{+} d f^{\prime}(u)
$$
where $x^{+}=x \vee 0$, and the integral with respect to $d f^{\prime}$ is ... | Solution. For $c>0$ we have
$$
\int_{-c}^{\infty}(x-u)^{+} d f^{\prime}(u)=\int_{-c}^{x}(x-u) d f^{\prime}(u)=\int_{-c}^{x} f^{\prime}(u) d u-(x+c) f^{\prime}(-c)
$$
From the non-negativity of the left-hand side of the last equality, it follows that for $x>-c$ the inequalities hold
$$
0 \leqslant(x+c) f^{\prime}(-c)... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,832 |
105. Let $\mathrm{E}|\xi|<\infty$. Show that
$$
\mathrm{E} \xi^{2} I(|\xi| \leqslant n)=o(n), \quad n \rightarrow \infty
$$ | Solution. We have
$$
\mathrm{E} \xi^{2} I(|\xi| \leqslant n) \leqslant 2 \int_{0}^{n} x \mathrm{P}(|\xi|>x) d x=o(n), \quad n \rightarrow \infty
$$
since $x \mathrm{P}(|\xi|>x) \leqslant \mathrm{E}|\xi| I(|\xi|>x) \rightarrow 0, x \rightarrow \infty$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,833 |
106. (See [50].) Let $\xi$ be a random variable. Also, let the function $f=f(x)$ be non-decreasing on $\mathbb{R}$, and the expression $x - f(x) + \mathrm{E} f(\xi) - \mathrm{E} \xi$ changes its sign exactly once from “-” to “+” as $x$ runs from $-\infty$ to $\infty$. Prove that for any continuous convex function $g$ o... | Solution. Let $h(x)=f(x)-\mathrm{E} f(\xi)+\mathrm{E} \xi$. From the convexity of the function $g$ it follows that
$$
g(x)-g(h(x)) \geqslant g^{\prime}(h(x))(x-h(x))
$$
where $g^{\prime}$ is the right derivative of the function $g$. By the condition, there exists $x_{0}$ for which
$$
x-h(x) \leqslant 0, \quad x<x_{0... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,834 |
107. Let $f$ be an arbitrary function for which
$$
|f(x)-f(y)| \leqslant|x-y|, \quad x, y \in \mathbb{R}
$$
(a) Prove that for any such random variable $\xi$ such that $E \xi^{2}<\infty$, the inequality
$$
\mathrm{D} f(\xi) \vee|E \xi f(\xi)-\mathrm{E} \xi \cdot \mathrm{E} f(\xi)| \leqslant \mathrm{D} \xi
$$
holds.... | Solution. (a) Let $\zeta$ be an independent copy of the random variable $\xi$. Then, as is easy to verify,
$$
\mathrm{D} \xi=\mathrm{E} \frac{(\xi-\zeta)^{2}}{2}, \quad \mathrm{E} \xi f(\xi)-\mathrm{E} \xi \cdot \mathrm{E} f(\xi)=\mathrm{E}(\xi-\zeta) \frac{f(\xi)-f(\zeta)}{2}
$$
Therefore,
$$
\begin{gathered}
\math... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,835 |
1. Let the random variable $X$ and the $\sigma$-algebra $\mathscr{G}$ be independent (collectively) of the $\sigma$-algebra $\mathscr{E}$, and suppose $\mathrm{E}|X|<\infty$. Show that almost surely
$$
\mathrm{E}(X \mid \mathscr{G} \vee \mathscr{E})=\mathrm{E}(X \mid \mathscr{G})
$$
Verify by example that the indepen... | Solution. Since $\mathrm{E}(X \mid \mathscr{G})$ is a $\mathscr{G} \vee \mathscr{E}$-measurable random variable, and the $\sigma$-algebra $\mathscr{G} \vee \mathscr{E}$ is generated by sets of the form $D=BC$, where $B \in \mathscr{G}$ and $C \in \mathscr{E}$, it suffices (for all such $D$) to establish the equality $\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,836 |
2. We will say that $\sigma$-algebras $\mathscr{A}$ and $\mathscr{B}$ are conditionally independent relative to $\sigma$-algebra $\mathscr{C}$ if $\mathrm{P}(A B \mid \mathscr{C})=\mathrm{P}(A \mid \mathscr{C}) \mathrm{P}(B \mid \mathscr{C})$ for all $A \in \mathscr{A}$ and $B \in \mathscr{B}$.
Show that the condition... | Solution. Obviously, (2) $\Rightarrow(1)$ (it follows to take $X=I_{A}$ ) and (1) $\Rightarrow(3)$. To verify the implication $(3) \Rightarrow(1)$, one needs to use the principle of suitable sets (applying the theorem about $\pi-\lambda$-systems, see theorem 2 in V1.II.2). The equivalence of condition (4) and the condi... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,837 |
3. Let $\left\{\left(X_{i}, Y_{i}\right)\right\}_{i=1}^{n}$ be independent random vectors (in $\mathbb{R}^{2}$ ). Prove that conditionally on $\mathscr{G}_{n}=\sigma\left(Y_{1}, \ldots, Y_{n}\right)$ the values $X_{1}, \ldots, X_{n}$ are also independent, and $\operatorname{Law}\left(X_{i} \mid \mathscr{G}_{n}\right)=\... | Solution. It is required to establish that for any Borel sets $B_{i}, 1 \leqslant i \leqslant n$, with probability one, the equalities hold:
$$
\mathrm{P}\left(\bigcap_{i}\left\{X_{i} \in B_{i}\right\} \mid \mathscr{G}_{n}\right)=\prod_{i} \mathrm{P}\left(X_{i} \in B_{i} \mid \mathscr{G}_{n}\right)=\prod_{i} \mathrm{P... | proof | Other | proof | Yes | Yes | olympiads | false | 33,838 |
4. Let $X^{n}=\left\{X_{i}\right\}_{i=1}^{n}$ be a random sample, i.e., a family of i.i.d. random variables. Suppose that
1) $\tau^{n}=\left(\tau_{1}, \ldots, \tau_{n}\right)$ is a random vector whose components are independent and uniformly distributed on $\{1, \ldots, n\}$;
2) $\sigma^{n}=\left(\sigma_{1}, \ldots, \s... | Solution. Let $f\left(X^{n}, i\right)=X_{i}, 1 \leqslant i \leqslant n$. Then the values $X_{\tau_{i}}=f\left(X^{n}, \tau_{i}\right)$ form a random sample conditionally on $X^{n}$, since due to the independence of $\tau^{n}$ and $X^{n}$ (see the remark to problem II.7.14) with probability one the equality holds
$$
\op... | proof | Other | proof | Yes | Yes | olympiads | false | 33,839 |
5. Let $f=f(x)$ be a convex function. Show that for any integrable quantity $\xi$, any $\sigma$-subalgebra $\mathscr{G} \subset \mathscr{F}$, and all $c>0$, the following inequality holds:
$$
\mathrm{E} f(\zeta) I(f(\zeta)>c) \leqslant \mathrm{E} 2 f(\xi) I(2 f(\xi)>c)
$$
where $\zeta=\mathrm{E}(\xi \mid \mathscr{G})... | Solution. The function $g(x)=(f(x)-c)^{+}$ is convex for any fixed $c>0$, so by Jensen's inequality for conditional expectations (see Problem II.7.10)
$$
\mathrm{E}(g(\zeta) \mid \mathscr{G}) \leqslant \mathrm{E}(g(\xi) \mid \mathscr{G}) \quad \text { a.s., } \quad \mathrm{E} g(\zeta) \leqslant \mathrm{E} g(\xi)
$$
F... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,840 |
7. Let $\xi$ and $\zeta$ be independent random variables.
(a) Assuming the existence of the expectation $\mathrm{E} \xi$, show that if the random variables $\xi, \zeta$ are identically distributed, then almost surely (a.s.)
$$
\mathrm{E}(\xi \mid \xi+\zeta)=\mathrm{E}(\zeta \mid \xi+\zeta)=\frac{\xi+\zeta}{2}
$$
(b)... | Solution. (a) Note that $(\xi, \zeta) \stackrel{d}{=}(\zeta, \xi)$. Therefore, $(\xi, \xi+\zeta) \stackrel{d}{=}$ $\stackrel{d}{=}(\zeta, \xi+\zeta)$ and $\xi \mathbf{1}_{A} \stackrel{d}{=} \zeta \mathbf{1}_{A}$ for any $A \in \sigma(\xi+\zeta)$. Consequently, $\mathrm{E} \xi \mathbf{1}_{A}=\mathrm{E} \zeta \mathbf{1}_... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,842 |
8. Let $\xi_{1}, \xi_{2}, \ldots$ be i.i.d. random variables, $\mathrm{E}\left|\xi_{1}\right|<\infty$. Show that a.s.
$$
\mathrm{E}\left(\xi_{1} \mid S_{n}, S_{n+1}, \ldots\right)=n^{-1} S_{n}
$$
where $S_{n}=\xi_{1}+\ldots+\xi_{n}$. | Solution. As in problem II.7.7(a), we should use the fact that $\mathrm{E} \xi_{i} \mathbf{1}_{A}=\mathrm{E} \xi_{i} \mathbf{1}_{A}$ for any $A \in \sigma\left(S_{n}, S_{n+1}, \ldots\right)$ and $i, j \leqslant n$. | proof | Other | proof | Yes | Yes | olympiads | false | 33,843 |
9. (See [92].) Let $\xi$ be a random variable with distribution function $F=F(x)$ such that $F(b)-F(a)>0$ for all $a, b, -\infty<a<b<\infty$. Express $\mathrm{E}(\xi \mid a<\xi \leqslant b)$ in terms of $F$.
Establish that $f(a, b)=\mathrm{E}(\xi \mid a<\xi \leqslant b)$ is an increasing function of $a, b$ (on the set... | Solution. According to the definition of conditional mathematical expectation
$$
\mathrm{E}(\xi \mid a\lambda \mathrm{E}(\xi \mid a<\xi \leqslant x)+(1-\lambda) \mathrm{E}(\xi \mid a<\xi & \leqslant x)= \\
& =\mathrm{E}(\xi \mid a<\xi \leqslant x)
\end{aligned}
$$
Similarly, we obtain $\mathrm{E}(\xi \mid a<\xi \leqs... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,844 |
10. Let $g=g(x)$ be a concave Borel function defined on $\mathbb{R}$ such that $\mathrm{E}|g(\xi)|<\infty$. Show that for conditional expectations a.s. the Jensen's inequality holds:
$$
g(\mathrm{E}[\xi \mid \mathscr{G}]) \leqslant \mathrm{E}[g(\xi) \mid \mathscr{G}]
$$
and, therefore,
$$
g(\mathrm{E} \xi) \leqslant... | Solution. Due to the convexity of the function $g$, there exists a Borel function $l=l(x)$ (as $l(x)$ one can take the left derivative of the function $g$ at the point $x$), such that
$$
g(y) \geqslant g(x)+l(x)(y-x), \quad x, y \in \mathbb{R}
$$
(for a strictly convex function $g$, the inequality is also strict when... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,845 |
11. Let $\xi$ be a random variable, $\mathrm{E}|\xi|<\infty$. Verify that if $\xi \leqslant \mathrm{E}[\xi \mid \mathscr{G}]$ a.s. for some $\sigma$-algebra $\mathscr{G}$, then $\xi=\mathrm{E}[\xi \mid \mathscr{G}]$ a.s. | Solution. Let's take an arbitrary strictly concave increasing function $g$ for which
$$
\lim _{x \rightarrow \infty} \frac{g(x)}{x}=\lim _{x \rightarrow-\infty} \frac{g(x)}{x / 2}=1
$$
Then $\mathrm{E} g(\xi) \leqslant \mathrm{E} g(\mathrm{E}[\xi \mid \mathcal{G}])$. Considering Jensen's inequality $\mathrm{E} g(\xi)... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,846 |
12. Let $\xi$ and $\zeta$ be random variables, $\mathrm{E}|\xi|, \mathrm{E}|\zeta|<\infty, \mathrm{E}[\xi \mid \mathscr{E}] \geqslant \zeta$ and $\mathrm{E}[\zeta \mid \mathscr{G}] \geqslant \xi$ a.s. for some (completed) $\sigma$-algebras $\mathscr{E}$ and $\mathscr{G}$. Show that $\xi$ is $\mathscr{E} \cap \mathscr{G... | Solution. Let the function $g$ be the same as in the solution of problem II.7.11. Then
$$
\mathrm{E} g(\mathrm{E}[\xi \mid \mathscr{E}]) \geqslant \mathrm{E} g(\zeta), \quad \mathrm{E} g(\mathrm{E}[\zeta \mid \mathscr{G}]) \geqslant \mathrm{E} g(\xi)
$$
Using Jensen's inequality, we get
$$
\mathrm{E} g(\mathrm{E}[\x... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,847 |
13. Let $X, Y$ be random variables, $\mathbf{E}|X|, \mathrm{E}|Y|<\infty$ and $\mathrm{E}(Y \mid X)=0$ a.s. Show that from the condition $\mathrm{E}(Y \mid X+Y)=0$ a.s. it follows that $Y=0$ with probability one. | Solution. Let $f(x)=|x|-\operatorname{arctg}|x|, x \in \mathbb{R}$. The function $f=f(x)$ is increasing on $\mathbb{R}_{+}$ and is an even, strictly convex function on $\mathbb{R}$, and $0 \leqslant f(x) \leqslant|x|$ for all $x \in \mathbb{R}$. By Jensen's inequality for conditional expectations, we have
$$
\begin{ga... | 0 | Algebra | proof | Yes | Yes | olympiads | false | 33,848 |
14. Let the random variable $\xi$ be independent of the $\sigma$-algebra $\mathscr{G}$. Suppose also that $\mathrm{E}|f(\xi, \zeta)|<\infty$ for some Borel function $f=f(x, y)$ and $\mathscr{G}$-measurable random variable $\zeta$. Prove that $\mathrm{E}[f(\xi, \zeta) \mid \mathscr{G}]=g(\zeta)$ a.s., where $g(x)=\mathr... | Solution. It is sufficient to prove that for any $A \in \mathscr{G}$ the conditions
$$
\mathrm{E} f(\xi, \zeta) I_{A}=\mathrm{E} g(\zeta) I_{A} \quad \text { and } \quad \mathrm{E} I_{A}|g(\zeta)|<\infty
$$
hold. The quantity $\xi$ is independent of ( $\left.\zeta, I_{A}\right)$. Moreover, $\mathrm{E}|f(\xi, \zeta)| ... | proof | Other | proof | Yes | Yes | olympiads | false | 33,849 |
15. Let $X_{1}, X_{2}, \ldots$ be a sequence of independent random variables, $S_{n}=\sum_{i=1}^{n} X_{i}$. Show that $\left(S_{1}, \ldots, S_{n-1}\right)$ and $\left(S_{n+1}, S_{n+2}, \ldots\right)$ are conditionally independent given the $\sigma$-algebra $\sigma\left(S_{n}\right)$. | Solution. Since
$$
\xi_{n}=\left(S_{1}, \ldots, S_{n}\right) \quad \text { and } \quad \zeta_{n}=\left(X_{n+1}, X_{n+1}+X_{n+2}, \ldots\right) \text { are independent, }
$$
by the remark after problem II.7.14, we have
$\mathrm{E}\left[I_{B}\left(\zeta_{n}+S_{n}\right) \mid \xi_{n}\right]=\mathrm{E}\left[I_{B}\left(\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,850 |
16. (See [97].) Let $\xi$ be a Bernoulli random variable taking values 0 and 1, and $X$ be some other random variable. Show that the following conditions are equivalent:
(1) the random variables $X$ and $\xi$ are conditionally independent given $g(X)$ (where $g=g(x)$ is a Borel function), i.e., with probability one
$... | Solution. Let condition (1) be satisfied initially. To prove the validity of condition (2), it is sufficient to establish that $\mathrm{D}(p(X) \mid g(X))=\mathrm{E}\left[p(X)^{2} \mid g(X)\right]-|\mathrm{E}[p(X) \mid g(X)]|^{2}=0$. The last equality follows from the relations
$$
\mathrm{E}[\xi \mid g(X)]=\mathrm{E}[... | proof | Other | proof | Yes | Yes | olympiads | false | 33,851 |
17. Show that a random variable $\xi$ and a $\sigma$-algebra $\mathscr{G}$ are independent, i.e., for any $A \in \mathscr{G}$, the variables $\xi$ and $I_{A}$ are independent if and only if $\mathrm{E}[g(\xi) \mid \mathscr{G}]=E g(\xi)$ for every Borel function $g=g(x)$ such that $\mathrm{E}|g(\xi)|<\infty$. | Solution. If $A \in \mathscr{G}$ and $B \in \mathscr{B}(\mathbb{R})$, then from the assumption of independence of $\xi$ and $\mathscr{G}$ we have $\mathrm{P}(A \cap\{g(\xi) \in B\})=\mathrm{P}(A) \mathrm{P}(g(\xi) \in B)$, and thus,
$\mathrm{E}[g(\xi) \mid \mathscr{G}]=\mathrm{E} g(\xi)$. Conversely, if this equality h... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,852 |
18. Let $\xi$ be a non-negative random variable and $\mathscr{G}$ a $\sigma$-subalgebra of the $\sigma$-algebra $\mathscr{F}$. Show that $\mathrm{E}[\xi \mid \mathscr{G}]<\infty$ a.s. if and only if the measure $\mathrm{Q}$, defined on sets $A \in \mathscr{G}$ by $\mathrm{Q}(A)=\int_{A} \xi d \mathrm{P}$, is $\sigma$-f... | Solution. To prove the necessity, let $A_{n}=\{\mathrm{E}[\xi \mid \mathscr{G}] \leqslant n\}$. Then $\mathrm{Q}\left(A_{n}\right)=\int_{A_{n}} \xi d \mathrm{P}=\int_{A_{n}} \mathrm{E}[\xi \mid \mathscr{G}] d \mathrm{P} \leqslant n$, from which we conclude that the measure $\mathrm{Q}$ is $\sigma$-finite, since $\bigcu... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,853 |
19. Show that conditional probabilities $\mathrm{P}(A \mid B)$ are continuous in the sense that if $\lim _{n} A_{n}=A, \lim _{n} B_{n}=B, \mathrm{P}\left(B_{n}\right)>0, \mathrm{P}(B)>0$, then $\lim _{n} \mathrm{P}\left(A_{n} \mid B_{n}\right)=\mathrm{P}(A \mid B)$ | Solution. It is easy to prove that under the given conditions $\lim A_{n} B_{n}=$ $=A B$ (see problem II.1.8). Therefore, according to problem II.1.16, we have $\mathrm{P}\left(A_{n} B_{n}\right) \rightarrow \mathrm{P}(A B), \mathrm{P}\left(B_{n}\right) \rightarrow \mathrm{P}(B)$, from which the required convergence fo... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,854 |
20. Prove the following version of Fatou's lemma for conditional mathematical expectations.
Let $(\Omega, \mathscr{F}, \mathrm{P})$ be a probability space and $\left(\xi_{n}\right)_{n \geqslant 1}$ be a sequence of random variables such that the expectations $\mathrm{E} \xi_{n}, n \geqslant 1$, and $\mathrm{E} \underl... | Solution. Let $\xi_{n}^{k}=\xi_{n}^{+}-\xi_{n}^{-} I\left(\xi_{n}^{-}<k\right), k \geqslant 1$. Then $\xi_{n}^{k} \geqslant \xi_{n}$,
$$
\frac{\lim }{n} \mathrm{E}\left(\xi_{n} \mid \mathscr{G}\right) \leqslant \frac{\lim }{n} \mathrm{E}\left(\xi_{n}^{k} \mid \mathscr{G}\right) \quad \text { and } \quad \mathrm{E}\lef... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,855 |
21. Let, as in Problem II.7.20, $\left(\xi_{n}\right)_{n \geqslant 1}$ be a sequence of random variables for which the expectations $\mathrm{E} \xi_{n}, n \geqslant 1$, are defined, and let $\mathscr{G}$ be a $\sigma$-subalgebra of events from $\mathscr{F}$ such that
$$
\varlimsup_{n}^{\lim ^{2}} \mathrm{E}\left(\left... | Solution. Since $\xi_{n}^{+}, \xi_{n}^{-} \leqslant\left|\xi_{n}\right|$ and the function $f(x)=x I(x \geqslant a)$, $a>0$, is monotonic, it follows from condition (*) that (a.s.)
$$
\lim _{k \rightarrow \infty} \sup _{n} \mathrm{E}\left(\xi_{n}^{*} I\left(\xi_{n}^{*} \geqslant k\right) \mid \mathscr{G}\right)=0
$$
w... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,856 |
23. If the family of random variables $\left\{\xi_{n}\right\}_{n \geqslant 1}$ is uniformly integrable and $\xi_{n} \rightarrow \xi$ a.s., then $\mathrm{E} \xi_{n} \rightarrow \mathrm{E} \xi$. At the same time, the a.s. convergence of conditional expectations $\mathrm{E}\left(\xi_{n} \mid \mathscr{G}\right) \rightarrow... | Solution. Let $U, V$ be two independent random variables with uniform distribution on $[0,1]$. Let $\mathscr{G}=\sigma(V)$ and
$$
X_{k n}=n I(n U \leqslant 1) I(k-1<n V \leqslant k), \quad 1 \leqslant k \leqslant n
$$
Then almost surely (a.s.)
$$
\mathrm{E}\left(X_{k n} \mid \mathscr{G}\right)=I(k-1<n V \leqslant k)... | \varlimsup_{n}\mathrm{E}(\xi_{n}\mid\mathscr{G})=1\neq0=\mathrm{E}(\xi\mid\mathscr{G}) | Other | proof | Yes | Yes | olympiads | false | 33,858 |
24. Let $\xi_{n} \xrightarrow{L^{p}} \xi$ for some $p \geqslant 1$. Show that then
$$
\mathrm{E}\left(\xi_{n} \mid \mathscr{G}\right) \xrightarrow{L^{p}} \mathrm{E}(\xi \mid \mathscr{G})
$$
for any $\sigma$-subalgebra $\mathscr{G} \subseteq \mathscr{F}$. | Solution. By the definition of convergence in $L^{p}$ space and using Jensen's inequality, we obtain
$$
\mathrm{E}\left|\mathrm{E}\left(\xi_{n} \mid \mathscr{G}\right)-\mathrm{E}(\xi \mid \mathscr{G})\right|^{p} \leqslant \mathrm{E}\left(\left|\xi_{n}-\xi\right|^{p} \mid \mathscr{G}\right)=\mathrm{E}\left|\xi_{n}-\xi\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,859 |
25. Show that $(X, Y) \stackrel{d}{=}(Z, Y)$ if and only if
$$
\mathrm{P}(X \in A \mid Y)=\mathrm{P}(Z \in A \mid Y)
$$
a.s. for every $A \in \mathscr{B}(\mathbb{R})$. | Solution. Necessity. Let $(X, Y) \stackrel{d}{=}(Z, Y)$. For each $C \in \sigma(Y)$, there exists a Borel set $B \in \mathscr{B}(\mathbb{R})$ such that $C = \{Y \in B\}$ and
$$
\begin{aligned}
\mathrm{EP}(X \in A \mid Y) I_{C}=\mathrm{E} I(X \in A, Y \in B)=\mathrm{E} I(Z \in A, Y & \in B)= \\
& =\mathrm{EP}(Z \in A \... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,860 |
26. Provide an example of independent random variables $X$ and $Y$ and a $\sigma$-algebra $\mathscr{G}$ such that on a set of positive measure for some sets $A$ and $B$ the inequality
$$
\mathrm{P}(X \in A, Y \in B \mid \mathscr{G}) \neq \mathrm{P}(X \in A \mid \mathscr{G}) \mathrm{P}(Y \in B \mid \mathscr{G})
$$
hol... | Solution. Let $X=\xi+\eta, Y=\xi-\eta$, where $\xi, \eta$ are independent random variables with distribution $\mathscr{N}(0,1)$. Then $X, Y$ are independent Gaussian variables (since $\operatorname{cov}(X, Y)=0$). Define $\mathscr{G}=\sigma(\eta)$. We have
$$
\begin{aligned}
\mathrm{P}(X \leqslant 0, Y \leqslant 0 \mi... | proof | Other | math-word-problem | Yes | Yes | olympiads | false | 33,861 |
27. Consider the probability space ( $\Omega, \mathscr{F}, \mathrm{P}$ ) as the space $([0,1], \mathscr{B}([0,1]), \lambda)$, where $\lambda$ is the Lebesgue measure. Provide an example of a $\sigma$-subalgebra $\mathscr{G} \subseteq \mathscr{B}([0,1])$ for which the conditional expectation $\mathrm{E}(1 \mid \mathscr{... | Solution. As $\mathscr{G}$, we can take the $\sigma$-subalgebra containing all $B$ from $\mathscr{B}([0,1])$ for which $\lambda(B)=0$ or 1. | proof | Other | math-word-problem | Yes | Yes | olympiads | false | 33,862 |
28. When determining the conditional probability $\mathrm{P}(B \mid \mathscr{G})(\omega)$ of an event $B \in \mathscr{F}$ relative to the $\sigma$-algebra $\mathscr{G} \subseteq \mathscr{F}$, it is usually not assumed that $\mathrm{P}(\cdot \mid \mathscr{G})(\omega)$ is a measure on $(\Omega, \mathscr{F})$ with probabi... | Solution. Let $(\Omega, \mathscr{F}, \mathrm{P})=([0,1], \mathscr{B}([0,1]), \lambda)$, where $\lambda$ is the Lebesgue measure, and
$$
\mathscr{G}=\{B \in \mathscr{F}: \lambda(B)=0 \text { or } 1\} .
$$
Furthermore, let $\mathrm{P}(B \mid \mathscr{G})(\omega) \equiv \lambda(B)$ when $B \in \mathscr{F}$ is not of the... | proof | Other | proof | Yes | Yes | olympiads | false | 33,863 |
30. Let $\eta$ be a $\mathscr{G}$-measurable random variable, $\xi$ be a $\mathscr{F}$-measurable random variable, and $\mathrm{E}|\eta|^{q}<\infty$, $1 / p + 1 / q = 1$. Show that then $\mathrm{E}(\xi \eta \mid \mathscr{G}) = \eta \mathrm{E}(\xi \mid \mathscr{G})$ a.s. | Solution. Let $\eta^{c}=\eta I(|\eta| \leqslant c), c>0$. Then $\eta^{c}-\mathscr{G}$-measurable and $\mathrm{E}\left(\xi \eta^{c} \mid \mathscr{G}\right)=\eta^{c} \mathrm{E}(\xi \mid \mathscr{G})$ a.s. (by the analogous property for bounded variables). Passing to the limit as $c \rightarrow \infty$, we see that $\eta^... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,865 |
31. Let $\Omega=\mathbb{Z}_{+}$ and
$$
\mathrm{P}_{\lambda}(\omega=k)=e^{-\lambda} \frac{\lambda^{k}}{k!}, \quad k \geqslant 0,
$$
be the Poisson distribution on $\Omega$ with parameter $\lambda>0$. Show that for the parameter $1 / \lambda$ there does not exist an unbiased estimator $T=T(\omega)$, i.e., such that for... | Solution. Let $T$ be an unbiased estimator of the parameter $1 / \lambda$. Then
$$
e^{\lambda} \lambda \mathrm{E}_{\lambda} T=\sum_{k=1}^{\infty} \frac{T(k-1)}{(k-1)!} \lambda^{k}=e^{\lambda}
$$
We have reached a contradiction. Indeed, since
$$
\mathrm{E}_{1}|T|=\sum_{k=1}^{\infty} \frac{|T(k-1)|}{(k-1)!}<\infty
$$
... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,866 |
32. (a) Let $\mathscr{P}=\{\mathrm{P}\}$ be a set of measures on $(\Omega, \mathscr{F})$, i.e., $\mathrm{P} \ll \nu$ for some probability measure $\nu$ and all $\mathrm{P} \in \mathscr{P}$. Show that there exist $\mathrm{P}_{1}, \mathrm{P}_{2}, \ldots \in \mathscr{P}$ such that for any $\mathrm{P} \in \mathscr{P}$ the ... | Solution. (a) As $\mathrm{P}_{n}$, one should take those elements of the class $\mathscr{P}$ for which
$$
\operatorname{ess} \sup _{\mathrm{P} \in \mathscr{P}} \frac{d \mathrm{P}}{d \nu}=\sup _{n} \frac{d \mathrm{P}_{n}}{d \nu}
$$
up to a set of $\nu$-measure zero (the definition of ess sup is given in problem II.4.2... | proof | Other | proof | Yes | Yes | olympiads | false | 33,867 |
33. Let $(E, \mathscr{E})$ be a Borel space, i.e., a measurable space for which there exists an injective mapping $\varphi: E \rightarrow \mathbb{R}$ such that $\varphi(E) \in \mathscr{B}(\mathbb{R})$ and $\{\varphi(E) \cap B: B \in \mathscr{B}(\mathbb{R})\}=\{\varphi(C): C \in \mathscr{E}\}$. Show that there exists a ... | Solution. Since $\varphi$ is an injection, we have $\varphi^{-1}(\mathscr{G})=\mathscr{E}$, where
$$
\mathscr{G}=\{\varphi(E) \cap B: B \in \mathscr{B}(\mathbb{R})\} \subseteq \mathscr{B}(\mathbb{R})
$$
Thus, it suffices to show that $\mathscr{G}=\sigma(\mathscr{A})$ for some countably generated algebra $\mathscr{A}$... | proof | Other | proof | Yes | Yes | olympiads | false | 33,868 |
34. Prove that
every Polish space $(S, \rho)$ with metric $\rho$ defines a Borel space $(S, \mathscr{B}(S))$, where $\mathscr{B}(S)$ is the Borel $\sigma$-algebra generated by the open sets.
In particular, the spaces $\left(\mathbb{R}^{n}, \mathscr{B}\left(\mathbb{R}^{n}\right)\right), 1 \leqslant n \leqslant \infty$... | Solution. We will divide the proof into several steps. | proof | Other | proof | Yes | Yes | olympiads | false | 33,869 |
35. Let $X$ be a random variable with a symmetric distribution $(X \stackrel{d}{=}-X)$ and the function $\varphi=\varphi(x), x \in \mathbb{R}$, such that $\mathbb{E}[\varphi(X)\|X\|]$ is defined. Show that
$$
\mathrm{E}[\varphi(X)|| X \mid]=\frac{1}{2}[\varphi(|X|)+\varphi(-|X|)] \quad \text { a.s. }
$$
Using the giv... | Solution. Clearly, we have $(X,|X|) \stackrel{d}{=}(-X,|X|)$, therefore
$$
\mathbb{E}[\varphi(X)\|X\|]=\mathbb{E}[\varphi(-X)\|X\|
$$
Moreover, $\varphi(|X|)+\varphi(-|X|)=\varphi(X)+\varphi(-X)$. Finally, we obtain
$$
2 \mathbb{E}[\varphi(X) \| X \mid]=\mathbb{E}[\varphi(X)+\varphi(-X) \||X|]=\varphi(|X|)+\varphi(-... | \frac{1}{2}[I(|X|\leqslantx)+1] | Algebra | proof | Yes | Yes | olympiads | false | 33,870 |
36. Let $X$ be a non-negative random variable. Find the conditional probabilities
$$
\mathrm{P}(X \leqslant x \mid \lfloor X \rfloor) \quad \text { and } \quad \mathrm{P}(X \leqslant x \mid \lceil X \rceil)
$$
where $\lfloor X \rfloor$ is the greatest integer not exceeding $X$ (this number is also denoted by $[X]$), ... | Solution. Let's find the first probability (the second is calculated similarly):
$$
\mathrm{P}(X \leqslant x \mid\lfloor X\rfloor)=\varphi(x,\lfloor X\rfloor)
$$
where $\varphi(x, n)$ is defined by the conditions
$$
\varphi(x, n)= \begin{cases}\frac{\mathrm{P}(X \leqslant x, n \leqslant X<n+1)}{\mathrm{P}(n \leqslan... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,871 |
37. Let $X$ be a random variable with a geometric distribution:
$$
\mathrm{P}(X=n)=p q^{n-1}, \quad n \in \mathbb{N}, \quad 0 \leqslant p \leqslant 1, \quad q=1-p
$$
Show that for $m, n \in \mathbb{N}$ the following equality holds:
$$
\mathrm{P}(X>n+m \mid X>n)=\mathrm{P}(X>m) .
$$
Also prove the converse statement... | Solution. The relation $(**)$ is established by direct verification. Let's turn to the second part of the problem. Denoting $f(\cdot)=\mathrm{P}(X>\cdot)$, we find that $f(n+m)=f(n) f(m)$ and $f(n)=f(1)^{n}, n \in \mathbb{N}$. Thus, for natural $n$ with $q=f(1)$ and $p=1-q$, we have
$$
\mathrm{P}(X>n+1)-\mathrm{P}(X>n... | proof | Other | proof | Yes | Yes | olympiads | false | 33,872 |
38. (a) If a random variable $X$ is exponentially distributed, then the lack of aftereffect property holds:
$$
\mathrm{P}(X>x+y \mid X>x)=\mathrm{P}(X>y), \quad x, y \geqslant 0
$$
Show that if for a non-negative extended (i.e., with values in $[0, \infty]$) random variable $X$ the stated property holds, then one of ... | Solution. (a) Denoting $f(x)=\mathrm{P}(X>x)$, we arrive at the equation $f(x+y)=f(x) f(y)$, which, in particular, guarantees that $f(n x)=f(x)^{n}$ and $f(q)=f(1)^{q}$ for $n \in \mathbb{N}, x \geqslant 0$ and $q \in \mathbb{Q}$. It is not difficult to prove that the specified equation in the class of right-continuous... | proof | Other | proof | Yes | Yes | olympiads | false | 33,873 |
39. Let random variables $X$ and $Y$ have finite second moments. Show that $\operatorname{cov}(X, Y)=\operatorname{cov}(X, \mathrm{E}(Y \mid X))$. Also prove that $\mathrm{D} X \leqslant \mathrm{D} X Y$, when $\mathrm{E}(Y \mid X)=1$. | Solution. Using the fact that $\mathrm{E} X Y=\mathrm{E} X \mathrm{E}(Y \mid X)$ and $\mathrm{E} Y=$ $=\mathrm{E}\mathrm{E}(Y \mid X)$, we obtain the first equality. Moreover, we have $\mathrm{E} X Y=\mathrm{E} X$ and
$$
\mathrm{E} X^{2} Y^{2}=\mathrm{E} X^{2} \mathrm{E}\left(Y^{2} \mid X\right) \geqslant \mathrm{E} X... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,874 |
1. Let $X_{1}, X_{2}, \ldots$ be i.i.d. random variables with the Cauchy distribution with density
$$
\frac{1}{\pi\left(1+x^{2}\right)}, \quad x \in \mathbb{R}
$$
Show that
$$
\frac{M_{n}}{n} \xrightarrow{d} \frac{1}{T}
$$
where $M_{n}=\max \left\{X_{1}, \ldots, X_{n}\right\}$, and the random variable $T$ has an ex... | Solution. Note that for all $x \leqslant 0$ the following relation holds
$$
\mathrm{P}\left(M_{n} \leqslant n x\right) \leqslant \mathrm{P}\left(M_{n} \leqslant 0\right)=\frac{1}{2^{n}} \rightarrow 0
$$
For $x>0$ we have
$$
\mathrm{P}\left(M_{n} \leqslant n x\right)=\left(\frac{\operatorname{arctg} n x}{\pi}+\frac{1... | proof | Other | proof | Yes | Yes | olympiads | false | 33,875 |
2. Let $X_{1}, \ldots, X_{n}, n \geqslant 2,$ be i.i.d. random variables with distribution function $F(x)$ (and density $f(x)$, if it exists)
and $M_{n}=\max \left\{X_{1}, \ldots, X_{n}\right\}, m_{n}=\min \left\{X_{1}, \ldots, X_{n}\right\}, R_{n}=M_{n}-m_{n}$. Show that
$$
\begin{aligned}
F_{m_{n}, M_{n}}(x, y) & =\... | Solution. Using the independence and identical distribution of random variables $X_{1}, \ldots, X_{n}$, we find that
$$
\begin{aligned}
F_{m_{n}, M_{n}}(x, y) & =\mathrm{P}\left(m_{n} \leqslant x, M_{n} \leqslant y\right)= \\
& =\mathrm{P}\left(M_{n} \leqslant y\right)-\mathrm{P}\left(xx \\
F(y)^{n}, & y \leqslant x\e... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,876 |
3. Let $X$ and $Y$ be independent Poisson random variables with parameters $\lambda > 0$ and $\mu > 0$ respectively. Show that
(a) $X+Y$ has a Poisson distribution with parameter $\lambda + \mu$,
(b) the distribution of $X$ conditional on $X+Y$ is binomial:
$$
\mathrm{P}(X=k \mid X+Y=n)=C_{n}^{k}\left(\frac{\lambda}... | Solution. Parts (a) and (b) are established by direct verification:
\[
\begin{aligned}
\mathrm{P}(X+Y=n) & =\sum_{k=0}^{n} \mathrm{P}(X=k, Y=n-k)=\sum_{k=0}^{n} \frac{\lambda^{k} e^{-\lambda}}{k!} \cdot \frac{\mu^{n-k} e^{-\mu}}{(n-k)!}= \\
& =\frac{e^{-(\lambda+\mu)}}{n!} \sum_{k=0}^{n} C_{n}^{k} \lambda^{k} \mu^{n-k... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,877 |
4. (See [66].) Let $X_{\lambda}$ be a Poisson random variable with parameter $\lambda>0$.
(a) Prove the inequalities
$$
\begin{array}{ll}
\mathrm{P}\left(X_{\lambda} \leqslant n\right) \leqslant \frac{\lambda}{\lambda-n} \cdot \mathrm{P}\left(X_{\lambda}=n\right), & 0 \leqslant n < \lambda \\
\mathrm{P}\left(X_{\lamb... | Solution. (a) We have
$$
\begin{aligned}
\mathrm{P}\left(X_{\lambda} \geqslant n\right) & =e^{-\lambda} \frac{\lambda^{n}}{n!}\left[1+\frac{\lambda}{n+1}+\frac{\lambda^{2}}{(n+1)(n+2)}+\ldots\right] \leqslant \\
& \leqslant e^{-\lambda} \frac{\lambda^{n}}{n!}\left[1+\frac{\lambda}{n+1}+\frac{\lambda^{2}}{(n+1)^{2}}+\l... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,878 |
6. Let $(\Omega, \mathscr{F}, \mathrm{P})$ be a probability space, $\mathscr{A}$ and $\mathscr{B}$ be two $\sigma$-subalgebras in $\mathscr{F}$. Let also $L^{2}(\Omega, \mathscr{G}, \mathrm{P})$ for any $\sigma$-algebra $\mathscr{G}$ be the space of $\mathscr{G}$-measurable random variables with finite second moments.
... | Solution. (a) The equality $\rho^{*}(\sigma(X), \sigma(Y))=\rho^{*}(X, Y)$ follows from the fact that any $\sigma(\xi)$-measurable random variable $\zeta$ can be represented as $\zeta=f(\xi)$ for some Borel function $f$.
(b) The relation is established by the same arguments as in Problem II.8.5. | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,880 |
8. Let $X$ and $Y$ be independent random variables uniformly distributed on $[0,1]$. Consider the variable $Z=|X-Y|$. Prove that the distribution $F_{Z}(z)$ has a density $f_{Z}(z)$ and $f_{Z}(z)=2(1-z)$, $z \in[0,1]$.
From this, in particular, we can conclude that $E Z=1 / 3$. | Solution. The joint density of the quantities $U$ and $Y$ is $f(u+y) f(y)$, where $U=X-Y$, and $f$ is the density of the quantities $X, Y$. We will compute the density $f_{U}(u)$ for $u \in[0,1]:$
$$
f_{U}(u)=\int_{\mathbb{R}} f(u+y) f(y) d y=\int_{0}^{1-u} d y=1-u
$$
The quantity $U$ is symmetric, and $|U|=Z \leqsla... | f_{Z}(z)=2(1-z),\quadz\in[0,1] | Other | proof | Yes | Yes | olympiads | false | 33,882 |
9. (Polar method.) Let the random vector $(U, V)$ have a uniform distribution on the unit circle centered at the origin. Set
$$
(X, Y)=(U, V) \sqrt{-\frac{2 \ln \left(U^{2}+V^{2}\right)}{U^{2}+V^{2}}}
$$
Show that $X$ and $Y$ are independent and have the distribution $\mathscr{N}(0,1)$. | Solution. Transitioning to polar coordinates
$$
(U, V)=\rho(\cos \theta, \sin \theta)
$$
we see that the joint density transforms to
$$
\pi^{-1} I\left(u^{2}+v^{2} \leqslant 1\right) \longrightarrow 2 r I(r \in[0,1]) \cdot(2 \pi)^{-1} I(\theta \in[0,2 \pi])
$$
i.e., $\theta$ is uniformly distributed on $[0,2 \pi]$,... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,883 |
11. Let the positive random variable $R$ have a Rayleigh distribution, i.e., for some $\sigma^{2}>0$ the density of $R$ is given by
$$
f_{R}(r)=\frac{r}{\sigma^{2}} e^{-\frac{r^{2}}{2 \sigma^{2}}}, \quad r>0
$$
Let also $\theta$ be a random variable independent of $R$ with a uniform distribution on ( $\alpha, \alpha+... | Solution. Note that
$$
\cos \theta=\cos 2 \pi\left\{\frac{\theta}{2 \pi}\right\}, \quad \sin \theta=\sin 2 \pi\left\{\frac{\theta}{2 \pi}\right\}
$$
and
$$
2 \pi\left\{\frac{\theta}{2 \pi}\right\} \stackrel{d}{=} 2 \pi\left\{\frac{\theta-\alpha}{2 \pi}\right\} \stackrel{d}{=} \varphi
$$
where $\varphi$ has a unifor... | proof | Other | proof | Yes | Yes | olympiads | false | 33,885 |
13. Let $F=F(x)$ be a distribution function. Show that for any $a>0$ the functions
$$
G(x)=\frac{1}{a} \int_{x}^{x+a} F(u) d u \quad \text { and } \quad H(x)=\frac{1}{2 a} \int_{x-a}^{x+a} F(u) d u
$$
are also distribution functions. | Solution. Let the quantity $\xi$ be distributed according to the distribution function $F$, and $\eta$ independently of $\xi$ and has a uniform distribution on $[0,1]$. Then $G$ is the distribution function of the quantity $\xi + a \eta$, and $H$ is the quantity $\xi + 2a \eta - a$ (verified using the convolution formu... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,887 |
14. Let $X \sim \operatorname{Exp}(\lambda), \lambda>0$.
(a) Find the density function of the random variable $Y=X^{1 / \alpha}, \alpha>0$, known as the Weibull distribution.
(b) Find the density function of the random variable $Y=\ln X$, known as the double exponential distribution.
(c) Show that the integer part $... | Solution. (a) The distribution function of the quantity $Y$ is
$$
\mathrm{P}(Y \leqslant y)=\mathrm{P}\left(X \leqslant y^{\alpha}\right)=1-e^{-\lambda y^{\alpha}}, \quad y>0
$$
Differentiating it, we obtain the density function of the quantity $Y$:
$$
f(y)=\lambda \alpha y^{\alpha-1} e^{-\lambda y^{\alpha}} I(y \ge... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,888 |
15. (a) Let random variables $X$ and $Y$ have a joint density function $f=f(x, y)$ of the form $f(x, y)=g\left(x^{2}+y^{2}\right)$. Let $R$ and $\theta$ be the polar coordinates:
$$
X=R \cos \theta, \quad Y=R \sin \theta
$$
Show that $R$ and $\theta$ are independent, and that the variable $\theta$ is uniformly distri... | Solution. (a) By switching to polar coordinates, it is easy to compute the density of the random vector $(R, \theta)$:
$$
h(r, \theta)= \begin{cases}r g\left(r^{2}\right), & r \geqslant 0, \\ 0 & \text { otherwise. }\end{cases}
$$
Since $h$ can be represented as $h(r, \theta)=p(r) q(\theta)$, the quantities $R$ and $... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,889 |
16. Let $\left(\xi_{1}, \ldots, \xi_{n}\right)$ be a random vector with density $f=f\left(x_{1}, \ldots\right)$,
$$
\mathrm{P}\left(\bigcup_{i<j}\left\{\xi_{i}=\xi_{j}\right\}\right)=\int_{\bigcup_{i<\left\{x_{i}=x_{j}\right\}}} f\left(x_{1}, \ldots, x_{n}\right) d x_{1} \ldots d x_{n}=0
$$
From this, it follows that... | Solution. Let the Borel set $B$ be such that for any $\left(x_{1}, \ldots, x_{n}\right) \in B$ the inequalities $x_{1}<x_{2}<\ldots<x_{n}$ hold. Then
$$
\begin{aligned}
& \mathrm{P}\left(X_{n} \in B\right)=\sum_{\left(i_{1}, \ldots, i_{n}\right)} \mathrm{P}\left(\left(\xi_{i_{1}}, \ldots, \xi_{i_{n}}\right) \in B\righ... | proof | Other | proof | Yes | Yes | olympiads | false | 33,890 |
17. Let $\xi_{1}, \ldots, \xi_{n}$ be independent and identically distributed (i.i.d.) random variables with a continuous distribution function $F = F(x)$. In this case, $\mathrm{P}\left(\xi_{i}=\xi_{j}\right)=0, i \neq j$ (see problem II.8.57), therefore
$\mathrm{P}\left(\xi_{i}=\xi_{j}\right.$ for some $\left.i \neq... | Solution. (a) Due to the continuity of the function $F(x)$, taking into account the independence and identical distribution of the values $\xi_{i}, 1 \leqslant i \leqslant n$, we obtain the formula for $F_{r: n}$:
$$
\begin{aligned}
F_{r: n}(x) & =\mathrm{P}\left(\xi_{r: n} \leqslant x\right)= \\
& =\mathrm{P}\left(\t... | proof | Other | proof | Yes | Yes | olympiads | false | 33,891 |
18. Let $X_{1}, X_{2}, \ldots$ be a sequence of exchangeable random variables,
$$
\mathrm{P}\left(X_{i}=X_{j}\right)=0, \quad i \neq j
$$
Let $A_{1}, A_{2}, \ldots$ be a sequence of events such that $A_{1}=\Omega$ and for $n \geqslant 2$ the equality
$$
A_{n}=\left\{X_{n}>X_{m} \text { for all } m<n\right\}
$$
hold... | Solution. Due to the exchangeability of $X_{n}$, we have
$$
\mathrm{P}\left(X_{1}<\ldots<X_{N}\right)=\mathrm{P}\left(B_{\mathrm{i}}\right), \quad B_{\mathrm{i}}=\left\{X_{i_{1}}<\ldots<X_{i_{N}}\right\}
$$
for any permutation $\mathbf{i}=\left(i_{1}, \ldots, i_{N}\right)$ of the indices $1, \ldots, N$. Taking into a... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,892 |
19. Let $\xi_{1}, \xi_{2}, \ldots$ be i.i.d. non-negative random variables and $\tau$ be a random variable independent of $\left(\xi_{n}\right)$ and taking values in $\mathbb{N}$.
(a) Establish that $\mathrm{E} S_{\tau}=\mathrm{E} \tau \mathrm{E} \xi_{1}$, where $S_{\tau}=\xi_{1}+\ldots+\xi_{\tau}$ and infinite values... | Solution. (a) Due to the independence of $(\xi_{n})$ and $\tau$, we have
$$
\mathrm{E} S_{\tau}=\mathrm{E} \sum_{n \geqslant 1} \xi_{n} I(\tau \geqslant n)=\mathrm{E} \xi_{1} \sum_{n \geqslant 1} \mathrm{P}(\tau \geqslant n)=\mathrm{E} \xi_{1} \mathrm{E} \tau
$$
(b) First, note that
$$
\mathrm{P}\left(S_{\tau}>x\rig... | proof | Other | proof | Yes | Yes | olympiads | false | 33,893 |
20. Let $\xi_{1}, \xi_{2}, \ldots$ be uncorrelated identically distributed random variables with finite second moments, and let $\tau$ be a random variable taking values in $\mathbb{N}$ such that $\mathrm{E} \tau^{2}<\infty$ and $\tau$ is independent of $\xi_{1}, \xi_{2}, \ldots$ Show that
$$
\mathrm{E} S_{\tau}=\math... | Solution. From the independence of the sequences $\left\{\xi_{n}\right\}_{n \geqslant 1}$ and $\tau$, it follows that
$$
\begin{gathered}
\mathrm{E} S_{\tau}^{2} \leqslant \mathrm{E} \tau \sum_{n=1}^{\tau} \xi_{n}^{2}=\mathrm{E} \tau \sum_{n=1}^{\tau} \mathrm{E}\left(\xi_{n}^{2} \mid \tau\right)=\mathrm{E} \tau^{2} \m... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,894 |
21. (Wiener.) Let $\xi_{1}, \xi_{2}, \ldots$ be i.i.d. random variables,
$$
\mathrm{P}\left(\xi_{1} \geqslant 0\right)=1 \quad \text { and } \quad \mathrm{E} \xi_{1} < \infty
$$
Define
$$
\tau=\inf \left\{n \geqslant 1: \sum_{i=1}^{n} \xi_{i} \geqslant a n\right\}
$$
Show that
$$
\mathrm{E} \tau < \infty
$$
Remar... | Solution. It is easy to see that
$$
\lim _{N} \mathrm{E} \frac{1}{N} \sum_{k=0}^{N-1} I_{A_{k, N}}=\lim _{N} \mathrm{P}\left(A_{0, N}\right)=\mathrm{P}\left(\sup _{n}\left[S_{n}-a n\right] \geqslant 0\right)
$$
It remains to establish that
$$
a \sum_{k=1}^{N} I_{A_{k}} \leqslant \sum_{n=1}^{N} \xi_{n}
$$
For any su... | proof | Other | proof | Yes | Yes | olympiads | false | 33,895 |
22. Let $\xi$ and $\eta$ be independent random variables with densities $f_{\xi}=f_{\xi}(x), x \in \mathbb{R}$, and $f_{\eta}(y)=I_{[0,1]}(y)$, i.e., $\eta$ has a uniform distribution on $[0,1]$. Find the densities of the variables $\xi \eta$ and $\xi / \eta$. | Solution. Due to the independence of $\xi$ and $\eta$, we have
$$
\begin{aligned}
\mathrm{P}(\xi \eta \leqslant z)=\int_{\{u, v: u v \leqslant z\}} f_{\xi}(u) I_{[0,1]}(v) d u d v=\int_{-\infty}^{z} d x & \int_{0}^{1} \frac{f_{\xi}(x / v)}{v} d v= \\
& =\int_{-\infty}^{z} d x \int_{x}^{\infty} \frac{f_{\xi}(y)}{y} d y... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,896 |
2. $\xi \wedge \eta$ is independent of $|\xi-\eta|$, and the density $f$ can be chosen to be positive and continuous on $\mathbb{R}_{+}$.
(b) Show that if $\xi$ and $\eta$ are independent exponentially distributed random variables with parameters $\lambda$ and $\mu$, and $\lambda \neq \mu$, then the density of the dis... | Solution. (a) 1. The density of the conditional distribution of $\xi$ given $\xi+\eta=z$ is
$$
\left(\int_{0}^{z} f(y) f(z-y) d y\right)^{-1} f(x) f(z-x), \quad 0 \leqslant x \leqslant z
$$
and is 0 otherwise. If $\xi$ is exponentially distributed, then the above ratio equals $1 / z$ and, consequently, $\xi$ given $\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,897 |
24. Let the random variable $C$ be distributed according to the Cauchy distribution. Show that
$$
C \stackrel{d}{=} \frac{X}{Y} \stackrel{d}{=} \frac{X}{|Y|}
$$
where $X$ and $Y$ are i.i.d. random variables, distributed according to $\mathscr{N}(0,1)$. | Solution. Due to the symmetry of the quantities $C(C \stackrel{d}{=}-C)$, as well as the quantities $X / Y$ and $X /|Y|$, it is sufficient to prove that
$$
|C| \stackrel{d}{=} \frac{|X|}{|Y|}
$$
Using the change of variables of the form $x=u y$, for all $z \geqslant 0$ we obtain the equalities
$$
\begin{aligned}
\ma... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,898 |
25. Let the random variable $\xi$ take a finite number of values $x_{1}, \ldots, x_{k} \geqslant 0$. Show that
$$
\lim _{n \rightarrow \infty}\left(E \xi^{n}\right)^{1 / n}=\max \left\{x_{1}, \ldots, x_{k}\right\}
$$ | Solution. Let $x=\max \left\{x_{1}, \ldots, x_{k}\right\}$, and $p=\mathrm{P}(\xi=x)>0$. Then
$$
x^{n} p \leqslant \mathrm{E} \xi^{n} \leqslant x^{n} \quad \text { and } \quad \lim _{n \rightarrow \infty} p^{1 / n}=1
$$
which leads to the desired equality. | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,899 |
26. Let $\xi$ and $\eta$ be independent random variables taking values in $\mathbb{N}$. Suppose that either $\mathrm{E} \xi < \infty$ or $\mathrm{E} \eta < \infty$. Show that
$$
\mathrm{E}(\xi \wedge \eta)=\sum_{n=1}^{\infty} \mathrm{P}(\xi \geqslant n) \mathrm{P}(\eta \geqslant n)
$$ | Solution. The statement follows from the equality
$$
\mathrm{E} \tau=\sum_{n \geqslant 1} n \mathrm{P}(\tau=n)=\sum_{n \geqslant 1} \mathrm{P}(\tau \geqslant n)
$$
valid for a quantity $\tau$ with values in $\mathbb{N}$, and also from the fact that
$$
\mathrm{P}(\xi \wedge \eta \geqslant n)=\mathrm{P}(\xi \geqslant ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,900 |
27. Let $\xi$ and $\eta$ be independent random variables having exponential distributions with parameters $\lambda$ and $\mu$ respectively. Find the distribution functions of the variables $\frac{\xi}{\xi+\eta}$ and $\frac{\xi+\eta}{\xi}$. | Solution. We have
$$
\frac{\xi+\eta}{\xi}=\frac{\lambda}{\mu}\left(\frac{\zeta+\eta}{\zeta}\right)-\frac{\lambda}{\mu}+1
$$
where $\zeta=\lambda \xi / \mu \stackrel{d}{=} \eta$. According to problem II.8.23(a), the quantity $(\zeta+\eta)^{-1} \zeta$ is uniformly distributed on the interval $[0,1]$. Therefore,
$$
\ma... | \begin{aligned}\mathrm{P}(\frac{\xi+\eta}{\xi}\leqslantx)&=1-(\frac{\mu}{\lambda}(x-1)+1)^{-1},\\\mathrm{P}(\frac{\xi}{\xi+\eta}\leqslantx)&=(\frac{\mu(1-x)}{\lambdax}+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,901 |
28. (See [73].) Let $X$ be a Bernoulli random variable,
$$
\mathrm{P}(X=1)=p, \quad \mathrm{P}(X=0)=1-p
$$
(a) Show that there exist random variables $Y$ independent of $X$ such that the distribution of their sum is symmetric, i.e., $X+Y \stackrel{d}{=}-(X+Y)$.
(b) Establish that the minimum variance $\mathrm{D} Y$ ... | Solution. (a) Let $Y$ be independent of $X$ and $Y \stackrel{d}{=}-X$. Then, obviously, the quantity $X+Y$ is symmetric.
(b) Due to the symmetry of the quantity $X+Y$, we have
$$
p+\mathrm{E} Y=\mathrm{E}(X+Y)=0
$$
i.e., $\mathrm{E} Y=-p$. Moreover, using the independence of the quantities $X$ and $Y$ and the oddnes... | \mathrm{D}Y\geqslantp(1-p) | Algebra | proof | Yes | Yes | olympiads | false | 33,902 |
29. Let $U$ be a random variable having a uniform distribution on $(0,1)$. Show that
(a) for any $\lambda>0$ the variable $-\frac{1}{\lambda} \ln U$ has an exponential distribution with parameter $\lambda$;
(b) the variable $\operatorname{tg}\left(\pi U-\frac{\pi}{2}\right)$ has a Cauchy distribution with density $\f... | Solution. (a) Let's compute the distribution function of the quantity $V = -\frac{1}{\lambda} \ln U \geqslant 0$:
$$
\mathrm{P}(V \leqslant v) = \mathrm{P}\left(U \geqslant e^{-\lambda v}\right) = \left(1 - e^{-\lambda v}\right) I(v \geqslant 0)
$$
Thus, $V$ has an exponential distribution with parameter $\lambda$.
... | proof | Other | proof | Yes | Yes | olympiads | false | 33,903 |
30. Let $\theta$ be a random variable uniformly distributed on $[0,2 \pi)$, and $C$ be a random variable with a Cauchy distribution. Show that
$$
\operatorname{ctg} \theta \stackrel{d}{=} \operatorname{ctg} \frac{\theta}{2} \stackrel{d}{=} C \quad \text { and } \quad \cos ^{2} \theta \stackrel{d}{=} \cos ^{2} \frac{\t... | Solution. Due to problem II.8.29, taking into account the symmetry of the quantity $C$, we have
$$
C \stackrel{d}{=}-C \stackrel{d}{=}-\operatorname{tg}\left(\pi U-\frac{\pi}{2}\right) \stackrel{d}{=} \operatorname{ctg}(\pi U) \stackrel{d}{=} \operatorname{ctg} \frac{\theta}{2},
$$
where $U$ is uniformly distributed ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,904 |
32. For a random variable $\xi$, having a $\mathscr{N}(0,1)$-distribution,
$$
\mathrm{P}(\xi \geqslant x) \sim \frac{\varphi(x)}{x}, \quad x \rightarrow \infty, \quad \text { where } \varphi(x)=\frac{1}{\sqrt{2 \pi}} e^{-\frac{x^{2}}{2}} .
$$
Find the corresponding asymptotic for a random variable $\zeta$, having a g... | Solution. The desired asymptotic is
$$
\mathrm{P}(\zeta \geqslant x)=\frac{1}{\Gamma(\alpha)} \int_{x}^{\infty} z^{\alpha-1} e^{-z} d z \sim f(x)=\frac{x^{\alpha-1} e^{-x}}{\Gamma(\alpha)}, \quad x \rightarrow \infty
$$
Indeed, by L'Hôpital's rule
$$
\begin{aligned}
\lim _{x \rightarrow \infty} \frac{\int_{x}^{\inft... | \frac{x^{\alpha-1}e^{-x}}{\Gamma(\alpha)},\quadxarrow\infty | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,906 |
33. Let $X$ and $Y$ be two non-degenerate independent random variables such that their product $X Y$ has a discrete distribution. Show that each of the variables $X$ and $Y$ has a discrete distribution. | Solution. Let us show, for example, that $X$ has a discrete distribution. Represent the distribution function $F=F(x)$ of the quantity $X$ in the form $F(x)=G(x)+H(x)$, where $x \in \mathbb{R}$,
$$
G(x)=\sum_{u \leqslant x}[F(u)-F(u-)]
$$
- the jump function of the function $F$, and $H(x)=F(x)-G(x)$ - the continuous ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,907 |
34. Let $I$ be an open set in $\mathbb{R}^{n}$ and $y=\varphi(x)$ be a function defined on $I$, with values in $\mathbb{R}^{n}$. (If $x=\left(x_{1}, \ldots, x_{n}\right) \in I$, then $y=\left(y_{1}, \ldots, y_{n}\right)$, where $y_{i}=\varphi_{i}\left(x_{1}, \ldots, x_{n}\right), i=1, \ldots, n$.) It is assumed that al... | Solution. To establish the desired formula, one should apply the multidimensional analogue of the integration by substitution formula (see the remark to Problem II.6.15). | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,908 |
35. Let $Y=A X+b$, where $X=\left(X_{1}, \ldots, X_{n}\right), Y=\left(Y_{1}, \ldots, Y_{n}\right)$, matrix $A$ of order $n \times n$ is such that $|\operatorname{det} A|>0$, and $b$ is an $n$-dimensional vector. Show that
$$
f_{Y}(y)=\frac{1}{|\operatorname{det} A|} f_{X}\left(A^{-1}(y-b)\right)
$$ | Solution. The statement follows from the result of Problem II.8.34 if we take $\varphi(x)=A x+b$ and note that $\left|J_{\varphi^{-1}}(y)\right|=1 /|\operatorname{det} A|$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,909 |
37. (a) Show that the following functions:
$$
\begin{aligned}
& F_{\mathrm{G}}(x)=\exp \left(-e^{-x}\right), \quad x \in \mathbb{R} ; \\
& F_{\mathrm{F}}(x, \alpha)=\left\{\begin{array}{ll}
0, & x \leq 0 ; \\
\exp \left(-x^{-\alpha}\right), & x > 0 ;\end{array}\right. \\
& F_{\mathrm{W}}(x, \alpha)=\left\{\begin{array... | Solution. Point (a) is obvious. Let's establish point (b):
$$
\begin{aligned}
\mathrm{P}(\ln X \leqslant x) & =F_{\mathrm{F}}\left(a e^{x}, \alpha\right)=F_{\mathrm{G}}(\alpha x+\alpha \ln a) \\
\mathrm{P}(-\ln (-Y) \leqslant x) & =F_{\mathrm{W}}\left(-a e^{-x}, \alpha\right)=F_{\mathrm{G}}(\alpha x-\alpha \ln a)
\end... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,911 |
38. Let $U, V$ be independent random variables, where $U$ has a uniform distribution on the interval $[0,1]$, and $\{u\}$ is the fractional part of the number $u$. Show that the random variable $\{U+V\}$ is uniformly distributed on $[0,1]$ and is independent of $V$. | Solution. First, let's note the following easily verifiable fact.
If a random variable $Z$ is uniformly distributed on $[z, 1+z]$, where $z \in \mathbb{R}$, then its fractional part $\{Z\}$ is uniformly distributed on $[0,1]$.
Due to the independence of $U$ and $V$, the variable $U+V$ conditioned on $V=v$ is uniforml... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,912 |
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