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class | __index_level_0__ int64 0 742k |
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39. Let $U, V$ be independent random variables with uniform distribution on $[0,1]$, and $\{u\}$ be the fractional part of a number $u \in \mathbb{R}$. Show that
$$
(\{U-V\},\{U+V\}) \stackrel{d}{=}(U, V) .
$$ | Solution. The random vector $(U+V, U-V)$ has a uniform distribution on the square
$$
\left\{(x, y) \in \mathbb{R}^{2}:|x-1|+|y| \leqslant 1\right\}
$$
From this, it follows that the joint density of $\{U+V\}$ and $U-V$ is
$$
f(x, y)=\frac{1}{2} I_{A}(x, y)+\frac{1}{2} I_{B}(x, y), \quad(x, y) \in \mathbb{R}^{2}
$$
... | proof | Other | proof | Yes | Yes | olympiads | false | 33,913 |
40. Let $\varphi$ and $\psi$ be two independent random variables uniformly distributed on $[0,2 \pi)$, and $\xi=\cos \varphi, \zeta=\cos \psi$. Show that
$$
\frac{\xi+\zeta}{2} \stackrel{d}{=} \xi \zeta
$$ | Solution. Due to the equality
$$
2 \xi \zeta=2 \cos \varphi \cos \psi=\cos (\varphi-\psi)+\cos (\varphi+\psi)
$$
it suffices to prove that
$$
\left(\left\{\frac{\varphi-\psi}{2 \pi}\right\},\left\{\frac{\varphi+\psi}{2 \pi}\right\}\right) \stackrel{d}{=}\left(\frac{\varphi}{2 \pi}, \frac{\psi}{2 \pi}\right)
$$
wher... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,914 |
41. Let $\xi_{1}, \ldots, \xi_{n}$ be independent copies of the random variable $\xi$, where
$$
\xi= \begin{cases}2 & \text { with probability } p \\ 1 & \text { with probability } q \\ 0 & \text { with probability } r\end{cases}
$$
and $p, q, r \geqslant 0, p+q+r=1$. Provide a general formula for the probabilities
... | Solution. Let $\zeta_{i}=\xi_{i}-1$. Then
$$
P_{n}(k, p, q)=\mathrm{P}\left(\zeta_{1}+\ldots+\zeta_{n}=k-n\right)
$$
First, consider the case $k \geqslant n$. We have
$$
\begin{aligned}
& P_{n}(k, p, q)= \\
& \quad=\quad \sum_{|k-n| \leqslant j \leqslant \frac{n+|k-n|}{2}} \mathrm{P}\left(\left|\left\{i: \zeta_{i}=1... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,915 |
42. Let $\xi$ and $\eta$ be two i.i.d. random variables with exponential distribution. Show that the random variables $\xi+\eta$ and $\xi / \eta$ are independent. | Solution. The statement follows from problem II.8.23(a), according to which, in particular, the quantity
$$
\zeta=\frac{\xi}{\xi+\eta}=\frac{1}{1+\xi / \eta}
$$
does not depend on $\xi+\eta$. Consequently, the quantity
$$
\frac{\xi}{\eta}=\frac{1}{\zeta}-1
$$
also does not depend on $\xi+\eta$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,916 |
43. Consider the matrix $\left\|\xi_{i j}\right\|$ of order $n \times n$, all (random) elements of which are independent and such that $\mathrm{P}\left(\xi_{i j}= \pm 1\right)=1 / 2$. Show that the mean value and variance of the determinant of this random matrix are respectively 0 and $n!$. | Solution. The statement follows from the equality $\mathrm{E} \xi \eta=\mathrm{E} \xi \mathrm{E} \eta$, which holds for independent integrable random variables $\xi$ and $\eta$, as well as from the fact that the determinant is a sum of $n$! non-coinciding products (up to -1) of different elements of the matrix. | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,917 |
44. Khinchin's Law of Large Numbers states that
$$
\frac{X_{1}+\ldots+X_{n}}{n}-\mathrm{E} X_{1} I\left(\left|X_{1}\right| \leqslant n\right) \xrightarrow{p} 0
$$
for any sequence of i.i.d. random variables $\left(X_{n}\right)_{n \geqslant 1}$ such that $n \mathrm{P}\left(\left|X_{1}\right|>n\right) \rightarrow 0$. A... | Solution. First of all, note that
$$
0 \leqslant \frac{X_{1}^{2}+\ldots+X_{n}^{2}}{\left(X_{1}+\ldots+X_{n}\right)^{2}} \leqslant 1-\frac{\sum_{i \neq j} X_{i} X_{j}}{\left(X_{1}+\ldots+X_{n}\right)^{2}} \leqslant 1
$$
Therefore (by the Lebesgue dominated convergence theorem), it is sufficient to check the convergenc... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,918 |
45. (See [20].) Let $X_{1}, X_{2}, \ldots$ be i.i.d. random variables, $\mathrm{P}\left(X_{1}>0\right)=1$. Let
$$
\phi(\lambda)=\mathrm{E} e^{-\lambda X_{1}}, \quad \lambda \geqslant 0,
$$
- the Laplace transform of the variable $X_{1}$. Express the quantity
$$
R_{n}(l, \alpha)=\mathrm{E} \frac{X_{1}^{l}+\ldots+X_{n... | Solution. By the Lebesgue dominated convergence theorem for all $\lambda>0$ we have
$$
\lim _{\Delta \lambda \rightarrow 0} \frac{\phi(\lambda+\Delta \lambda)-\phi(\lambda)}{\Delta \lambda}=\lim _{\Delta \lambda \rightarrow 0} \mathrm{E} e^{-\lambda X_{1}} \frac{e^{-X_{1} \Delta \lambda}-1}{\Delta \lambda}=-\mathrm{E}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,919 |
46. (See [52].) Let $X_{1}, \ldots, X_{n} (n \geqslant 1)$ be i.i.d. non-negative random variables with density $f=f(x)$,
$$
S_{n}=X_{1}+\ldots+X_{n} \quad \text { and } \quad M_{n}=\max \left\{X_{1}, \ldots, X_{n}\right\}
$$
Prove that the Laplace transform $\phi_{n}=\phi_{n}(\lambda), \lambda \geqslant 0$, of the r... | Solution. Due to the continuity of the variables $X_{i}, 1 \leqslant i \leqslant n$, the probability
$$
\mathrm{P}\left(X_{i}=X_{j} \text { for some } 1 \leqslant i<j \leqslant n\right)
$$
is equal to zero. Therefore,
$$
\phi_{n}(\lambda)=\mathrm{E} e^{-\lambda S_{n} / M_{n}}=n e^{-\lambda} \mathrm{E} e^{-\lambda S_... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,920 |
47. Let $X_{1}, \ldots, X_{n}$ be positive i.i.d. random variables, $\mathrm{E} X_{1}, \mathrm{E} X_{1}^{-1}<\infty$ and $S_{m}=X_{1}+\ldots+X_{m}, 1 \leqslant m \leqslant n$. Show that
$$
E S_{n}^{-1} \leqslant \frac{E X_{1}^{-1}}{n}, \quad E X_{1} S_{n}^{-1}=\frac{1}{n},
$$
and also that for $m \leqslant n$ the fol... | Solution. The distribution of the random vector $\left(X_{1}, \ldots, X_{n}\right)$ does not change under any permutation of its components. Therefore, taking into account the convexity of the function $f(x)=x^{-1}$ for $x>0$, we have
$$
\mathrm{E} n S_{n}^{-1}=\mathrm{E} f\left(\frac{1}{n} \sum_{i=1}^{n} X_{i}\right)... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,921 |
48. The concentration function of a random variable $X$ is defined as the function
$$
Q(X ; l)=\sup _{x \in \mathbb{R}} \mathrm{P}(x<X \leqslant x+l), \quad l \geqslant 0
$$
Show that
(a) if $X$ and $Y$ are independent random variables, then
$$
Q(X+Y ; l) \leqslant \min (Q(X ; l), Q(Y ; l)) \quad \text { for all } ... | Solution. (a) Note that if $F_{X}$ and $F_{Y}$ are the distribution functions of $X$ and $Y$, then
$$
\mathrm{P}(z<X+Y \leqslant z+l)=\int_{\mathbb{R}}\left[F_{X}(z+l-y)-F_{X}(z-y)\right] d F_{Y}(y) .
$$
Therefore,
$$
Q(X+Y ; l) \leqslant Q(X ; l) \cdot \int_{\mathbb{R}} d F_{Y}(y)=Q(X ; l)
$$
Similarly, it is esta... | proof | Other | proof | Yes | Yes | olympiads | false | 33,922 |
50. Give an example of a density $f=f(x)$, which is not an even function, for which, nevertheless,
$$
\int_{-\infty}^{0}|x|^{n} f(x) d x=\int_{0}^{\infty}|x|^{n} f(x) d x, \quad n \geqslant 1
$$ | Solution. As $f=f(x)$, one can take, for example,
$$
\frac{g(-x)}{2}+\frac{f_{\zeta}(x)}{2}
$$
where $f_{\zeta}(x)$ and $g(x)$ are featured in the solution to problem II.8.49(b). | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,924 |
51. Let $\left(\xi_{n}\right)_{n \geqslant 1}$ be a sequence of symmetrically distributed (both among themselves and with respect to zero) random variables,
$$
S_{n}=\xi_{1}+\ldots+\xi_{n}, \quad n \geqslant 1
$$
Define the sequence of partial maxima $M=\left(M_{n}\right)_{n \geqslant 1}$ by
$$
M_{n}=\max \left(0, S... | Solution. It is sufficient to use the easily established relations
$$
\left(S_{n}-S_{n-k} ; k \leqslant n\right) \stackrel{d}{=}\left(S_{k} ; k \leqslant n\right), \quad\left(-S_{k} ; k \leqslant n\right) \stackrel{d}{=}\left(S_{k} ; k \leqslant n\right)
$$
for any $n$. | proof | Other | proof | Yes | Yes | olympiads | false | 33,925 |
52. Let $\xi_{1}, \ldots, \xi_{n}$ be i.i.d. random variables with a continuous distribution function. Show that
$$
\mathrm{P}\left(\xi_{\max }=\xi_{1}\right)=n^{-1}
$$
where $\xi_{\max }=\max \left\{\xi_{1}, \ldots, \xi_{n}\right\}$.
Also, establish that the random variables $\xi_{\max }$ and $I\left(\xi_{\max }=\x... | Solution. Due to the symmetry of the values $\xi_{1}, \ldots, \xi_{n}$ for all Borel sets $B$ we have
$$
\mathrm{P}\left(\xi_{\max } \in B\right)=\sum_{i=1}^{n} \mathrm{P}\left(\xi_{\max } \in B, \xi_{\max }=\xi_{i}\right)=n \mathrm{P}\left(\xi_{\max } \in B, \xi_{\max }=\xi_{1}\right)
$$
where the fact (see problem ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,926 |
54. (Bernoulli scheme with random success probability.) Let the random variables $\xi_{1}, \ldots, \xi_{n}$ and $\pi$ be such that $\pi$ has a uniform distribution on $(0,1)$, and the variables $\xi_{i}, i=1, \ldots, n$, take two values 1 and 0 with conditional probabilities
$$
\mathrm{P}\left(\xi_{i}=1 \mid \pi=p\rig... | Solution. (a) The equality holds
$$
\begin{aligned}
& \mathrm{P}\left(\xi_{1}=x_{1}, \ldots, \xi_{n}=x_{n}\right)=\int_{0}^{1} \mathrm{P}\left(\xi_{1}=x_{1}, \ldots, \xi_{n}=x_{n} \mid \pi=p\right) d p= \\
& =\int_{0}^{1} \prod_{i=1}^{n} p^{x_{i}}(1-p)^{1-x_{i}} d p=\int_{0}^{1} p^{x}(1-p)^{n-x} d p= \\
& =\mathrm{B}(... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,928 |
55. Let $\xi$ be a non-negative random variable with a distribution function $F=F(x)$ such that $F(0)<1$ and the right derivative $F_{+}^{\prime}(0)=\lambda$ exists. Show that
$$
\text { the variable } \xi \text { is exponential } \Leftrightarrow \xi \wedge \eta \stackrel{d}{=} \xi / 2,
$$
where $\eta-$ is an indepen... | Solution. We will prove the sufficiency of the condition $\xi \wedge \eta \stackrel{d}{=} \xi / 2$ (its necessity is verified trivially). For all $x \geqslant 0$ we have
$$
\mathrm{P}(\xi>x)^{2}=\mathrm{P}(\xi \wedge \eta>x)=\mathrm{P}(\xi>2 x)
$$
and, consequently, for $n \geqslant 1$ we obtain
$$
\mathrm{P}(\xi>x)... | proof | Other | proof | Yes | Yes | olympiads | false | 33,929 |
56. (See [90].) Let $X$ be a bounded random variable with distribution function $F=F(x)$ and smooth density $f=f(x)$.
(a) Suppose $f(x)$ is non-increasing for $x>0$ and non-decreasing for $x \leqslant 0$, in other words, the variable $X$ is unimodal with mode at $x=0$. Prove that $X \stackrel{d}{=} U Z$, where $U$ and... | Solution. (a) First, note that $G$ is indeed a distribution function. Indeed, $G(\infty)=1, G(-\infty)=0$ (since $f(x)=0$ outside some interval), and for all $x \in \mathbb{R}$, the equality
$$
G^{\prime}(x)=f(x)-x f^{\prime}(x)-f(x)=-x f^{\prime}(x) \geqslant 0
$$
holds due to the unimodality (with mode at $x=0$) of... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,930 |
57. Let $X$ and $Y$ be independent random variables with distribution functions $F$ and $G$ respectively. Show that
$$
\mathrm{P}(X=Y)=\sum_{-\infty<x<\infty}[F(x)-F(x-)] \cdot[G(x)-G(x-)]
$$ | Solution. By Fubini's theorem
\[
\begin{aligned}
& \mathrm{P}(X=Y)=\left.\mathrm{E}P(x=Y)\right|_{x=X}=\mathrm{E}[G(X)-G(X-)]= \\
&=\sum_{-\infty<x<\infty}[F(x)-F(x-)] \cdot[G(x)-G(x-)]
\end{aligned}
\] | proof | Other | proof | Yes | Yes | olympiads | false | 33,931 |
58. Let $X$ and $Y$ be independent random variables. Show that the distribution function of the random variable $X+Y$
(a) is continuous if the variable $Y$ is continuous, i.e., $\mathrm{P}(Y=y)=0$ for all $y \in \mathbb{R}$;
(b) has the density $g(z)=\mathrm{E} f(z-X)$, if $Y$ has the density $f=f(y)$
(c) is singula... | Solution. (a) The statement follows from Problem II.8.57.
(b) By Fubini's theorem
$$
\mathrm{E} I_{B}(X+Y)=\left.\operatorname{EP}(x+Y \in B)\right|_{x=X}=\mathrm{E} \int_{B-X} f(y) d y=\int_{B} \mathrm{E} f(z-X) d z
$$
(c) If $X \in\left\{x_{1}, x_{2}, \ldots\right\}$ a.s., then
$$
G(z)=\mathrm{P}(X+Y \leqslant z)... | proof | Other | proof | Yes | Yes | olympiads | false | 33,932 |
59. (See [94].) Let $X, Y$ be independent symmetric random variables with densities $f=f(x)$ and $g=g(x)$ respectively. Prove that
(a) $X+Y$ is a symmetric random variable;
(b) $X+Y$ is a symmetric unimodal variable when each of the variables $X$ and $Y$ has a unimodal distribution with mode at zero. | Solution. (a) By the condition $X \stackrel{d}{=}-X$ and $Y \stackrel{d}{=}-Y$. Due to independence $(X, Y) \stackrel{d}{=}(-X,-Y)$, and thus $f(X, Y) \stackrel{d}{=} f(-X,-Y)$ for any Borel function $f=f(x, y)$. Setting $f(x, y)=x+y$, we obtain the required property.
(b) As established in problem II.8.58(b), the dens... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,933 |
60. Show that the following inclusion-exclusion formula for the maximum of random variables $X_{1}, \ldots, X_{n}$ is valid:
$$
\begin{aligned}
X_{1} \vee \ldots \vee X_{n}=\sum_{i=1}^{n} X_{i} & -\sum_{1 \leqslant i<j \leqslant n} X_{i} \wedge X_{j}+ \\
& +\sum_{1 \leqslant i<j<k \leqslant n} X_{i} \wedge X_{j} \wedg... | Solution. Due to the symmetry of the considered expressions with respect to any permutation of $X_{n}$, we can assume that $X_{1} \geqslant \ldots \geqslant X_{n}$. In this case, the required equality can be rewritten as
$$
\begin{aligned}
X_{1}=\sum_{i=1}^{n} X_{i}-\sum_{j=2}^{n} C_{j-1}^{1} X_{j}+\sum_{k=3}^{n} C_{k... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,934 |
61. Let the random variable $X$ be distributed according to the binomial law with parameters $n \in \mathbb{N}$ and $p \in [0,1]$, i.e.,
$$
\mathrm{P}(X \leqslant m)=\sum_{k=0}^{m} C_{n}^{k} p^{k} q^{n-k}, \quad 0 \leqslant k \leqslant n, \quad q=1-p,
$$
Prove that $\mathrm{P}(X \leqslant m)$ can be expressed in term... | Solution. Integrating by parts, we find that
$$
\begin{array}{rl}
\int_{p}^{1} x^{m}(1-x)^{n-m-1} & d x=\frac{p^{m} q^{n-m}}{n-m}+\frac{m}{n-m} \int_{p}^{1} x^{m-1}(1-x)^{n-m} d x= \\
= & \frac{p^{m} q^{n-m}}{n-m}+\frac{m p^{m-1} q^{n-m+1}}{(n-m)(n-m+1)}+ \\
& \quad+\frac{m(m-1)}{(n-m)(n-m+1)} \int_{p}^{1} x^{m-1}(1-x... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,935 |
63. Let $\xi_{1}, \ldots, \xi_{n}$ be i.i.d. random variables with skewness parameter
$$
\operatorname{skw}\left(\xi_{1}\right)=\frac{\mathrm{E}\left(\xi_{1}-\mathrm{E} \xi_{1}\right)^{3}}{\left(\mathrm{D} \xi_{1}\right)^{3 / 2}}
$$
and kurtosis parameter
$$
\operatorname{kur}\left(\xi_{1}\right)=\frac{\mathrm{E}\le... | Solution. Let $S_{n}=\xi_{1}+\ldots+\xi_{n}$. Without loss of generality, we can assume that $\mathrm{E} \xi_{1}=0$. From the independence and identical distribution of the variables $\xi_{i}, 1 \leqslant i \leqslant n$, we obtain that
\[
\begin{gathered}
\mathrm{D} S_{n}=n \mathrm{D} \xi_{1} \\
\mathbb{E} S_{n}^{3}=n... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,937 |
64. Let $X$ be a binomial random variable with parameters $n \in \mathbb{N}$ and $p \in [0,1]$. Show that for it, the skewness parameter is given by the formula (where $q=1-p$)
$$
\operatorname{skw}(X)=\frac{\mathrm{E}(X-\mathrm{E} X)^{3}}{(\mathrm{D} X)^{3 / 2}}=\frac{q-p}{\sqrt{n p q}}
$$
In particular, if $0<p<1$,... | Solution. Using the result of problem II.8.63, as well as the equality $X \stackrel{d}{=} \sum_{i=1}^{n} \xi_{i}$, where $\left(\xi_{i}\right)_{i=1}^{n}-$ are i.i.d. Bernoulli random variables, $\mathrm{P}\left(\xi_{1}=1\right)=p$ and $\mathrm{P}\left(\xi_{1}=0\right)=q$, we find that
$$
\operatorname{skw}(X)=\frac{q-... | \operatorname{skw}(X)=\frac{q-p}{\sqrt{npq}},\quad\operatorname{kur}(X)=3+\frac{1}{npq}-\frac{6}{n} | Algebra | proof | Yes | Yes | olympiads | false | 33,938 |
65. When a binomial distribution is introduced, the number of trials $n$ is fixed, and the probability $\mathrm{P}_{n}(\nu=r)$ that the number of "successes" $\nu$ in these $n$ trials equals $r$ is considered. These probabilities $\mathrm{P}_{n}(\nu=r)=C_{n}^{r} p^{r} q^{n-r}, 0 \leqslant r \leqslant n$, form the binom... | Solution. Let $\left(\xi_{n}\right)_{n \geqslant 1}$ be independent Bernoulli random variables,
$$
\mathrm{P}\left(\xi_{n}=1\right)=p \quad \text { and } \quad \mathrm{P}\left(\xi_{n}=0\right)=q
$$
and $S_{n}=\xi_{1}+\ldots+\xi_{n}, n \geqslant 1$. Then
$$
\mathrm{P}^{r}(\tau=k)=\mathrm{P}\left(S_{k-1}=r-1, \xi_{k}=... | \frac{r}{p} | Combinatorics | proof | Yes | Yes | olympiads | false | 33,939 |
66. (a) It is said that a random variable $\xi$ with natural values is distributed according to the discrete Pareto law with parameter $\rho>0$, if
$$
\mathrm{P}(\xi=n)=\frac{1}{\zeta(\rho+1) n^{\rho+1}}, \quad n \in \mathbb{N}
$$
where $\zeta(s)=\sum_{n \geqslant 1} n^{-s}$ is the Riemann zeta function. Show that
$... | Solution. (a) The statement follows from the definition of $\mathrm{E} \xi$.
(b) First of all, note that (by Fubini's theorem)
$$
I=\int_{[0,1]^{2}} \frac{d x d y}{1-x y}=\int_{[0,1]^{2}} \sum_{n=1}^{\infty}(x y)^{n-1} d x d y=\int_{0}^{1} \sum_{n=1}^{\infty} \frac{y^{n-1}}{n} d y=\sum_{n=1}^{\infty} \frac{1}{n^{2}}
... | \frac{\pi^{2}}{6} | Algebra | proof | Yes | Yes | olympiads | false | 33,940 |
67. Let $\xi_{1}, \ldots, \xi_{n}$ be independent and identically distributed (i.i.d.) random variables with distribution function $F(x \mid \theta)$, depending on some (random) parameter $\theta$, which has a prior distribution $\Pi(\theta)$ from some class $\mathscr{K}$.
Let $\Pi\left(\theta \mid x_{1}, \ldots, x_{n... | Solution. The conjugacy property of the specified distributions is established by direct computation using the formula
$$
\pi\left(\theta \mid x_{1}, \ldots, x_{n}\right)=\frac{\pi(\theta) f\left(x_{1} \mid \theta\right) \ldots f\left(x_{n} \mid \theta\right)}{\int_{\mathbb{R}} \pi(z) f\left(x_{1} \mid z\right) \ldots... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,941 |
69. (Gamma and Beta functions, [56].) In this problem, we are to obtain an integral representation for $n!$ and thereby introduce the gamma function as L. Euler did. First, show that
$$
\int_{0}^{1} x^{\alpha-1}(1-x)^{n} d x=\frac{n!}{\alpha(\alpha+1) \ldots(\alpha+n)}, \quad \alpha>0, \quad n=0,1,2, \ldots
$$
From t... | Solution. To establish the relation (*), it is sufficient to integrate by parts $n$ times. By transitioning to the new variable $z=x^{\alpha}$ and setting $\alpha=1/a$, we get
$$
\int_{0}^{1} x^{1/a-1}(1-x)^{n} d x=a^{n} \int_{0}^{1}\left(\frac{1-z^{a}}{a}\right)^{n} \frac{d z}{1/a}
$$
On the other hand,
$$
\frac{1}... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,942 |
70. Let $X$ be a random variable with a gamma distribution with density
$$
f(x ; \alpha, \lambda)=\frac{\lambda^{\alpha}}{\Gamma(\alpha)} x^{\alpha-1} e^{-\lambda x} I(x>0)
$$
where $\alpha, \lambda>0$ and
$$
\Gamma(\alpha)=\int_{0}^{\infty} x^{\alpha-1} e^{-x} d x
$$
Show that
$$
\mathrm{E} X=\frac{\alpha}{\lambd... | Solution. If $X$ has a gamma distribution with parameters $(\alpha, \lambda)$, then $\lambda X$ has a gamma distribution with parameters $(\alpha, 1)$. For simplicity, we will assume $\lambda=1$ from now on. We will find $\mathrm{E} X^{p}$:
$$
\mathrm{E} X^{p}=\int_{0}^{\infty} x^{p} \frac{x^{\alpha-1} e^{-x}}{\Gamma(... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,943 |
71. Let $X$ be a random variable having a beta distribution with density
$$
f(x ; \alpha, \beta)=\frac{x^{\alpha-1}(1-x)^{\beta-1}}{\mathrm{~B}(\alpha, \beta)} I(0<x<1)
$$
where $\alpha>0$, $\beta>0$ and
$$
\mathrm{B}(\alpha, \beta)=\frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha+\beta)}=\int_{0}^{1} x^{\alpha-1}(... | Solution. Let's find $E X^{p}$:
$$
\mathrm{E} X^{p}=\int_{0}^{1} x^{p} \frac{x^{\alpha}(1-x)^{\beta-1}}{\mathrm{~B}(\alpha, \beta)} d x=\frac{\int_{0}^{1} x^{\alpha+p}(1-x)^{\beta-1} d x}{\mathrm{~B}(\alpha, \beta)}=\frac{\mathrm{B}(\alpha+p, \beta)}{\mathrm{B}(\alpha, \beta)}
$$
In particular,
$$
\mathrm{E} X=\frac... | \mathrm{E}X=\frac{\alpha}{\alpha+\beta},\quad\mathrm{D}X=\frac{\alpha\beta}{(\alpha+\beta+1)(\alpha+\beta)^{2}},\quad\mathrm{E}X^{p}=\frac{\mathrm{B}(\alpha+p,\beta)}{\mathrm{B}(\alpha,\beta)} | Algebra | proof | Yes | Yes | olympiads | false | 33,944 |
72. The multidimensional analog of the beta distribution is the Dirichlet distribution, introduced as a distribution on the set
$$
\Delta_{n-1}=\left\{\left(x_{1}, \ldots, x_{n-1}\right): x_{i} \in \mathbb{R}_{+}, 0 \leqslant x_{1}+\ldots+x_{n-1} \leqslant 1\right\}, \quad n \geqslant 2
$$
defined by the density
$$
... | Solution. Point (a) can be derived from point (b) taking into account the equality $\Gamma(\alpha+1)=\alpha \Gamma(\alpha)$, valid for $\alpha>0$. Point (b) is a direct consequence of the fact that for all $\gamma_{1}, \ldots, \gamma_{n}>0$ the function
$$
f\left(x_{1}, \ldots, x_{n} \mid \gamma_{1}, \ldots, \gamma_{n... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,945 |
73. Let $X_{1}, \ldots, X_{n}$ be independent random variables uniformly distributed on the interval $[0,1]$. Denote by $X_{1: n}, \ldots, X_{n: n}$ the corresponding order statistics (see problem II.8.17).
Prove that for any $1 \leqslant r_{1}<\ldots<r_{k} \leqslant n$ the random vector
$$
\left(X_{r_{1}: n}, X_{r_{... | Solution. The density of the vector ( $X_{r_{1}: n}, X_{r_{2}: n}, \ldots, X_{r_{k}: n}$ ) is derived in the same way as in problem II.8.17(c). Next, one should use the formula for the transformation of density under a smooth change of coordinates (see problem II.8.34). | proof | Other | proof | Yes | Yes | olympiads | false | 33,946 |
75. (See [105].) Let $\xi$ and $\zeta$ be independent random variables, where $\xi$ has a gamma distribution with parameters $(\gamma, \lambda)$, and $\zeta$ has a beta distribution with parameters $(\alpha, \gamma-\alpha)$ for some $\alpha, \gamma, 0<\alpha<\gamma, \lambda>0$.
Using the fact that the Laplace transfor... | Solution. For simplicity, we will assume that $\lambda=1$ (otherwise, we multiply $\xi$ by $\lambda$). For any $s>0$, the following equalities hold (see problems II.8.70 and II.8.71):
$$
\begin{aligned}
\phi(s)=\mathrm{E}(\zeta \xi)^{s}=\mathrm{E} \xi^{s} \mathrm{E} \zeta^{s} & =\frac{\Gamma(s+\gamma)}{\Gamma(\gamma)}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,948 |
76. Let $X_{\alpha}, \alpha>0,$ be independent random variables having a gamma distribution with parameters $(\alpha, \lambda)$ for some positive $\lambda$. Using the Legendre duplication formula for the gamma function
$$
\Gamma(s) \Gamma\left(s+\frac{1}{2}\right)=2^{1-2 s} \Gamma(2 s) \Gamma\left(\frac{1}{2}\right), ... | Solution. Without loss of generality, we will assume that $\lambda=1$ (otherwise, we divide all $X_{\alpha}$ by $\lambda$). Reasoning as in the solution to problem II.8.75, using the duplication formula of Legendre for any $s>0$, we find
$$
\begin{aligned}
\mathrm{E} 4^{s} X_{\alpha} X_{\alpha+1 / 2}^{s} & =4^{s} \mat... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,949 |
77. (See [67].) Let $X, Y_{1}, Y_{2}, \ldots$ be independent random variables, where $X$ has a gamma distribution with parameters $(\alpha, 1)$, and $Y_{n}$, $n \geqslant 1$, are exponentially distributed with $\mathrm{P}\left(Y_{n}>x\right)=e^{-x}$, $x>0$. Using the result from problem II.8.74 and the elementary repre... | Solution. It is easy to show that $Z^{1 / \alpha}$ has a uniform distribution when $Z$ is distributed according to the beta law with parameters $(\beta, \beta+1)$, $\beta>0$. Therefore, by problem II.8.74 (parts (a) and (b)), the quantities
$U_{0}=\left(\frac{X}{X+S_{1}}\right)^{1 / \alpha}, \quad \ldots, \quad U_{n-1... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,950 |
78. Based on the representation of random variables from problem II.8.77, provide another proof of the main result of problem II.8.76:
$$
4 X_{\alpha} X_{\alpha+1 / 2} \stackrel{d}{=} X_{2 \alpha}^{2}
$$
without using the Legendre duplication formula. Derive this formula as a consequence of the stated property, there... | Solution. Without loss of generality, we can assume that $X_{\alpha}$ has a gamma distribution with parameters $(\alpha, 1)$ (otherwise, we multiply $X_{\alpha}$ by $\lambda^{-1}$). We have
\[
\begin{aligned}
& 2 \ln X_{2 \alpha} \stackrel{d}{=}-2 \gamma+2 \sum_{n=0}^{\infty}\left[\frac{1}{n+1}-\frac{Y_{n}}{n+2 \alpha... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,951 |
80. (See [36].) Let the conditions of problem II.8.79 be satisfied.
(a) Applying the Lévy formula (see the remark to problem II.13.49)
$$
\int_{0}^{\infty} e^{-\lambda t} \frac{a e^{-\frac{a^{2}}{2 t}} d t}{\sqrt{2 \pi t^{3}}}=e^{-a \sqrt{2 \lambda}}, \quad a, \lambda>0
$$
find the density of the quantity $Z_{\alpha... | Solution. (a) According to problem II.11.17, the Laplace transform of a non-negative quantity with a density uniquely determines this density. Therefore, we only need to find non-negative functions $f_{\alpha}=f_{\alpha}(y)$ and $g_{\alpha}=g_{\alpha}(z)$, for which for all $\lambda>0$ the following equalities hold:
$... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,953 |
81. Let $\theta$ be an absolutely continuous random variable taking values in the interval $[0, \pi]$. Show that if
$$
\cos 2 \theta \stackrel{d}{=} \cos \theta,
$$
then $\theta$ has a uniform distribution on $[0, \pi]$. | Solution. Suppose for simplicity that $\theta$ has a density $f$ that is continuous on $[0, \pi]$. Let's illustrate the main idea of the proof by calculating the probability $\mathrm{P}(\theta \in[0, \pi / 2])$. Since $\cos (2 \theta) \stackrel{d}{=} \cos \theta$, we have
$$
\begin{aligned}
& \mathrm{P}(\theta \in[0, ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,954 |
82. Let $\xi_{1}, \ldots, \xi_{n}$ be independent random variables uniformly distributed on $[0,1]$. Using the inclusion-exclusion formula from problem II.8.60, prove Laplace's formula
$$
\mathrm{P}\left(\xi_{1}+\ldots+\xi_{n} \leqslant x\right)=\frac{1}{n!} \sum_{k=0}^{[x]}(-1)^{k} C_{n}^{k}(x-k)^{n}, \quad 0 \leqsla... | Solution. We need the formula for the volume of an $n$-dimensional simplex
$$
V\left(S_{z}\right)=\frac{z^{n}}{n!}, \quad S_{z}=\left\{x=\left(x_{1}, \ldots, x_{n}\right) \in \mathbb{R}_{+}^{n}: \sum_{i=1}^{n} x_{i} \leqslant z\right\}
$$
Let's establish it. The volume of the $n$-dimensional cube $B=[0, z]^{n}$ is $z... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,955 |
1. Let $(\Omega, \mathscr{F})=((0,1], \mathscr{B}((0,1]))$, and $\mathrm{P}$ be the Lebesgue measure on $(0,1]$. Show that $(\Omega, \mathscr{F}, \mathrm{P})$ is a universal probability space in the sense that for any distribution function $F=F(x)$, one can define a random variable $\xi=\xi(\omega), \omega \in \Omega$,... | Solution. Let $U(\omega)=\omega, \omega \in(0,1]$. Set $\xi=F^{-1}(U)$ for $U \in(0,1)$, and take $\xi(1)$ arbitrarily; here
$$
F^{-1}(u)=\inf \{x: u \leqslant F(x)\}, \quad u \in(0,1)
$$
Then for all $u \in(0,1)$ and $x \in \mathbb{R}$ we have
$$
F^{-1}(u) \leqslant x \Leftrightarrow u \leqslant F(x)
$$
therefore ... | proof | Other | proof | Yes | Yes | olympiads | false | 33,956 |
2. Continuing from the previous task, construct on the space $((0,1], \mathscr{B}((0,1]), \mathrm{P})$
(a) a countable system of independent Bernoulli random variables $\left\{\xi_{n}\right\}_{n \geqslant 1}$ such that
$$
\mathrm{P}\left(\xi_{1}=0\right)=\mathrm{P}\left(\xi_{1}=1\right)=\frac{1}{2}
$$
(b) a sequence... | Solution. (a) As $\xi_{n}(\omega)$, one should take the $n$-th digit in the non-terminating (i.e., with an infinite number of ones) binary expansion of the number $\omega$.
(b) Reorder $\left\{\xi_{n}\right\}_{n \geqslant 1}$ from part (a) into a matrix $\left\{\xi_{i n}\right\}_{i, n \geqslant 1}$ and set
$$
U_{n}=\... | proof | Other | proof | Yes | Yes | olympiads | false | 33,957 |
3. In solving problem II.9.1, it was established that the probability space ( $\Omega, \mathscr{F}, \mathrm{P}$ ) is universal if and only if a random variable $U=U(\omega)$ can be constructed on it, having a uniform distribution on the interval $[0,1]$.
Show that the property of universality of ( $\Omega, \mathscr{F}... | Solution. We will prove the necessity. Suppose that on $(\Omega, \mathscr{F}, \mathrm{P})$ a random variable $U=U(\omega)$ is defined with a uniform distribution on $[0,1]$, but the measure $\mathrm{P}$ has an atom $A(\mathrm{P}(A)>0)$. By the definition of an atom, if $\mathrm{P}(A)>\mathrm{P}(B)$ and $B \subset A(B \... | proof | Other | proof | Yes | Yes | olympiads | false | 33,958 |
4. Consider the so-called renewal equation
$$
m(t)=F(t)+\int_{0}^{t} m(t-s) d F(s)
$$
where $F(t)$ is the distribution function of a non-degenerate non-negative random variable $\xi$. Show that in the class of functions bounded on finite intervals, the unique solution to the renewal equation is the function defined b... | Solution. Iterating the equality $(*)$, we get $m(t)=\sum_{i=1}^{n} F_{i}(t)+R(t)$,
Now we will show that the series $\sum_{n \geqslant 1} F_{n}(t)$ converges for all $t \geqslant 0$ (in ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,959 |
5. Let $T$ be an arbitrary set.
(a) Suppose that for each $t \in T$ a probability space $\left(\Omega_{t}, \mathscr{F}_{t}, \mathrm{P}_{t}\right)$ is given. Let
$$
\Omega=\prod_{t \in T} \Omega_{t}, \quad \mathscr{F}=\otimes_{t \in T} \mathscr{F}_{t}
$$
Then
on $(\Omega, \mathscr{F})$ there exists a unique probabil... | Solution. (a) We have
$$
\mathscr{F}=\bigcup_{t_{\infty}} \mathscr{F}_{t_{\infty}}
$$
where the union is taken over all $t_{\infty}=\left(t_{n}\right)_{n \geqslant 1} \in T^{\infty}$, and the $\sigma$-algebras $\mathscr{F}_{\infty}$ consist of all sets $B \subseteq \Omega$ such that
$$
B=\left\{\left(\omega_{t}\righ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,960 |
1. Let $\left\{X_{n}\right\}_{n \geqslant 1}$ be a uniformly integrable family of non-negative random variables, and $M_{n}=\max \left\{X_{1}, \ldots, X_{n}\right\}$. Prove that
$$
\frac{\mathrm{E} M_{n}}{n} \rightarrow 0, \quad n \rightarrow \infty
$$
Establish that if in addition $X_{n} \preccurlyeq X$ for some non... | Solution. For $a>0$ we have
$$
M_{n} I\left(M_{n}>a\right)=\max _{1 \leqslant k \leqslant n} X_{k} I\left(X_{k}>a\right) \leqslant \sum_{k=1}^{n} X_{k} I\left(X_{k}>a\right)
$$
therefore
$$
\mathrm{E} \frac{M_{n}}{n} \leqslant \frac{a}{n}+\frac{1}{n} \sum_{k=1}^{n} \mathrm{E} X_{k} I\left(X_{k}>a\right) \leqslant \f... | proof | Other | proof | Yes | Yes | olympiads | false | 33,961 |
2. Let $X_{1}, X_{2}, \ldots$ be some random variables. Consider the conditions
(1) $\mathrm{P}\left(X-X_{n}>\varepsilon\right) \rightarrow 0$ for all $\varepsilon>0$;
(2) $\mathrm{P}\left(X-X_{n}<-\varepsilon\right) \rightarrow 0$ for all $\varepsilon>0$.
Establish that among all extended variables $X$ satisfying p... | Solution. Let $\mathscr{X}$ be the set of quantities $X$ satisfying property 1. Note that $\mathscr{X}$ is non-empty, as $-\infty$ can be taken as $X$. If $X, Y \in \mathscr{X}$, then $X \wedge Y \in \mathscr{X}$, since
$$
\mathrm{P}\left(X \wedge Y - X_{n} > \varepsilon\right) \leqslant \mathrm{P}\left(X - X_{n} > \v... | proof | Other | proof | Yes | Yes | olympiads | false | 33,962 |
3. Let $\left\{X_{n}\right\}_{n \geqslant 1}$ be i.i.d. random variables, and
$$
S_{0}=0, \quad S_{n}=X_{1}+\ldots+X_{n}, \quad n \geqslant 1
$$
Let also $\mathscr{R}_{n}$ be the range, i.e., the number of distinct values taken by the variables $S_{0}, S_{1}, \ldots, S_{n}$. Let
$$
\sigma(0)=\inf \left\{n>0: S_{n}=0... | Solution. Due to the boundedness of $\mathscr{R}_{n} / n$ (for example, by two), it is sufficient to prove the almost sure (a.s.) convergence in formula $(*)$, as $L^{p}$ convergence will then be ensured by the dominated convergence theorem of Lebesgue. For any $k \in \mathbb{N}$, by the strong law of large numbers, we... | proof | Other | proof | Yes | Yes | olympiads | false | 33,963 |
4. Prove that the space $L^{\infty}$ is complete. | Solution. Let $\left(\xi_{n}\right)_{n \geqslant 1}$ be a fundamental sequence in $L^{\infty}$, i.e., $\left\|\xi_{m}-\xi_{n}\right\|_{\infty} \rightarrow 0$ as $m, n \rightarrow \infty$. Set
$$
\xi=\varlimsup_{n} \xi_{n} \cdot I\left(\varlimsup_{n} \xi_{n}<\infty\right)
$$
The $\xi$ defined in this way is a random v... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,964 |
5. Show that if $\xi_{n} \xrightarrow{p} \xi$ and at the same time $\xi_{n} \xrightarrow{p} \zeta$, then $\xi$ and $\zeta$ are equivalent, i.e., $\mathrm{P}(\xi=\zeta)=1$. | Solution. The statement follows from the inequality
$$
\mathrm{P}(|\xi-\zeta| \geqslant 2 \varepsilon) \leqslant \mathrm{P}\left(\left|\xi_{n}-\xi\right| \geqslant \varepsilon\right)+\mathrm{P}\left(\left|\xi_{n}-\zeta\right| \geqslant \varepsilon\right)
$$
satisfied for any $\varepsilon>0$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,965 |
6. Let $\xi_{n} \xrightarrow{p} \xi, \zeta_{n} \xrightarrow{p} \zeta$ and the random variables $\xi$ and $\zeta$ are equivalent. Show that then for any $\varepsilon>0$ the condition
$$
\mathrm{P}\left(\left|\xi_{n}-\zeta_{n}\right| \geqslant \varepsilon\right) \rightarrow 0, \quad n \rightarrow \infty
$$
holds. | Solution. The statement follows from the inequality
$$
\mathrm{P}\left(\left|\xi_{n}-\zeta_{n}\right| \geqslant 3 \varepsilon\right) \leqslant \mathrm{P}\left(\left|\xi_{n}-\xi\right| \geqslant \varepsilon\right)+\mathrm{P}\left(\left|\zeta_{n}-\zeta\right| \geqslant \varepsilon\right)+\mathrm{P}(|\xi-\zeta| \geqslant... | proof | Other | proof | Yes | Yes | olympiads | false | 33,966 |
7. Prove that if $\xi_{n} \xrightarrow{p} \xi$ and $\zeta_{n} \xrightarrow{p} \zeta$, then $\left(\xi_{n}, \zeta_{n}\right) \xrightarrow{p}(\xi, \zeta)$, i.e., for all $\varepsilon>0$ the following equality holds:
$$
\lim _{n} \mathrm{P}\left(\left|\left(\xi_{n}, \zeta_{n}\right)-(\xi, \zeta)\right|>\varepsilon\right)... | Solution. This property is easily verified using the relation
$$
\left\{\left|\left(\xi_{n}, \zeta_{n}\right)-(\xi, \zeta)\right|>\varepsilon\right\} \subset\left\{\left|\xi_{n}-\xi\right|>\varepsilon / 2\right\} \cup\left\{\left|\zeta_{n}-\zeta\right|>\varepsilon / 2\right\}
$$ | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,967 |
8. Let $\xi, \xi_{1}, \xi_{2}, \ldots$ be random vectors in $\mathbb{R}^{d}$. Prove that $\xi_{n} \xrightarrow{p} \xi$, if and only if for any increasing sequence $\left(n^{\prime}\right) \subset(n)$ there exists a subsequence $\left(n^{\prime \prime}\right) \subset\left(n^{\prime}\right)$, for which
$$
\xi_{n^{\prime... | Solution. If $\xi_{n} \xrightarrow{p} \xi$, then by definition $\xi_{n^{\prime}} \xrightarrow{p} \xi$ and, therefore, for some subsequence $\left(n^{\prime \prime}\right)=\left(n_{1}, n_{2}, \ldots\right) \subset\left(n^{\prime}\right)$ the inequality holds
$$
\sum_{k} \mathrm{P}\left(\left|\xi_{n_{k}}-\xi\right|>2^{-... | proof | Other | proof | Yes | Yes | olympiads | false | 33,968 |
9. Let $\xi_{n} \xrightarrow{p} \xi$, where $\xi, \xi_{1}, \xi_{2}, \ldots$ are random vectors in $\mathbb{R}^{d}$, and the function $\varphi=\varphi(x)$, defined on $\mathbb{R}^{d}$, is Borel measurable. Prove that
$$
D_{\varphi}=\left\{x \in \mathbb{R}^{d}: \varphi \text { is discontinuous at } x\right\} \in \mathsc... | Solution. The generalized Slutsky's lemma is a consequence of the equivalent definition of convergence in probability from problem II.10.8. We will prove that $D_{\varphi} \in \mathscr{B}\left(\mathbb{R}^{d}\right)$. We have $D_{\varphi}=\bigcup_{m=1}^{\infty} \bigcap_{n=1}^{\infty} D_{m n}$, where
\[
\begin{aligned}
... | proof | Other | proof | Yes | Yes | olympiads | false | 33,969 |
10. Suppose that $\zeta_{n} \xrightarrow{d} c$, where $c$ is a constant. Prove that then $\zeta_{n} \xrightarrow{p} c$.
Show also that if $\xi_{n} \xrightarrow{d} \xi$ for some $\xi, \xi_{1}, \xi_{2} \ldots$, then $\left(\xi_{n}, \zeta_{n}\right) \xrightarrow{d}(\xi, c)$, i.e., for any continuous bounded function $f$ ... | Solution. Fix $\varepsilon>0$ and consider the function
$$
f_{\varepsilon}(x)=\left(1-\frac{|x-c|}{\varepsilon}\right)^{+}
$$
Then
$$
\mathrm{P}\left(\left|\zeta_{n}-c\right| \leqslant \varepsilon\right) \geqslant \mathrm{E} f_{\varepsilon}\left(\zeta_{n}\right) \rightarrow \mathrm{E} f_{\varepsilon}(c)=1
$$
Now le... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,970 |
11. Show that $\xi_{n} \rightarrow 0$ a.s., if
$$
\sum_{n} \mathrm{E}\left|\xi_{n}\right|<\infty
$$ | Solution. Let $\xi=\sum_{n}\left|\xi_{n}\right|$ be a non-negative random variable with values in $\overline{\mathbb{R}}$. By Fubini's theorem, $\mathrm{E} \xi<\infty$, which, in particular, means that $\xi$ is finite almost surely. From this, it follows that with probability one, $\left|\xi_{n}\right| \rightarrow 0$. ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,971 |
12. Let $\left(\xi_{n}\right)_{n \geqslant 1}$ be a sequence of identically distributed random variables. Prove that the following implications hold:
$$
\begin{aligned}
& \mathrm{E}\left|\xi_{1}\right|\varepsilon n\right)0 \Leftrightarrow \\
& \Leftrightarrow \sum_{n \geqslant 1} \mathrm{P}\left(\left|\xi_{n}\right|>\... | Solution. The statement follows from the easily verifiable inequalities
$$
\varepsilon \sum_{n \geqslant 1} \mathrm{P}\left(\left|\xi_{1}\right|>\varepsilon n\right) \leqslant \mathrm{E}\left|\xi_{1}\right| \leqslant \varepsilon+\varepsilon \sum_{n \geqslant 1} \mathrm{P}\left(\left|\xi_{1}\right|>\varepsilon n\right)... | proof | Other | proof | Yes | Yes | olympiads | false | 33,972 |
13. Prove that if $\sum \mathrm{P}\left(\xi_{n}>\varepsilon\right)<\infty$ for any $\varepsilon > 0$, then $\lim _{n} \xi_{n} \leqslant \varepsilon$ a.s.[^4] | Solution. By the Borel-Cantelli lemma, $\mathrm{P}\left(\xi_{n}>\varepsilon\right.$ i.o.) $=0$. This means that $\overline{\lim } \xi_{n} \leqslant \varepsilon$ a.s. | proof | Other | proof | Yes | Yes | olympiads | false | 33,973 |
15. Let's define the "d-metric" in the set of random variables by setting
$$
d(\xi, \eta)=\mathrm{E} f(|\xi-\eta|)
$$
for a bounded, increasing, and concave function \( f \) such that \( f(0)=0 \), and identifying random variables that coincide almost surely (in particular, one can take \( f(x)=x /(1+x) \)). Show tha... | Solution. Due to the concavity of the function $f$, we have
$$
f(x+y)-f(x) \leqslant f(y)-f(0)=f(y), \quad x, y \geqslant 0,
$$
which, taking into account the monotonicity of $f$, ensures the triangle inequality for $d$. Furthermore, for all $\varepsilon>0$ we have
$f(\varepsilon) \mathrm{P}\left(\left|\xi_{n}-\xi\r... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,975 |
16. Show that there is no such metric in the set of random variables that convergence in it is equivalent to convergence a.s. | Solution. Suppose there exists a metric $\rho$ that defines convergence in probability. Consider some sequence $\left(\xi_{n}\right)_{n \geqslant 1}$ for which $\xi_{n} \xrightarrow{p} 0$, but $\xi_{n} \nrightarrow 0$ almost surely. Then there exists some $\varepsilon>0$ such that for some subsequence $\left(n_{k}\righ... | proof | Other | proof | Yes | Yes | olympiads | false | 33,976 |
17. Let $\xi_{(n)} \preccurlyeq \xi_{n+1}, n \geqslant 1$, and $\xi_{n} \xrightarrow{p} \xi$, where $\xi_{(n)}=\xi_{1} \vee \ldots \vee \xi_{n}$, and «ঞ» denotes stochastic dominance (see problem II.10.1). Prove that then $\xi_{n} \rightarrow \xi$ a.s. | Solution. Let $\eta=\lim _{n} \xi_{(n)}$ (the limit is defined everywhere, since $\xi_{(n)}(\omega)$ is non-decreasing in $n$ for each outcome $\omega$). The following implications hold:
$$
\xi_{(n)} \rightarrow \eta \text { a.s. } \Rightarrow \xi_{(n)} \stackrel{d}{\rightarrow} \eta \quad \text { and } \quad \xi_{n+1... | proof | Other | proof | Yes | Yes | olympiads | false | 33,977 |
18. Let $\left(X_{n}\right)_{n \geqslant 1}$ be a sequence of random variables, and $S_{n}=$ $=X_{1}+\ldots+X_{n}$.
(a) Assuming that $p \geqslant 1$, establish that
$$
\begin{aligned}
& X_{n} \rightarrow 0 \text { a.s. } \Rightarrow \frac{S_{n}}{n} \rightarrow 0 \quad \text { a.s. } \quad \text { as } n \rightarrow ... | Solution. For numerical sequences $\left\{x_{n}\right\}_{n \geqslant 1}$ we have
$$
x_{n} \rightarrow 0 \Rightarrow \frac{1}{n} \sum_{i=1}^{n} x_{i} \rightarrow 0
$$
This, together with the inequality
$$
\mathrm{E}\left|\frac{S_{n}}{n}\right|^{p} \leqslant \frac{1}{n} \sum_{i=1}^{n} \mathrm{E}\left|X_{i}\right|^{p}
... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,978 |
19. Let $(\Omega, \mathscr{F}, \mathrm{P})$ be a probability space and $\xi_{n} \xrightarrow{p} \xi$. Show that if the measure P is atomic (see problem II.3.35), then $\xi_{n} \rightarrow \xi$ with probability one. Also, establish that if the notions of convergence in probability and almost sure convergence for $(\Omeg... | Solution. Suppose that $\xi_{n} \xrightarrow{p} \xi$, but $\xi_{n} \nrightarrow \xi$ a.s., i.e., the set on which there is no convergence has a positive measure. Let us select an atom $A$ of positive measure in this set. Note that $\mathrm{P}\left(\left\{\left|\xi_{n}-\xi\right|>\varepsilon\right\} \cap A\right) \right... | proof | Other | proof | Yes | Yes | olympiads | false | 33,979 |
20. (To the Borel-Cantelli lemma.) (a) Let $\Omega=(0,1), \mathscr{B}=$ $\mathscr{B}((0,1)), \mathrm{P}-$ be the Lebesgue measure. Consider the events $A_{n}=(0,1 / n)$. Show that $\sum \mathrm{P}\left(A_{n}\right)=\infty$, but each $\omega$ from $(0,1)$ can belong to only a finite number of sets $A_{1}, \ldots, A_{[1 ... | Solution. (a) The statement is established by direct verification.
(b) It suffices to take $\xi_{n}=n I_{A_{n}}$, where events $A_{n}, n \geqslant 1$, and the corresponding probability space are given in part (a). | proof | Other | proof | Yes | Yes | olympiads | false | 33,980 |
21. (See [60].) Let $A_{1}, A_{2}, \ldots$ be an arbitrary sequence of events. Prove the following variant of the second Borel-Cantelli lemma (weighted version of the Erdős-Rényi theorem):
(a) If $\sum_{n \geqslant 1} \alpha_{n} \mathrm{P}\left(A_{n}\right)=\infty$ for some $\alpha_{n} \geqslant 0$, then
$$
\mathrm{P... | Solution. (a) According to the weighted Chung-Erdős inequality (see problem I.1.11) we have
$\mathrm{P}\left(A_{k}\right.$ i.o. $)=\lim _{k} \mathrm{P}\left(\bigcup_{i=k}^{\infty} A_{i}\right) \geqslant \lim _{k} \varlimsup_{n} \frac{\left[\sum_{i=k}^{n} \alpha_{i} \mathrm{P}\left(A_{i}\right)\right]^{2}}{\sum_{i, j=k... | proof | Other | proof | Yes | Yes | olympiads | false | 33,981 |
22. Let $A_{1}, A_{2}, \ldots$ be pairwise independent events and $\sum_{n=1}^{\infty} \mathrm{P}\left(A_{n}\right)=$ $=\infty$. Prove that for $S_{n}=\sum_{k=1}^{n} I\left(A_{k}\right)$ the following second Borel-Cantelli lemma holds:
$$
\lim _{n} \frac{S_{n}}{E S_{n}}=1 \quad \text { a.s. }
$$ | Solution. Letting $n_{k}=\inf \left\{n \geqslant 1: \mathrm{ES}_{n} \geqslant k^{2}\right\}$, taking into account that $\mathrm{D} S_{n} \leqslant \mathrm{E} S_{n}$, we have
$$
\sum_{k=1}^{\infty} \mathrm{E}\left|\frac{S_{n_{k}}}{\mathrm{E} S_{n_{k}}}-1\right|^{2} \leqslant \sum_{k=1}^{\infty} \frac{1}{\mathrm{E} S_{n... | proof | Other | proof | Yes | Yes | olympiads | false | 33,982 |
23. Let $\left(\xi_{n}\right)_{n \geqslant 1}$ and $\left(\eta_{n}\right)_{n \geqslant 1}$ be two sequences of random variables, whose finite-dimensional distributions coincide, i.e., $\left(\xi_{1}, \ldots, \xi_{n}\right) \stackrel{d}{=}\left(\eta_{1}, \ldots, \eta_{n}\right), n \in \mathbb{N}$. Suppose $\xi_{n} \xrig... | Solution. Since $\xi_{n} \xrightarrow{p} \xi$, the quantities $\xi_{n}$ form a fundamental sequence in probability, and thus the same property holds for $\eta_{n}$. This ensures the existence of the quantity $\eta$. Moreover, $\xi_{n} \stackrel{d}{=} \eta_{n}$, so $\xi_{n} \xrightarrow{d} \xi$ and $\eta_{n} \xrightarro... | proof | Other | proof | Yes | Yes | olympiads | false | 33,983 |
24. Let $\left(X_{n}\right)_{n \geqslant 1}$ be a sequence of pairwise independent random variables such that $X_{n} \xrightarrow{p} X$ for some random variable $X$. Prove that $X$ is a degenerate random variable. | Solution. By the condition $X_{2 n} \xrightarrow{p} X$ and $X_{2 n+1} \xrightarrow{p} X$. Therefore, in view of problems II. 10.7 and II. 10.9, taking into account the continuity of the function $x \vee y$, we have
$$
\left(X_{2 n}, X_{2 n+1}\right) \xrightarrow{p}(X, X) \quad \text { and } \quad X_{2 n} \vee X_{2 n+1... | proof | Other | proof | Yes | Yes | olympiads | false | 33,984 |
25. Show that for each sequence of random variables $\xi_{1}, \xi_{2}, \ldots$ one can find a sequence of constants $\varepsilon_{1}, \varepsilon_{2}, \ldots>0$, such that $\varepsilon_{n} \xi_{n} \rightarrow 0$ a.s. | Solution. We should choose $\varepsilon_{n}$ from the condition
$$
\sum_{n=1}^{\infty} \mathrm{P}\left(\left|\varepsilon_{n} \xi_{n}\right| \geqslant 1 / n\right)<\infty
$$
and use the Borel-Cantelli lemma. | proof | Other | proof | Yes | Yes | olympiads | false | 33,985 |
26. Let $\left\{\xi_{n}\right\}_{n \geqslant 1}$ be a sequence of random variables. Show that the set $\left\{\xi_{n} \rightarrow\right\}$, i.e., the set of those $\omega \in \Omega$ where $\xi_{n}(\omega)$ converges, can be represented in the following form:
$$
\left\{\xi_{n} \rightarrow\right\}=\bigcap_{n=1}^{\infty... | Solution. The given representations are nothing but a rephrasing of the logical expression
$$
\forall n \geqslant 1 \exists m \geqslant 1 \quad \forall k \geqslant m \forall l \geqslant k:\left|\xi_{k}(\omega)-\xi_{l}(\omega)\right| \leqslant n^{-1} .
$$
The latter is equivalent to the fundamental nature of the seque... | proof | Other | proof | Yes | Yes | olympiads | false | 33,986 |
28. ([32]) Prove the following version of the first Borel-Cantelli lemma: if the sequence of events $A_{1}, A_{2}, \ldots$ is such that
$\lim _{n} \mathrm{P}\left(A_{n}\right)=0$ and
$$
\sum_{n=1}^{\infty} \mathrm{P}\left(A_{n} \backslash A_{n+1}\right)<\infty
$$
then $\mathrm{P}\left(A_{n}\right.$ i.o. $)=0$ | Solution. By the first Borel-Cantelli lemma from condition (*), it follows that
$$
\mathrm{P}\left(A_{n} \backslash A_{n+1} \text { i.o. }\right)=0
$$
On the other hand, due to the condition $\mathrm{P}\left(A_{n}\right) \rightarrow 0$ we have
$$
\mathrm{P}\left(\underline{\lim } A_{n}\right)=0
$$
To complete the p... | proof | Other | proof | Yes | Yes | olympiads | false | 33,988 |
29. In connection with the second Borel-Cantelli lemma, show that $\mathrm{P}\left(A_{n}\right.$ i.o. $)=1$, if and only if $\sum_{n} \mathrm{P}\left(A_{n} \cap A\right)=\infty$ for every set $A$ such that $\mathrm{P}(A)>0$. | Solution. From the condition $\mathrm{P}\left(A_{n}\right.$ i.o. $)=1$ it follows that
$$
\mathrm{P}\left(\bigcup_{k \geqslant n} A_{k}\right)=1, \quad n \geqslant 1
$$
If $\sum_{n} \mathrm{P}\left(A_{n} \cap A\right)<\infty$, then
$$
\mathrm{P}(A)=\mathrm{P}\left(A \cap \bigcup_{k \geqslant n} A_{k}\right) \leqslan... | proof | Other | proof | Yes | Yes | olympiads | false | 33,989 |
30. (
citePtak63) Let events $A_{1}, A_{2}, \ldots$ be such that $\mathrm{P}\left(A_{n}\right.$ i.o. $)>0$. Then
$$
\mathrm{P}\left(\bigcap_{k=1}^{K} A_{n_{k}}\right)>0 \quad \text { for all } K \geqslant 1
$$
for some subsequence of indices $\left(n_{k}\right)$. | Solution. Let $A=\left\{A_{n}\right.$ i.o. $\}$. Then there exists an $n_{1}$ such that $\mathrm{P}\left(A_{n_{1}} A\right)>0$, otherwise we would have $\mathrm{P}(A)=0$. Next, $A_{n_{1}} A=\left\{A_{n_{1}} A_{n}\right.$ i.o. $\}$. Applying a similar argument to $A_{n_{1}} A$, we find $n_{2}$ such that $\mathrm{P}\left... | proof | Other | proof | Yes | Yes | olympiads | false | 33,990 |
32. Let $\xi_{1}, \xi_{2}, \ldots$ be independent random variables,
$$
\mathrm{P}\left(\xi_{n}=0\right)=n^{-1} \quad \text { and } \quad \mathrm{P}\left(\xi_{n}=1\right)=1-n^{-1} .
$$
Let $A_{n}=\left\{\xi_{n}=0\right\}$. From the properties $\sum_{n=1}^{\infty} \mathrm{P}\left(A_{n}\right)=\infty, \sum_{n=1}^{\infty... | Solution. By the Borel-Cantelli lemma, with probability one, the sequence $\xi_{1}, \xi_{2}, \ldots$ contains an infinite number of both zeros and ones. This implies the desired property. | proof | Other | proof | Yes | Yes | olympiads | false | 33,992 |
33. Prove Egorov's Theorem.
Let $\xi, \xi_{1}, \xi_{2}, \ldots$ be a sequence of random variables such that $\xi_{n} \rightarrow \xi$ almost surely. Then for any $\varepsilon>0$, there exists a set $A_{\varepsilon} \in \mathscr{F}$ with $\mathrm{P}\left(A_{\varepsilon}\right) \geqslant 1-\varepsilon$, on which the uni... | Solution. By the definition of convergence with probability one
$$
\xi_{n} \rightarrow \xi \text { a.s. } \Leftrightarrow \sup _{i \geqslant n}\left|\xi_{i}-\xi\right| \rightarrow 0 \text { a.s. }
$$
Therefore, $\sup _{i \geqslant n}\left|\xi_{i}-\xi\right| \xrightarrow{p} 0$ and, consequently, there exists a sequenc... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,993 |
34. The statement of Egorov's theorem makes it natural to introduce the following concept. A sequence of measurable functions $\left\{f_{n}\right\}_{n \geqslant 1}$ on a space $(\Omega, \mathscr{F}, \mu)$ with finite measure $\mu$ is said to converge almost uniformly to a function $f$ if for every $\varepsilon>0$ there... | Solution. Due to the finiteness of $\mu$, from the $\mu$-a.s. convergence follows the convergence in measure. Therefore, it is sufficient to prove that
$$
f_{n} \rightrightarrows f \Rightarrow f_{n} \xrightarrow{\mu-\text{a.s.}} f
$$
Thus, assume that $f_{n} \rightrightarrows f$. By definition, on the set $\bigcup_{\... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,994 |
35. Prove Lusin's theorem.
Any $\overline{\mathscr{B}}[a, b]$-measurable function $f=f(x)$ on $[a, b]$ can be made continuous by removing suitable open intervals from $[a, b]$ with arbitrarily small total length.
From this, deduce that for any $\varepsilon>0$ there exists a continuous function $f_{\varepsilon}=f_{\va... | Solution. Consider step functions
$$
f_{n}(x)=\sum_{k=-\infty}^{\infty} k 2^{-n} I\left([k-1] 2^{-n} \leqslant f(x) < k 2^{-n}\right)
$$
where $I$ is the indicator function. Due to problem II.3.32, we can find closed sets
$$
F_{n k} \subseteq\left\{x: f_{n}(x)=k 2^{-n}\right\}
$$
and $K_{n} \geqslant 1$, such that ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,995 |
37. Let the function $f=f(x), x \in \mathbb{R}$, be such that $\int_{0}^{\infty}|f(x)|^{p} d x < \infty$. Prove that the following continuity property in $L^{p}$ holds:
$$
\int_{\mathbb{R}}|f(x+h)-f(x)|^{p} d x \rightarrow 0, \quad h \rightarrow 0
$$ | Solution. Fix an arbitrary $\varepsilon>0$, for which we will choose $a>0$ such that
$$
\int_{|x|>a}|f(x)|^{p} d x<\varepsilon, \quad \int_{|x|>a}|g(x)|^{p} d x<\varepsilon
$$
where $g(x)$ is a continuous function on $\mathbb{R}$, and $c_{p}$ is a constant depending only on $p$. For any $b>0$, we have
$$
\int_{\math... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,997 |
38. Let $\xi, \xi_{1}, \xi_{2}, \ldots$ be a sequence of random variables defined on a complete probability space such that $\xi_{n} \xrightarrow{p} \xi$. Show that
$$
\sigma\left(\xi_{1}, \xi_{2}, \ldots\right)=\sigma\left(\xi_{1}, \xi_{2}, \ldots, \xi\right)
$$
where $\sigma(X, Y, \ldots)$ is the $\sigma$-algebra g... | Solution. For some subsequence $n_{k}$, we have $\xi_{n_{k}} \rightarrow \xi$ a.s., therefore
$$
\{\xi \leqslant x\} \in \sigma\left(\xi_{n_{1}}, \xi_{n_{2}}, \ldots\right) \subset \sigma\left(\xi_{1}, \xi_{2}, \ldots\right), \quad x \in \mathbb{R}
$$
Now, using the monotone class theorem, it is easy to derive the de... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,998 |
39. Let $\xi_{1}, \xi_{2}, \ldots$ be independent and identically distributed random variables such that $n \mathrm{P}\left(\left|\xi_{1}\right|>\varepsilon f(n)\right)=o(1), n \rightarrow \infty$, for some $f(n) \rightarrow \infty$ and any $\varepsilon>0$. Show that
$$
\frac{\xi_{n: n}}{f(n)} \xrightarrow{p} 0
$$
wh... | Solution. For $\varepsilon>0$ we have
$$
\begin{aligned}
& \mathbf{P}\left(\xi_{n: n}>\varepsilon f(n)\right)=1-\left[\mathrm{P}\left(\xi_{1} \leqslant \varepsilon f(n)\right)\right]^{n} \\
& \quad=1-\left[1+o\left(n^{-1}\right)\right]^{n} \rightarrow 0, \quad n \rightarrow \infty .
\end{aligned}
$$
Next, for suffici... | proof | Other | proof | Yes | Yes | olympiads | false | 33,999 |
40. Suppose that
$$
\xi_{n} \xrightarrow{p} \xi, \quad \eta_{n} \xrightarrow{p} \eta, \quad \mathrm{P}\left(\xi_{n} \leqslant \eta_{n}\right) \rightarrow 1 .
$$
Show that then $\mathrm{P}(\xi \leqslant \eta)=1$. | Solution. From $\xi_{n} \xrightarrow{p} \xi, \eta_{n} \xrightarrow{p} \eta$, it follows that $\xi_{n_{k}} \rightarrow \xi, \eta_{n_{k}} \rightarrow \eta$ with probability one for some subsequence $n_{k}$. From this, we obtain the desired equality. | proof | Other | proof | Yes | Yes | olympiads | false | 34,000 |
41. Let $\left(\xi_{n}\right)_{n \geqslant 1},\left(\eta_{n}\right)_{n \geqslant 1}$ be two sequences of random variables, such that $\xi_{n}$ is independent of $\eta_{n}$ for all $n$ and $\mathrm{P}\left(\eta_{n} \geqslant 0\right)=1$. Prove the implication
$$
\xi_{n} \xrightarrow{d} \xi, \quad \xi_{n} \eta_{n} \xrig... | Solution. We will show that the family $\eta_{n}$ is relatively compact, i.e.,
$$
\lim _{k} \sup _{n} \mathrm{P}\left(\left|\eta_{n}\right|>k\right)=0
$$
Assume the contrary: $\mathrm{P}\left(\left|\eta_{n_{k}}\right|>k\right) \geqslant \delta$ for some $n_{k} \rightarrow \infty$ and $\delta>0$. Choose $\varepsilon>0... | proof | Other | proof | Yes | Yes | olympiads | false | 34,001 |
42. Let $\xi_{1}, \xi_{2}, \ldots$ be non-negative random variables and $\sigma$-algebras $\mathscr{F}_{1}, \mathscr{F}_{2}, \ldots$ such that $\mathrm{E}\left(\xi_{n} \mid \mathscr{F}_{n}\right) \xrightarrow{p} 0$. Show that then $\xi_{n} \xrightarrow{p} 0$. | Solution. For all $\varepsilon>0$, by the Lebesgue's dominated convergence theorem we have
$$
\mathrm{P}\left(\xi_{n}>\varepsilon\right)=\mathrm{EP}\left(\xi_{n}>\varepsilon \mid \mathscr{F}_{n}\right) \leqslant \mathrm{E} 1 \wedge\left[\varepsilon^{-1} \mathrm{E}\left(\xi_{n} \mid \mathscr{F}_{n}\right)\right] \right... | proof | Other | proof | Yes | Yes | olympiads | false | 34,002 |
43. ([64]) Show that in the space $C[0,1]$ of continuous functions on the interval $[0,1]$, there does not exist a metric $\rho$ such that the convergence $\rho\left(f_{n}, f\right) \rightarrow 0$ is equivalent to the pointwise convergence of $f_{n}$ to $f$. (Compare with problem II.10.16.) | Solution. We will need a certain set of functions $\{f_{nm}\}_{m, n \geqslant 1}$ continuous on $[0,1]$, for the construction of which it will be necessary to introduce several auxiliary constructions.
We call
$$
\left\{\left[x_{n-1}, x_{n}\right): x_{0}=a, x_{n}=\frac{x_{n-1}+b}{2}, n \geqslant 1\right\}
$$
a norma... | proof | Calculus | proof | Yes | Yes | olympiads | false | 34,003 |
44. Provide an example showing that one can find random variables $\xi_{1}, \xi_{2}, \ldots$, such that $\xi_{n} \rightarrow 0$ everywhere, yet $\mathrm{E} \xi_{n}=1$ for all $n \geqslant 1$. | Solution. Consider the quantities
$$
\xi_{n}=n \cdot I_{[0,1]}(n \xi)
$$
where $\xi$ is a random variable uniformly distributed on $[0,1]$. Then $\xi_{n} \rightarrow 0$ everywhere and $\mathrm{E} \xi_{n}=1$. | proof | Other | proof | Yes | Yes | olympiads | false | 34,004 |
45. Let $\xi$ be a random variable, the median $\mu$ of which is uniquely determined by the inequality
$$
\mathrm{P}(\xi<\mu) \leqslant \frac{1}{2} \leqslant \mathrm{P}(\xi \leqslant \mu)
$$
Show that $\mu_{n} \rightarrow \mu$ for any medians $\mu_{n}$ of random variables $\xi_{n} \xrightarrow{d} \xi$, satisfying the... | Solution. The set of continuity points $C_{F}$ of the distribution function $F(x)=\mathrm{P}(\xi \leqslant x)$ is everywhere dense (since the set of discontinuity points is at most countable). Therefore, one can choose an arbitrarily small $\delta>0$ such that $\mu \pm \delta \in C_{F}$ and (due to the uniqueness of th... | proof | Other | proof | Yes | Yes | olympiads | false | 34,005 |
46. Let $\xi_{1}, \xi_{2}, \ldots$ be a sequence of independent and identically distributed (i.i.d.) random variables, $\mathrm{P}\left(\xi_{1}>x\right)=\exp (-x), x \geqslant 0$ (i.e., the variables are exponentially distributed). Show that
$$
\mathrm{P}\left(\xi_{n}>\alpha \ln n \text { i.o. }\right)= \begin{cases}1... | Solution. According to the Borel-Cantelli lemmas
$$
\mathrm{P}\left(\xi_{n}>c_{n} \text { i.o. }\right)= \begin{cases}1, & \text { if } \sum_{n} \exp \left(-c_{n}\right)=\infty, \\ 0, & \text { if } \sum_{n} \exp \left(-c_{n}\right)<\infty\end{cases}
$$
From here, by checking the convergence of the series $\sum_{n} \... | proof | Other | proof | Yes | Yes | olympiads | false | 34,006 |
47. Let $\xi_{1}, \xi_{2}, \ldots$ be positive i.i.d. random variables with a distribution function $F=F(x)$ such that
$$
F(x)=\lambda x+o(x), \quad x \rightarrow 0
$$
for some $\lambda>0$. Denoting $\xi_{\min }=\xi_{1} \wedge \ldots \wedge \xi_{n}$, show that
$$
n \xi_{\text {min }} \xrightarrow{d} \eta, \quad n \r... | Solution. For a fixed $x>0$ we have
$$
\mathrm{P}\left(n \xi_{\min }>x\right)=\left[\mathrm{P}\left(n \xi_{1}>x\right)\right]^{n}=\left(1-\frac{\lambda x+o(1)}{n}\right)^{n} \rightarrow e^{-\lambda x}
$$ | proof | Other | proof | Yes | Yes | olympiads | false | 34,007 |
49. Let $\xi_{1}, \xi_{2}, \ldots$ be independent random variables uniformly distributed on $[0,1]$. Set $\zeta_{n}=\xi_{1} \ldots \xi_{n}$ and consider the series $\sum_{n \geqslant 1} z^{n} \zeta_{n}$. Show that the radius of convergence $R$ of this series is $e$ with probability one. | Solution. It is known from the course of analysis that
$$
1 / R=\varlimsup_{n}\left|\zeta_{n}\right|^{1 / n}
$$
It remains to use the relations $\left|\zeta_{n}\right|=\zeta_{n}$ and
$$
\frac{\ln \zeta_{n}}{n}=\frac{\ln \xi_{1}+\ldots+\ln \xi_{n}}{n} \rightarrow \mathrm{E} \ln \xi_{1}=-1 \text { a.s., }
$$
the latt... | e | Algebra | proof | Yes | Yes | olympiads | false | 34,009 |
51. (On the metric properties of continued fractions, [21].) Let
$$
(\Omega, \mathscr{F}, P)=([0,1), \mathscr{B}([0,1)), \lambda)
$$
where $\lambda$ is the Lebesgue measure. Let $\omega \in [0,1)$ and $\omega = [0; a_1, a_2, \ldots]$ be the continued fraction expansion of $\omega$ (see problem II.10.50). Denote $X_n(... | Solution. (a) The first relation follows from the considerations featured in the solution of part (b) of problem II.10.50. The second relation is verified by induction, taking into account the following. If for a given $n-1$ $(n \geqslant 2)$, $F_{n-1}$ has a bounded derivative $f_{n-1}$ on $[0,1)$, then the series def... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,011 |
52. (See [2].) Let $\xi_{1}, \xi_{2}, \ldots$ be independent Bernoulli random variables, $\mathrm{P}\left(\xi_{n}=1\right)=p$ and $\mathrm{P}\left(\xi_{n}=0\right)=1-p, n \geqslant 1$, where $p$ is some fixed number from $(0,1)$. Show that the series
$$
\sum_{n=1}^{\infty} \frac{\xi_{n}}{2^{n}}
$$
converges almost su... | Solution. By Fubini's theorem, the given series converges almost surely, since all its terms are non-negative and
$$
\mathrm{E} \sum_{n \geqslant 1} 2^{-n} \xi_{n} \leqslant \sum_{n \geqslant 1} 2^{-n}<\infty
$$
Let us first consider $p=1 / 2$. We need to establish that the quantity $U=\sum_{n \geqslant 1} 2^{-n} \xi... | proof | Other | proof | Yes | Yes | olympiads | false | 34,012 |
55. As proven in B1.I.5.5, the Bernstein polynomials
$$
B_{n}(x, f)=\sum_{k=0}^{n} f\left(\frac{k}{n}\right) C_{n}^{k} x^{k}(1-x)^{n-k}
$$
provide a uniform approximation of a continuous function $f=f(x)$ on the interval $[0,1]$. If we consider not an interval but the half-line $\mathbb{R}_{+}$, then the Szász-Miraky... | Solution. Let $X_{1}, X_{2}, \ldots$ be independent Poisson random variables with parameter $x \geqslant 0$. Then $Z_{n}=X_{1}+\ldots+X_{n}(n \geqslant 1)$ is a Poisson random variable with parameter $n x$. Consequently,
$$
S_{n}(x, f)=e^{-n x} \sum_{k=0}^{\infty} f\left(\frac{k}{n}\right) \frac{(n x)^{k}}{k!}=\mathrm... | proof | Calculus | proof | Yes | Yes | olympiads | false | 34,015 |
1. Let $X, Y$ be independent random variables, and $h=h(x, y)$ be a Borel function, $\mathrm{E}|h(X, Y)|^{2}<\infty$. Prove that almost surely there is a representation
$$
h(X, Y)=\mathrm{E} h(X, Y)+\sum_{m \geqslant 1} a_{m} f_{m}(X)+\sum_{n \geqslant 1} b_{n} g_{n}(Y)+\sum_{m, n \geqslant 1} c_{m n} f_{m}(X) g_{n}(Y... | Solution. Let $\left(f_{n}\right)_{n \geqslant 0}$ and $\left(g_{m}\right)_{m \geqslant 0}$ be complete orthonormal systems of functions (i.e., they form a basis) in the spaces
$$
L^{2}(X)=\left\{f: \mathbb{R} \rightarrow \mathbb{R}: \mathbb{E}|f(X)|^{2}<\infty\right\}
$$
and
$$
L^{2}(Y)=\left\{g: \mathbb{R} \righta... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,016 |
2. Let $X_{1}, \ldots, X_{n}$ be exchangeable random variables, i.e., the joint distribution of any group of random variables consisting of $j$ elements with different indices $X_{i_{1}}, \ldots, X_{i}$ depends only on $j$ and not on the specific choice of values $i_{1}, \ldots, i_{j}$. Show that
$$
\mathrm{D} X_{1} \... | Solution. The statement of the first part of the problem follows from the relations
$$
\begin{gathered}
\mathrm{D}\left(\sum_{i=1}^{n} X_{i}\right)=n \mathrm{D} X_{1}+n(n-1) \operatorname{cov}\left(X_{1}, X_{2}\right) \geqslant 0, \\
\frac{\mathrm{D}\left(X_{1}-X_{2}\right)}{2} \geqslant \mathrm{D} X_{1}-\operatorname... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,017 |
3. Show that if $\xi=1$. i.m. $\xi_{n}$, then $\left\|\xi_{n}\right\| \rightarrow\|\xi\|$.
---
Note: The phrase "i.m." typically stands for "in measure" in the context of convergence in measure in probability theory and measure theory. However, the context of the statement is not entirely clear without additional inf... | Solution. The statement follows from the triangle inequality
$$
\left|\|\xi\|-\left\|\xi_{n}\right\|\right| \leqslant\left\|\xi-\xi_{n}\right\|
$$ | proof | Calculus | proof | Yes | Yes | olympiads | false | 34,018 |
6. Let $\left\{\xi_{1}, \ldots, \xi_{n}\right\}$ be a family of orthogonal random variables. Show that for them, the Pythagorean theorem holds:
$$
\left\|\sum_{i=1}^{n} \xi_{i}\right\|^{2}=\sum_{i=1}^{n}\left\|\xi_{i}\right\|^{2}
$$ | Solution. Obviously,
$$
\mathrm{E}\left|\sum_{i=1}^{n} \xi_{i}\right|^{2}=\sum_{i=1}^{n} \mathrm{E} \xi_{i}^{2}+\sum_{i \neq j} \mathrm{E} \xi_{i} \xi_{j}=\sum_{i=1}^{n} \mathrm{E} \xi_{i}^{2}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,021 |
7. Let $\xi, \xi_{1}, \xi_{2}, \ldots, \xi_{n}$ be complex-valued random variables with finite second moments. Prove Selberg's inequality
$$
\sum_{i=1}^{n} \frac{\left|\left(\xi, \xi_{i}\right)\right|^{2}}{\sum_{j=1}^{n}\left|\left(\xi_{i}, \xi_{j}\right)\right|} \leqslant \mathrm{E}|\xi|^{2}
$$
from which, in partic... | Solution. For any $\alpha_{1}, \ldots, \alpha_{n} \in \mathbb{C}$, by the Cauchy-Bunyakovsky inequality we have
$$
\mathrm{E}|\xi|^{2} \geqslant \frac{\left|\left(\xi, \sum_{i=1}^{n} \alpha_{i} \xi_{i}\right)\right|^{2}}{\mathrm{E}\left|\sum_{i=1}^{n} \alpha_{i} \xi_{i}\right|^{2}} \geqslant \frac{\left|\sum_{i=1}^{n}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,022 |
8. (See [58].) Let $X, Y, Z$ be complex-valued random variables with finite second moments, and $\mathrm{E}|Z|^{2}=1$. Prove the following variant of Blatter's inequality (see [38]):
$$
\left|(X, Z)(Z, Y)-\frac{(X, Y)}{2}\right| \leqslant \frac{\|X\|\|Y\|}{2}
$$
Derive from this Brudzynski's inequality
$$
|(X, Y)|^{... | Solution. By the Cauchy-Bunyakovsky inequality
$$
|(X-2(X, Z) Z, Y)|^{2} \leqslant \mathrm{E}|X-2(X, Z) Z|^{2} \cdot \mathrm{E}|Y|^{2}
$$
To complete the proof, it is sufficient to use the relations
$$
(X-2(X, Z) Z, Y)=(X, Y)-2(X, Z)(Z, Y), \quad \mathrm{E}|X-2(X, Z) Z|^{2}=\mathrm{E}|X|^{2}
$$
From Blatter's inequ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,023 |
9. (Schur's Lemma.) Let $c=c(x, y)$ be a bounded Borel function on $\mathbb{R}^{2}$, and let $X, Y$ be independent random variables, $\mathrm{E} X^{2}=\mathrm{E} Y^{2}=1$. Prove that
$$
|\mathrm{E} c(X, Y) X Y|^{2} \leqslant \sup _{x} \mathrm{E}|c(x, Y)| \sup _{y} \mathrm{E}|c(X, y)|
$$ | Solution. By the Cauchy-Bunyakovsky inequality
$$
|\mathrm{E} c(X, Y) X Y|^{2} \leqslant \mathrm{E}|c(X, Y)| X^{2} \cdot \mathrm{E}|c(X, Y)| Y^{2}
$$
Moreover, by Fubini's theorem
$\mathrm{E}|c(X, Y)| X^{2}=\left.\mathrm{E}\left(x^{2} \mathrm{E}|c(x, Y)|\right)\right|_{x=X} \leqslant \sup _{x} \mathrm{E}|c(x, Y)| \m... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,024 |
10. Let $X, Y, Z$ be independent random variables, and $f=f(x, y)$, $g=g(y, z)$, $h=h(z, x)$ be bounded Borel functions on $\mathbb{R}^{2}$. Prove that
$$
|\mathrm{E} f(X, Y) g(Y, Z) h(Z, X)|^{2} \leqslant \mathrm{E} f^{2}(X, Y) \mathrm{E} g^{2}(Y, Z) \mathrm{E} h^{2}(Z, X)
$$ | Solution. For brevity, let $\xi=f(X, Y), \eta=g(Y, Z)$ and $\zeta=$ $=h(Z, X)$. Then (taking into account the conditional Cauchy-Bunyakovsky inequality and problem II.7.14)
$$
\begin{aligned}
|\mathrm{E} \xi \eta \zeta|^{2} & =|\mathrm{E}(\xi \mathrm{E}[\eta \zeta \mid X, Y])|^{2} \leqslant \mathrm{E} \xi^{2} \mathrm{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,025 |
11. Let $\xi_{1}, \xi_{2}, \ldots$ be orthogonal random variables and $S_{n}=$ $=\xi_{1}+\ldots+\xi_{n}$.
(a) Show that if $\sum \mathrm{E} \xi_{n}^{2}<\infty$, then $S_{n} \rightarrow S$ in mean square, where $S=\sum_{n=1}^{\infty} \xi_{n}$.
(b) If $\sum \mathrm{E} \xi_{n}^{2}=\infty$, show that $\mathrm{E} S_{n}^{2... | Solution. (a) It is sufficient to use the fact that, according to problem II.11.6, the equality holds
$$
\left\|S_{n}-S_{m}\right\|^{2}=\sum_{i=m+1}^{n}\left\|\xi_{i}\right\|^{2}, \quad m < n
$$
(b) Let us show that for any $\varepsilon > 0$
$$
\begin{aligned}
\sum_{n \geqslant 0} \mathrm{P}\left(\max _{2^{n}<k \leq... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,026 |
13. Prove that for $\xi \in L^{2}(\Omega, \mathscr{F}, \mathrm{P})$ and any $\sigma$-subalgebra $\mathscr{G} \subseteq \mathscr{F}$, the inequality
$$
\|\xi\| \geqslant\|\mathrm{E}(\xi \mid \mathscr{G})\|
$$
holds, and equality holds if and only if $\xi=\mathrm{E}(\xi \mid \mathscr{G})$ a.s. | Solution. The desired inequality follows from the fact that
$$
\mathrm{E} \xi^{2}-\mathrm{E}[\mathrm{E}(\xi \mid \mathscr{G})]^{2}=\mathrm{E}[\xi-\mathrm{E}(\xi \mid \mathscr{G})]^{2}
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,028 |
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