problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
26. (Vitali Set.) Introduce on $[0,1]$ the equivalence relation
$$
x \sim y \Leftrightarrow x-y \in \mathbb{Q}.
$$
Assuming the validity of the Axiom of Choice:
"for any set $S$ there exists a choice function that assigns to each subset $B \subseteq S$ some element $b \in B$",
we select from each equivalence class ... | Solution. Assuming that $\mathscr{V} \in \overline{\mathscr{B}}([0,1])$, we find that
$$
[0,1] \subset \bigcup_{q \in \mathbb{Q} \cap[-1,1]}(\mathscr{V}+q) \subset[-1,2]
$$
Considering that the sets $\mathscr{V}+q$ do not intersect for different $q \in \mathbb{Q} \cap[-1,1]$, we obtain the following relations:
$$
\b... | proof | Other | proof | Yes | Yes | olympiads | false | 33,690 |
27. (Probabilistic proof of Euler's formula for the Riemann zeta function.) Let $\zeta(s)=\sum_{n=1}^{\infty} n^{-s}-$ be the Riemann zeta function $(1<s<\infty)$. Euler's formula states that $\zeta(s)$ admits the following representation:
$$
\zeta(s)=\prod_{n=1}^{\infty}\left(1-\frac{1}{p_{n}^{s}}\right)^{-1}
$$
whe... | Solution. Let $\mathbb{N}=\{1,2, \ldots\}$ be the set of natural numbers with the Borel $\sigma$-algebra $2^{\mathbb{N}}$ of its subsets. Define a probability measure $\mathrm{P}=\mathrm{P}(\cdot)$ on $(\mathbb{N}, 2^{\mathbb{N}})$ for subsets $A \subseteq \mathbb{N}$ by
$$
\mathrm{P}(A)=\frac{1}{\zeta(s)} \sum_{n \in... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 33,691 |
28. (Probabilistic proof of Euler's formula for the number of prime numbers not exceeding a given number.) Let $\varphi(n)$ be the Euler's function - the number of prime numbers $p$ such that $1<p \leqslant n$. Using probabilistic arguments, prove the following Euler's formula:
$$
\frac{\varphi(n)}{n}=\prod_{p \mid n}... | Solution. On the set $\{1, \ldots, n\}$, we define a uniform probability distribution $\mathrm{P}(\{k\})=1 / n, 1 \leqslant k \leqslant n$. For a fixed $p$, let
$$
A_{p}=\{k \leqslant n: p \text { divides } k\} .
$$
Let $p_{1}, p_{2}, \ldots$ be the distinct prime divisors of the number $n$. Then
$$
\mathrm{P}\left(... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 33,692 |
29. Show that the Lebesgue measure $\lambda$ on $\left(\mathbb{R}^{n}, \mathscr{B}\left(\mathbb{R}^{n}\right)\right)$ is invariant under rotation for $n>1$ and translation for $n \geqslant 1$. $\langle$ See problem II.3.32. $\rangle$ | Solution. According to problem II. 3.32, for any $B \in \mathscr{B}\left(\mathbb{R}^{n}\right)$ we have
$$
\lambda(B)=\inf \{\lambda(G): B \subseteq G \text { and } G \text { is open }\},
$$
which allows us to reduce the question of measure transformations to open sets $G$. For example, if the measure $\lambda$ is in... | proof | Other | proof | Yes | Yes | olympiads | false | 33,693 |
30. Provide examples showing that the assumption that the system of sets $\mathscr{A}$ involved in the Carathéodory theorem is an algebra is essential both for the fact of the possibility of extending a probability measure $\mathrm{P}$ defined on the system $\mathscr{A}$ to a measure on the $\sigma$-algebra $\mathscr{F... | Solution. (a) The desired example is constructed as follows: $\Omega=\{1,2,3\}, \mathscr{E}=\{\varnothing,\{1\},\{1,2\},\{1,3\}, \Omega\}, \mathscr{F}$ is the $\sigma$-algebra of all subsets of the set $\Omega$ and $\mathrm{P}(\Omega)=\mathrm{P}(\{1,2\})=\mathrm{P}(\{1,3\})=1, \mathrm{P}(\{1\})=1 / 2$, $\mathrm{P}(\var... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,694 |
31. Let $F=F(x)$ be a distribution function on $\mathbb{R}$. Show that for all $a \in \mathbb{R}$, the following equality holds:
$$
\int_{\mathbb{R}}[F(x+a)-F(x)] d x=a .
$$ | Solution. The first proof is based on considering the area between two graphs
$$
\Gamma=\{(x, y): y \in[F(x-), F(x)]\}
$$
and
$$
\Gamma_{a}=\{(x, y): y \in[F(x+a-), F(x+a)]\}
$$
This area is $\int_{\mathbb{R}}[F(x+a)-F(x)] d x$. If we interchange the axes $x$ and $y$, then the graph $\Gamma_{a}$ will be obtained by... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,695 |
32. Let $S$ be a topological space with the Borel $\sigma$-algebra $\mathscr{E}$ generated by all open sets. Suppose also that for any closed set $F$ in $S$, there exists a decreasing sequence of open sets $G_{1} \supset G_{2} \supset \ldots$ such that $F=\bigcap_{n} G_{n}$. Show using the principle of suitable sets th... | Solution. Define the system of sets
$$
\mathscr{A}=\left\{A \in \mathscr{E}: \mu(A)=\mu^{*}(A)=\mu_{*}(A)\right\}
$$
All open and closed sets lie in $\mathscr{A}$. This, in particular, follows from the fact that by condition, for any open set $G$ and its closed complement $F=\bar{G}$, there exist a decreasing system ... | proof | Other | proof | Yes | Yes | olympiads | false | 33,696 |
34. (Continuation of problem II.3.33.) The solutions provided in problem II.3.33 are correct, but, as follows from the exposition, they actually pertain to different probabilistic problems. Specifically, we will define a chord by two parameters $(r, \theta) \in [0,1] \times [0, 2\pi)$ - the polar coordinates of the poi... | Solution. (a) Let $A=\left(\cos \theta_{A}, \sin \theta_{A}\right)$ and $B=\left(\cos \theta_{B}, \sin \theta_{B}\right)$, where $\theta_{A}, \theta_{B} \in[0,2 \pi)$. Then, by condition, $\theta_{A}$ and $\theta_{B}$ are independent and uniformly distributed on $[0,2 \pi)$. We have $r=\left|\cos \theta_{-}\right|=\sin... | proof | Geometry | proof | Yes | Yes | olympiads | false | 33,698 |
1. Prove that a random variable $X$ is discrete, absolutely continuous, or singular if and only if $\xi X$ is also such, where $\xi$ is a variable independent of $X$, $\mathrm{P}(\xi=1)=\mathrm{P}(\xi=-1)=$ $=1 / 2$. | Solution. In the case of discrete variables, the statement is obvious. If now the variable $X$ is singular, then for some Borel set $D$ of Lebesgue measure zero, we have $\mathrm{P}(X \in D)=1$. But then $\mathrm{P}(\xi X \in D \cup\{x:-x \in D\})=1$, and the Lebesgue measure of the set $D \cup\{x:-x \in D\}$ is zero. ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,700 |
2. Let $\left\{X_{\alpha}\right\}$ be a family of extended random variables, where $\alpha$ runs through an arbitrary set of indices. Prove that one can select a countable subset $\left\{Y_{n}\right\} \subseteq\left\{X_{\alpha}\right\}$ such that
$$
\mathrm{P}\left(\sup _{n} Y_{n} \leqslant X\right)=1 \Rightarrow \mat... | Solution. Let countable families $\left\{Y_{n k}\right\} \subseteq\left\{X_{\alpha}\right\}$, $k=1,2, \ldots$, be such that
$$
\lim _{k} \mathrm{E} \operatorname{arctg} \sup _{n} Y_{n k}=\sup _{\left\{X_{\alpha_{n}}\right\}} \mathrm{E} \operatorname{arctg} \sup _{n} X_{\alpha_{n}}
$$
where the second sup is taken ove... | proof | Other | proof | Yes | Yes | olympiads | false | 33,701 |
3. (See [15].) Let $f=f(x)$ be an increasing Borel function on $\mathbb{R}$. Verify that the inf $\mathrm{E} f(|X|)$ over all variables $X$ with density not exceeding $1 / 2$, is attained at the variable $U$ with a uniform distribution on $[-1,1]$. | Solution. Let $p=p(x)$ be an arbitrary density not exceeding $1 / 2$. Then
$$
\begin{aligned}
& \int_{\mathbb{R}} f(|x|) p(x) d x= \\
& \quad=f(1)+\int_{|x|>1}\{f(|x|)-f(1)\} p(x) d x+\int_{-1}^{1}\{f(|x|)-f(1)\} p(x) d x \geqslant \\
& \quad \geqslant f(1)+\frac{1}{2} \int_{-1}^{1}\{f(|x|)-f(1)\} d x=\frac{1}{2} \int... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,702 |
4. Show that a random variable $\xi$ is continuous if and only if $\mathrm{P}(\xi=x)=0$ for all $x \in \mathbb{R}$. | Solution. If $F$ is the distribution function of the random variable $\xi$, then
$$
\mathrm{P}(\xi=x)=F(x)-F(x-)
$$
which, by the condition of the problem, implies the continuity of the function $F$ (and thus, of $\xi$). | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,703 |
5. Is it true that if the magnitude $|\xi|$ is $\mathscr{F}$-measurable, then $\xi$ is also $\mathscr{F}$-measurable? | Solution. No, incorrect. Indeed, let $\Omega=\{0,1\}, \mathscr{F}=$ $=\{\varnothing, \Omega\}$, and $\xi(0)=-1, \xi(1)=1$. Then $|\xi| \equiv 1$ is a $\mathscr{F}$-measurable quantity, while $\xi$ is not. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,704 |
6. Prove that the functions
$$
x^{n}, x^{+}=x \vee 0, x^{-}=(-x) \vee 0, \quad|x|=x^{+}+x^{-}
$$
are Borel functions. Also prove the more general result: any continuous function \( f=f(x), x \in \mathbb{R}^{n}, n \geqslant 1 \), is a Borel function. | Solution. According to the topological definition of a continuous function,
a function $f$ is continuous if the preimage of any open set is also open.
Thus, the set $f^{-1}(G)$ is open when $G$ is open. Therefore,
$$
\begin{aligned}
& f^{-1}(\mathscr{B}(\mathbb{R}))=f^{-1}(\sigma(G: G \text { open }))= \\
& \quad=\s... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,705 |
7. Show that if $\xi$ and $\eta$ are $\mathscr{F}$-measurable quantities, then $\{\omega: \xi(\omega)=$ $=\eta(\omega)\} \in \mathscr{F}$. | Solution. The statement follows from the representation
$$
\{\xi=\eta\}=\bigcap_{n=1}^{\infty} \bigcup_{q}\left\{\xi, \eta \in\left(q-\frac{1}{n}, q+\frac{1}{n}\right)\right\}
$$
where the union is taken over all rational $q$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,706 |
8. Let $\xi$ and $\eta$ be two random variables on $(\Omega, \mathscr{F})$ and set $A$ belongs to $\mathscr{F}$. Then
$$
\zeta=\xi I_{A}+\eta I_{\bar{A}}
$$
is also a random variable. | Solution. For any Borel set $B \in \mathscr{B}(\mathbb{R})$ we have
$$
\{\zeta \in B\}=(A \cap\{\xi \in B\}) \cup(\bar{A} \cap\{\eta \in B\})
$$
from which the required statement follows. | proof | Other | proof | Yes | Yes | olympiads | false | 33,707 |
9. Let $\xi_{1}, \ldots, \xi_{n}$ be random variables and $\varphi\left(x_{1}, \ldots, x_{n}\right)$ be a Borel function. Show that the function $\varphi\left(\xi_{1}(\omega), \ldots, \xi_{n}(\omega)\right)$ is a random variable. | Solution. It is sufficient to show that the mapping
$$
\omega \mapsto\left(\xi_{1}(\omega), \ldots, \xi_{n}(\omega)\right) \in \mathbb{R}^{n}
$$
is $\mathscr{F} / \mathscr{B}\left(\mathbb{R}^{n}\right)$-measurable, since in this case the mapping
$$
\omega \mapsto \varphi\left(\xi_{1}(\omega), \ldots, \xi_{n}(\omega)... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,708 |
10. Let $\xi$ and $\zeta$ be arbitrary random variables. Determine when $\sigma(\xi)=\sigma(\zeta)$. | Solution. If $\sigma(\xi)=\sigma(\zeta)$, then by Theorem 3 from V1.II. 4, there exist Borel functions $\varphi$ and $\psi$ such that $\xi=\varphi(\zeta)$ and $\zeta=\psi(\xi)$. The latter, in particular, means that $\varphi$ is a bijective mapping from $\{\zeta(\omega): \omega \in \Omega\}$ to $\{\xi(\omega): \omega \... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,709 |
11. Provide an example of a density $f=f(x)$ (of some random variable $\xi$) for which
\[
\lim _{x \rightarrow \infty} f(x) \text { does not exist }
\]
and, consequently, the function $f(x)$ does not vanish at infinity.
Show, however, that for almost all $x \in \mathbb{R}$ (in the sense of Lebesgue measure) the equa... | Solution. It is sufficient to consider the density
$$
f(x)=\sum_{n=1}^{\infty} I_{\left[n, n+2^{-n}\right]}(x)
$$
For such a function $f$, we have
$$
0=\lim _{x \rightarrow \infty} f(x)0$ the equality holds
$$
\lambda\left(x \in[0,1]: \varlimsup_{n} f(x+n)>\varepsilon\right)=0
$$
where $\lambda$ is the Lebesgue me... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,710 |
12. Let $E$ be a subset of the space $\mathbb{R}$ that is at most countable, and let $\xi$ be a mapping $\Omega \rightarrow E$. Prove that $\xi$ is a random variable on $(\Omega, \mathscr{F})$ if and only if $\{\xi=x\} \in \mathscr{F}$ for each $x \in E$. | Solution. The statement follows from the fact that for each Borel set $B$ in $\mathbb{R}$, the equality
$$
\{\xi \in B\}=\bigcup_{x \in B \cap E}\{\xi=x\}
$$
holds, where the union is obviously taken over no more than a countable number of points $x$. | proof | Other | proof | Yes | Yes | olympiads | false | 33,711 |
13. Let $\xi$ be a non-degenerate random variable such that for constants $a>0$ and $b \in \mathbb{R}$, the distribution of the variable $a \xi + b$ coincides with the distribution of the variable $\xi$. Show that then necessarily $a=1$ and $b=0$. | Solution. Without loss of generality, we can assume that $a \leqslant 1$, otherwise we will transition from the triplet $(a, b, \xi)$ to $\left(1 / a, b / a^{2}, \xi / a\right)$. Let $F=F(x)$ be the distribution function of the random variable $\xi$. Then for all $x \in \mathbb{R}$ we have
$$
F(x)=F\left(\frac{x-b}{a}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,712 |
14. Let $(\Omega, \mathscr{F}, \mathrm{P})$ be a probability space and $(\Omega, \overline{\mathscr{F}}, \mathrm{P})$ the completion of $(\Omega, \mathscr{F}, \mathrm{P})$ with respect to the measure $\mathrm{P}$. Suppose that $\bar{\xi}=\bar{\xi}(\omega)$ is a random variable on $(\Omega, \overline{\mathscr{F}}, \math... | Solution. We will assume that $\mathrm{P}(\bar{\xi} \geqslant 0)=1$, otherwise we will transition to $\exp \bar{\xi}$. Discretizing $\bar{\xi}$, we find a sequence $\bar{\xi}_{n}$ such that $\bar{\xi} \uparrow \bar{\xi}$ everywhere, specifically
$$
\bar{\xi}_{n}=\frac{1}{n} \sum_{k \geqslant 0} k I(k \leqslant n \bar{... | proof | Other | proof | Yes | Yes | olympiads | false | 33,713 |
15. A growth point of the distribution function $F=F(x)$ is any $x \in \mathbb{R}$ for which $F(x-\varepsilon) < F(x+\varepsilon)$ for any $\varepsilon > 0$. Show that the random variable $\xi$ with distribution function $F$ almost surely lies in the set of growth points of the function $F$. | Solution. Introduce an equivalence relation $\sim$ on the set of points that are not growth points of the function $F$,
$$
x \sim y, x \leqslant y \Leftrightarrow \mathrm{P}(\xi \in[x, y])=0
$$
Obviously, if $x \sim z$ and $x \leqslant y \leqslant z$, then $y$ is not a growth point of the function $F, x \sim y$ and $... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,714 |
16. Show that the support supp $F$ of a continuous distribution function $F$ is a perfect set (i.e., closed and having no isolated points), coinciding with the set $G_{F}$ of growth points of $F$.
Provide an example of a discrete distribution function $F$ whose support supp $F=\mathbb{R}$, i.e., coincides with the ent... | Solution. Any growth point $x$ of the function $F$ necessarily lies in $\operatorname{supp} F$. Indeed, if $x \notin \operatorname{supp} F$, then $(x-\varepsilon, x+\varepsilon) \cap \operatorname{supp} F = \varnothing$ for small $\varepsilon > 0$ due to the closedness of $\operatorname{supp} F$. However, in this case,... | proof | Other | proof | Yes | Yes | olympiads | false | 33,715 |
17. Let $\xi_{1}, \xi_{2}, \ldots$ be i.i.d. random variables with distribution function $F=F(x)$. Show that the closure of the set
$$
\left\{\xi_{1}(\omega), \xi_{2}(\omega), \ldots\right\}, \quad \omega \in \Omega
$$
with probability one coincides with some set that does not depend on $\omega$. Describe the structu... | Solution. The considered closure belongs to the closure of the set $A$ of growth points of the distribution function $F$ (see problem II.4.15).
Choose a countable set $B \subset A$ that is dense in $A$ and show that
$$
\mathrm{P}\left(\bigcap_{k \geqslant 1, x \in B}\left\{\exists n: \xi_{n} \in(x-1 / k, x+1 / k)\rig... | proof | Other | proof | Yes | Yes | olympiads | false | 33,716 |
18. Let $\xi_{1}, \xi_{2}, \ldots$ be a sequence of random variables. Show that
$\left\{\omega: \lim _{n} \xi_{n}(\omega)\right.$ does not exist $\}$,
$\left\{\omega: \lim _{n} \xi_{n}(\omega)\right.$ exists and is finite $\} \in \mathscr{F}$. | Solution. The statement follows from the fact that $\frac{\lim }{n} \xi_{n}$ and $\varlimsup_{n} \xi_{n}$ are random variables with values in $[-\infty, \infty]$, as well as from the equalities
$\left\{\lim _{n} \xi_{n}\right.$ does not exist $\}=\left\{\underline{\lim } \xi_{n}<\overline{\lim } \xi_{n}\right\}=$
$$
... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,717 |
19. Let $\xi$ and $X$ be independent random variables, such that
$$
\mathrm{P}(\xi=1)=\mathrm{P}(\xi=-1)=\frac{1}{2}
$$
Show the equivalence of the following statements:
(a) $X \stackrel{d}{=}-X$, i.e., $X$ has a symmetric distribution;
(b) $X \stackrel{d}{=} \xi X$;
(c) $X \stackrel{d}{=} \xi|X|$. | Solution. It is sufficient to prove that (a) $\Leftrightarrow$ (b) and (a) $\Leftrightarrow$ (c). For an arbitrary Borel set $B$, the equality $\mathrm{P}(X \in B)=\mathrm{P}(-X \in B)$ is equivalent to the following:
$$
\mathrm{P}(X \in B)=\frac{\mathrm{P}(X \in B)}{2}+\frac{\mathrm{P}(-X \in B)}{2}=\mathrm{P}(\xi X ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,718 |
20. Show that the joint distribution of two Bernoulli random variables $X$ and $Y$, taking values 0 and 1 (for simplicity), is completely determined by the parameters
$$
\mathrm{P}(X=1), \quad \mathrm{P}(Y=1), \quad \operatorname{cov}(X, Y)
$$
In particular, $X$ and $Y$ are independent if $\operatorname{cov}(X, Y)=0$... | Solution. By the definition of $\operatorname{cov}(X, Y)$ we have
$$
\begin{aligned}
\mathrm{P}(X=1, Y=1)=\mathrm{E} X Y=\operatorname{cov}(X, Y) & +\mathrm{E} X \mathrm{E} Y= \\
& =\operatorname{cov}(X, Y)+\mathrm{P}(X=1) \mathrm{P}(Y=1) .
\end{aligned}
$$
Moreover,
$\mathrm{P}(X=0, Y=1)=\mathrm{E}(1-X) Y=\mathrm{P... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,719 |
1. The set $\{\omega: X(\omega)=Y(\omega)\}$ is an event if $X$ and $Y$ are random variables (see problem II.4.7). Verify by example that this is not generally the case when $X$ and $Y$ are random elements. | Solution. Let $(\Omega, \mathscr{F})=\left(\mathbb{R}^{[0,1]}, \mathscr{B}\left(\mathbb{R}^{[0,1]}\right)\right), X(\omega) \equiv \omega$ for $\omega \in \Omega$ and $Y \equiv(0)_{0 \leqslant t \leqslant 1}-\mathscr{F}$-measurable mappings into $\left(\mathbb{R}^{[0,1]}, \mathscr{B}\left(\mathbb{R}^{[0,1]}\right)\righ... | proof | Other | proof | Yes | Yes | olympiads | false | 33,720 |
2. Let $\xi_{1}, \ldots, \xi_{n}$ be discrete random variables. Show that they are independent if and only if for any real numbers $x_{1}, \ldots, x_{n}$ the equality
$$
\mathrm{P}\left(\xi_{1}=x_{1}, \ldots, \xi_{n}=x_{n}\right)=\prod_{i=1}^{n} \mathrm{P}\left(\xi_{i}=x_{i}\right)
$$
holds. | Solution. The necessity of the given equality is obvious. Let's prove its sufficiency. For any not more than countable sets $B_{1}, \ldots, B_{n}$, consisting of the values of random variables $\xi_{1}, \ldots, \xi_{n}$ respectively, the following relations hold:
$$
\begin{aligned}
& \mathrm{P}\left(\xi_{1} \in B_{1},... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,721 |
3. Prove that any random function $X=\left(X_{t}\right)_{t \in T}$, considered as a random element with values in the space ( $\mathbb{R}^{T}, \mathscr{B}\left(\mathbb{R}^{T}\right)$ ), is a stochastic process, i.e., a collection of random variables $X_{t}$, and vice versa. | Solution. If $X=X(\omega)$ is a $\mathscr{F} / \mathscr{B}\left(\mathbb{R}^{T}\right)$-measurable function, then for $t \in T$ and $B \in \mathscr{B}(\mathbb{R})$ we have
$$
\left\{\omega: X_{t}(\omega) \in B\right\}=\{\omega: X(\omega) \in C\} \in \mathscr{F}, \quad \text { where } C=\left\{x \in \mathbb{R}^{T}: x_{t... | proof | Other | proof | Yes | Yes | olympiads | false | 33,722 |
4. Let $X_{1}, \ldots, X_{n}$ be random elements with values in $\left(E_{1}, \mathscr{E}_{1}\right), \ldots, \left(E_{n}, \mathscr{E}_{n}\right)$ respectively. Let, further, $\left(G_{1}, \mathscr{G}_{1}\right), \ldots, \left(G_{n}, \mathscr{G}_{n}\right)$ be measurable spaces and $g_{1}, \ldots, g_{n}$ be $\mathscr{E... | Solution. It is sufficient to note that for any $B_{i} \in \mathscr{\varphi}_{i}, i=1, \ldots, n$, the equality holds
$$
\begin{aligned}
\mathrm{P}\left(g_{1}\left(X_{1}\right) \in B_{1}, \ldots, g_{n}\left(X_{n}\right) \in\right. & \left.B_{n}\right)= \\
& =\mathrm{P}\left(X_{1} \in g_{1}^{-1}\left(B_{1}\right), \ldo... | proof | Other | proof | Yes | Yes | olympiads | false | 33,723 |
5. Let $\xi_{1}, \ldots, \xi_{m}$ and $\eta_{1}, \ldots, \eta_{n}$ be random variables. Form the random vectors $X=\left(\xi_{1}, \ldots, \xi_{m}\right)$ and $Y=\left(\eta_{1}, \ldots, \eta_{n}\right)$. Suppose the following conditions are satisfied:
1) the random variables $\xi_{1}, \ldots, \xi_{m}$ are independent;
2... | Solution. For any $B_{1}, \ldots, B_{m}, C_{1}, \ldots, C_{n} \in \mathscr{B}(\mathbb{R})$ the following relations hold
$$
B=B_{1} \times \ldots \times B_{m} \in \mathscr{B}\left(\mathbb{R}^{m}\right), \quad C=C_{1} \times \ldots \times C_{n} \in \mathscr{B}\left(\mathbb{R}^{n}\right)
$$
and by conditions $1-3$ we ha... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,724 |
6. Given random vectors $X=\left(\xi_{1}, \ldots, \xi_{m}\right)$ and $Y=\left(\eta_{1}, \ldots, \eta_{n}\right)$, for which it is known that the random variables $\xi_{1}, \ldots, \xi_{m}, \eta_{1}, \ldots, \eta_{n}$ are independent.
(a) Prove that the random vectors $X$ and $Y$, considered as random elements, are in... | Solution. (a) We will use the principle of suitable sets. For a fixed $C \in \mathscr{B}\left(\mathbb{R}^{n}\right)$, denote
$$
\mathscr{B}(C)=\left\{B \in \mathscr{B}\left(\mathbb{R}^{m}\right): \mathrm{P}(X \in B, Y \in C)=\mathrm{P}(X \in B) \mathrm{P}(Y \in C)\right\}
$$
By the condition, for all $C=C_{1} \times ... | proof | Other | proof | Yes | Yes | olympiads | false | 33,725 |
8. Let $(\Omega, \mathscr{F})$ be a measurable space and $(E, \mathscr{E}, \rho)$ be a metric space with metric $\rho$ and $\sigma$-algebra $\mathscr{E}$ of Borel subsets generated by open sets. Let $X_{1}(\omega), X_{2}(\omega), \ldots$ be a sequence of $\mathscr{F} / \mathscr{E}$-measurable functions (random elements... | Solution. Any closed set $F \in \mathscr{E}$ can be represented as
$$
F=\bigcap_{k=1}^{\infty} F^{(1 / k)}
$$
where $F^{(\delta)}=\left\{x \in E: \inf _{y \in F} \rho(x, y) < \delta\right\}$, the open $\delta$-neighborhood of the set $F$. By the definition of $X$ we have
$$
\{X \in F\}=\bigcap_{k=1}^{\infty} \bigcap... | proof | Other | proof | Yes | Yes | olympiads | false | 33,727 |
9. Let $\xi_{1}, \xi_{2}, \ldots$ be a sequence of independent identically distributed random variables, $\mathscr{F}_{n}=\sigma\left(\xi_{1}, \ldots, \xi_{n}\right), n \geqslant 1$, and $\tau-$ a stopping time with respect to $\left(\mathscr{F}_{n}\right)_{n \geqslant 1}$. Define
$$
\eta_{n}(\omega)=\xi_{n+\tau(\omeg... | Solution. Due to the independence and identical distribution of the variables $\xi_{n}, n \geqslant 1$, the distributions of the sequences $\xi$ and $\xi^{t}=$ $=\left(\xi_{1+t}, \xi_{2+t}, \ldots\right)$ coincide for any $t=0,1, \ldots$ Moreover, since $\{\tau=t\} \in \sigma\left(\xi_{1}, \ldots, \xi_{t}\right)$, the ... | proof | Other | proof | Yes | Yes | olympiads | false | 33,728 |
1. Let $X$ be a random variable, $\mathrm{E}|X|<\infty$, and $\varphi(x)=\mathrm{E}|X-x|$, $x \in \mathbb{R}$. Prove that the distribution of the variable $X$ is uniquely determined by the function $\varphi=\varphi(x)$. | Solution. Since
$$
\frac{\|X-x|-| X-y\|}{|x-y|} \leqslant 1, \quad x \neq y
$$
by the Lebesgue's dominated convergence theorem, the right derivative of the function $\varphi$ exists everywhere and is equal to
$$
\mathrm{E}[-I(x<X)+I(x \geqslant X)]=2 \mathrm{P}(X \leqslant x)-1
$$
In other words, $\mathrm{P}(X \leq... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,729 |
2. Let $\xi$ be a non-negative random variable, and $S$ be the set of simple non-negative functions $s$ that do not exceed $\xi(s \leqslant \xi)$. Prove that
$$
\sup _{s \in S} \mathrm{E} s=\mathrm{E} \xi
$$ | Solution. If $s \in S$ and $\left(\xi_{n}\right)_{n \geqslant 1}$ is a sequence of simple random variables such that $\xi_{n} \uparrow \xi$, then $\max \left(\xi_{n}, s\right) \uparrow \xi$ and $\mathrm{E} s \leqslant \max \left(\xi_{n}, s\right)$,
which means $\mathrm{E} s \leqslant \mathrm{E} \xi$ and $\sup _{s \in S... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,730 |
3. Let $\xi$ and $\eta$ be random variables for which $E \xi$ and $E \eta$ are defined, with at least one of them being infinite and the expression $E \xi + E \eta$ making sense (not of the form $\infty - \infty$ or $-\infty + \infty$). Then $\mathrm{E}(\xi + \eta) = \mathrm{E} \xi + \mathrm{E} \eta$. Prove this proper... | Solution. To establish the desired property, we need to consider different possibilities arising from the representations $\xi=\xi^{+}-\xi^{-}$ and $\eta=\eta^{+}-\eta^{-}$. Suppose, for example, $\mathrm{E} \xi^{+}=\infty, \mathrm{E} \xi^{-}2 \eta^{-}\right) \xi^{+} / 2 \leqslant(\xi+\eta)^{+}$. Let us choose simple n... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,731 |
4. Show that if $\xi=\eta$ a.s. and $\mathrm{E} \xi$ exists, then $\mathrm{E} \eta$ also exists and $\mathrm{E} \eta=\mathrm{E} \xi$. | Solution. From the equality $\xi=\eta$ a.s., it follows that $\mathrm{P}\left(\xi^{-}=\eta^{-}\right)=$ $=\mathrm{P}\left(\xi^{+}=\eta^{+}\right)=1$. Thus (by the definition of mathematical expectation), it is sufficient to consider the case when $\xi$ and $\eta$ are non-negative. Then
$$
\sum_{k=1}^{n^{2}} \frac{k-1}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,732 |
5. Let $\xi$ be an extended random variable, and $\mu$ be a measure such that $I=\int_{\Omega}|\xi| d \mu<\infty$. Show that then $\mu(|\xi|=\infty)=0$. | Solution. The desired equality follows from the fact that
$$
\mu(|\xi|=\infty) \leqslant \mu(|\xi|>m) \leqslant \frac{1}{m} \int_{\Omega}|\xi| d \mu \leqslant \frac{I}{m}, \quad m \rightarrow \infty
$$ | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,733 |
6. Let $\mu-$ be an arbitrary measure, and $\xi$ and $\eta-$ be extended random variables for which $\int_{\Omega}|\xi| d \mu$ and $\int_{\Omega}|\eta| d \mu$ are finite. Then if for all sets $A \in \mathscr{F}$ the inequality $\int_{A} \xi d \mu \leqslant \int_{A} \eta d \mu$ holds, then
$\mu(\xi>\eta)=0$. | Solution. From the finiteness of the integral $\int_{\Omega}(|\xi|+|\eta|) d \mu$ it follows that $\mu\left(A_{n}\right)1 / n\}, n \geqslant 1$. On each $A_{n}$ the inequality $\xi>\eta$ is satisfied with $\mu$-measure zero. The latter follows from an analogous property for finite measures (by proper normalization of w... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,734 |
7. Let $X$ be an arbitrary random variable. Prove that for all $t \in \mathbb{R}$ the following equality holds:
$$
\mathrm{E} \operatorname{ch}(t X)=\sum_{n \geqslant 1} \frac{t^{2 n}}{(2 n)!} \mathbb{E} X^{2 n}
$$
in the sense that the left-hand side of the equation is finite if and only if all $\mathrm{E} X^{2 n}$ ... | Solution. The statement follows from Fubini's theorem and the expansion
$$
\operatorname{ch} x=\frac{e^{x}+e^{-x}}{2}=\sum_{n \geqslant 1} \frac{x^{2 n}}{(2 n)!}
$$ | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,735 |
10. The Dirichlet function, defined on $[0,1]$ by the formula
$$
d(x)= \begin{cases}1, & x \text { irrational } \\ 0, & x \text { rational }\end{cases}
$$
is integrable in the sense of Lebesgue, but not integrable in the sense of Riemann. Why? | Solution. Consider the Riemann integral sum
$$
\begin{gathered}
I_{n}=I_{n}\left(x_{1}, \ldots, x_{n}, \omega_{1}, \ldots, \omega_{n-1}\right)=\sum_{i=1}^{n-1} d\left(\omega_{i}\right)\left(x_{i+1}-x_{i}\right) \\
0=x_{1}<\ldots<x_{n}=1
\end{gathered}
$$
where $\omega_{i} \in\left[x_{i}, x_{i+1}\right]$ for all $1 \l... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,738 |
11. Provide an example of a sequence of Riemann integrable functions $\left(f_{n}\right)_{n \geqslant 1}$, defined on $[0,1]$ and such that $\left|f_{n}\right| \leqslant 1, f_{n} \rightarrow f$ everywhere, but $f$ is not Riemann integrable. | Solution. Consider the functions $f_{n}(x)=\sum_{i=1}^{n} I_{\{q i\}}(x)$, where $q_{n}$ runs through all rational numbers on $[0,1]$. Then the Riemann integrable functions $f_{n}$ converge everywhere to the Dirichlet function, which is not integrable on the interval $[0,1]$ (see problem II.6.10). | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,739 |
12. Let $\left\{a_{i j}\right\}_{i, j \geqslant 1}$ be a family of real numbers such that $\sum_{i, j}\left|a_{i j}\right|<\infty$. Derive from Fubini's theorem that
$$
\sum_{i, j} a_{i j}=\sum_{i} \sum_{j} a_{i j}=\sum_{j} \sum_{i} a_{i j}
$$ | Solution. Consider an arbitrary sequence of positive numbers $p_{1}, p_{2}, \ldots, \sum_{i=1}^{\infty} p_{i}=1$, and define a probability measure $\mathrm{P}$ on $\Omega=\mathbb{N}=\{1,2, \ldots\}$ by the formula $\mathrm{P}(A)=\sum_{i \in A} p_{i j}$. Next, take the function $f(i, j)=\frac{a_{i j}}{p_{i} p_{j}}$, for... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,740 |
13. Provide an example of a family $\left\{a_{i j}\right\}_{i, j \geqslant 1}$ for which $\sum_{i, j}\left|a_{i j}\right|=\infty$ and the equality in formula (*) from problem II.6.12 does not hold. | Solution. As $\left(a_{i j}\right)$, one can take
$$
a_{i j}= \begin{cases}0, & i=j \\ (i-j)^{-3}, & i \neq j\end{cases}
$$ | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,741 |
14. Prove the validity of the following result on integration by substitution in $\mathbb{R}$.
Let $\varphi=\varphi(y)$ be a smooth function on $[a, b]$, $\varphi(a)<\varphi(b)$, and let $f=f(x)$ be a Borel function that is Lebesgue integrable on $[\varphi(a), \varphi(b)]$. Then the function $f(\varphi(y)) \varphi^{\p... | Solution. If the function $f$ is continuous, then by the Newton-Leibniz formula
$$
(\mathrm{R}) \int_{\varphi(a)}^{\varphi(b)} f(x) d x=F(\varphi(b))-F(\varphi(a))
$$
where (R) $\int$ is the Riemann integral, and
$$
F(z)=(\mathrm{R}) \int_{\varphi(a)}^{z} f(x) d x, \quad \varphi(a) \leqslant z \leqslant \varphi(b)
$... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,742 |
15. (See [79].) Prove the validity of the following result on integration by substitution in $\mathbb{R}^{n}$.
Let $\varphi=\left(\varphi_{1}(y), \ldots, \varphi_{n}(y)\right)$ be a smooth (up to the second order) mapping from $\mathbb{R}^{n}$ to $\mathbb{R}^{n}, \varphi(y)=y$ outside some ball, and
$f=f(x)$ be a Leb... | Solution. First, consider the case when the function $f$ is continuous. For continuous functions with bounded support, the Riemann and Lebesgue integrals coincide (this can be verified in the same way as in the one-dimensional case), so it is sufficient to establish the desired equality for the Riemann integral.
By th... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,743 |
16. Let $(X, \mathscr{X}, \mu)$ be a measurable space with measure $\mu$, and let $f=f(x)$, $g=g(x)$, and $h=h(x)$ be $\mathscr{X}$-measurable functions on $X$. Prove that for any such $p, q, r \geqslant 1$, where $p^{-1}+q^{-1}+r^{-1}=1$, the following inequality holds:
$$
\int_{X}|f g h| d \mu \leqslant\left(\int_{X... | Solution. To derive the required inequality, it is necessary to apply Hölder's inequality twice for arbitrary measurable spaces (it is proved in the same way as in the case of probability spaces)
$$
\int_{X}|\varphi \psi| d \mu \leqslant\left(\int_{X}|\varphi|^{\alpha} d \mu\right)^{\frac{1}{\alpha}}\left(\int_{X}|\ps... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,744 |
17. Let $f, g$ and $h$ be functions in $L^{p}, L^{q}$ and $L^{r}$, respectively, where $p, q, r \geqslant 1$ such that $p^{-1}+q^{-1}+r^{-1}=2$. Denote by $f * g$ the convolution $\int_{\mathbb{R}} f(y) g(x-y) d y$. Using problem II.6.16, prove that
$$
\left|\int_{\mathbb{R}}(f * g)(x) h(x) d x\right| \leqslant\|f\|_{... | Solution. Let's determine $\alpha, \beta, \gamma$ from the following equations:
$$
\alpha^{-1}+\beta^{-1}=p^{-1}, \quad \beta^{-1}+\gamma^{-1}=q^{-1}, \quad \gamma^{-1}+\alpha^{-1}=r^{-1}
$$
Then $\alpha, \beta, \gamma \geqslant 1$ and
$$
\alpha^{-1}+\beta^{-1}+\gamma^{-1}=\frac{p^{-1}+q^{-1}+r^{-1}}{2}=1
$$
Theref... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,745 |
18. Let $X$ be a random variable, $\varphi(x)=\frac{x}{1+x}, x>0$. Prove the validity of the following inequalities:
$$
\mathrm{E}[\varphi(|X|)-\varphi(x)] \leqslant \mathrm{P}(|X| \geqslant x) \leqslant \frac{\mathrm{E} \varphi(|X|)}{\varphi(x)}
$$ | Solution. The statement follows from the inequalities
$$
\begin{aligned}
\varphi(|X|)-\varphi(x)=\frac{|X|-x}{(1+x)(1+|X|)} \leqslant \frac{|X|-x}{|X|+x} \leqslant I(|X| & \geqslant x) \leqslant \\
& \leqslant \frac{\varphi(|X|)}{\varphi(x)} \cdot I(|X| \geqslant x)
\end{aligned}
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,746 |
19. Let $\xi$ and $\zeta$ be bounded random variables. Prove that if
$$
\mathrm{E} \xi^{m} \zeta^{n}=\mathrm{E} \xi^{m} \cdot \mathrm{E} \zeta^{n}
$$
for all $m, n \geqslant 1$, then $\xi$ and $\zeta$ are independent. | Solution. Without loss of generality, we can assume that
$$
\mathrm{P}(|\xi| \leqslant 1)=\mathrm{P}(|\zeta| \leqslant 1)=1
$$
Due to the condition of the problem,
$$
\mathrm{E} f(\xi) g(\zeta)=\mathrm{E} f(\xi) \cdot \mathrm{E} g(\zeta)
$$
for any polynomials $f$ and $g$. Using Bernstein's theorem on the uniform a... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,747 |
20. Let $X=X_{t}(\omega)$ and $Y=Y_{t}(\omega),(\omega, t) \in \Omega \times[0,1]-\mathscr{F} \times \mathscr{B}[0,1]$-measurable functions, and $\int_{0}^{1}\left|X_{t}\right| d t, \int_{0}^{1}\left|Y_{t}\right| d t<\infty$ a.s. Suppose that
$$
\left(X_{t_{1}}, \ldots, X_{t_{n}}\right) \quad \text { and } \quad\left(... | Solution. Since a.s.
\[
\int_{0}^{1} X_{t} I\left(\left|X_{t}\right| \leqslant n\right) d t \rightarrow \int_{0}^{1} X_{t} d t, \quad \int_{0}^{1} Y_{t} I\left(\left|Y_{t}\right| \leqslant n\right) d t \rightarrow \int_{0}^{1} Y_{t} d t \quad \text{as } n \rightarrow \infty,
\]
it suffices to prove the desired propert... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,748 |
21. (See [59].) Let $f=f(x, \theta)$ for each fixed $\theta \in (a, b), a<\theta<b, a>0\}
$$
does not depend on $\theta$. Assuming the possibility of differentiating under the Lebesgue integral sign, prove the equivalence of the following representations of Fisher information $I(\theta)$:
$$
\begin{aligned}
I(\theta)... | Solution. Since $\int_{\mathbb{R}} f(x, \theta) d x \equiv 1$, we have
$$
\frac{d}{d \theta} \int_{\mathbb{R}} f(x, \theta) d x=\int_{\mathbb{R}} f_{\theta}^{\prime}(x, \theta) d x=\int_{\mathbb{R}} s(x, \theta) f(x, \theta) d x \equiv 0
$$
Moreover,
$$
\frac{d^{2}}{d \theta^{2}} \int_{\mathbb{R}} f(x, \theta) d x=\... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,749 |
22. Let $\xi, \eta$ be non-negative random variables.
(a) If $\mathrm{E} \xi^{p}, \mathrm{E} \eta^{q}1, 1 / p + 1 / q = 1$, then by Hölder's inequality
$$
\mathrm{E} \xi \eta \leqslant \left(\mathrm{E} \xi^{p}\right)^{1 / p} \left(\mathrm{E} \eta^{q}\right)^{1 / q}
$$
Prove that for $q > 0$
$$
(b) If $\mathrm{E} \x... | Solution. (a) If $\mathrm{E} \xi \eta=\infty$, then the inequality is obvious. Let, further, $\mathrm{E} \xi \eta<\infty$. By Hölder's inequality, we have
$$
\mathrm{E} \xi^{p}=\mathrm{E}(\xi \eta)^{p} \eta^{-p} \leqslant(\mathrm{E} \xi \eta)^{p}\left(\mathrm{E} \eta^{-p /(1-p)}\right)^{1-p}
$$
from which the desired... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,750 |
23. (See [84].) Let a concave function $f=f(x)$ be defined on $(0, \infty)$ such that the function $f\left(x^{-1}\right)$ is convex. Prove that the function $x f(x)$ is also convex and that $f$ is either strictly increasing or a constant.
Show that in the first case, for any positive random variable $\xi$, the followi... | Solution. First, we will show that the function $x f(x)$ is convex on $(0, \infty)$. The function $f\left(x^{-1}\right)$ is convex on $(0, \infty)$, which is equivalent to the inequality
$$
\frac{f\left(y^{-1}\right)-f\left(x^{-1}\right)}{y-x} \leqslant \frac{f\left(z^{-1}\right)-f\left(y^{-1}\right)}{z-y}
$$
satisfi... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,751 |
24. Let $0 \leqslant \xi \leqslant 1$ a.s., and $\zeta$ be a random variable with values in $\mathbb{C}$ and $\mathrm{E}|\zeta|<\infty$. Prove Abel's inequality
$$
|\mathrm{E} \xi \zeta| \leqslant \sup _{0 \leqslant x \leqslant 1}|\mathrm{E} \zeta I(\xi \geqslant x)|
$$ | Solution. By Fubini's theorem
$$
\mathrm{E} \xi \zeta=\mathrm{E} \zeta \int_{0}^{1} I(\xi \geqslant x) d x=\int_{0}^{1} \mathrm{E} \zeta I(\xi \geqslant x) d x
$$
from which the required equality obviously follows. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,752 |
25. Let's call functions $f=f(x)$ and $g=g(x), x \in \mathbb{R}^{d}$, one-monotonic if
$$
[f(y)-f(x)][g(y)-g(x)] \geqslant 0
$$
for all $x, y \in \mathbb{R}^{d}$. Suppose that the distribution of the random vector $(\xi, \zeta)$ coincides with the distribution of the pair $(\zeta, \xi)$, where $\xi, \zeta$ are random... | Solution. To derive the desired inequality, we should use the easily verifiable formula
$$
\mathrm{E} f(\xi) g(\xi)-\mathrm{E} f(\xi) g(\zeta)=\frac{1}{2} \mathrm{E}[f(\xi)-f(\zeta)][g(\xi)-g(\zeta)]
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,753 |
26. (Equivalent definition of uniform integrability; see definition 4 from V1.I.6.4.) Prove that
$$
\sup _{n} \mathrm{E}\left|\xi_{n}\right| I\left(\left|\xi_{n}\right|>c\right) \rightarrow 0 \Leftrightarrow \varlimsup_{n} \mathrm{E}\left|\xi_{n}\right| I\left(\left|\xi_{n}\right|>c\right) \rightarrow 0, \quad c \righ... | Solution. By the absolute continuity of the Lebesgue integral
$$
\max _{i \leqslant n} \mathrm{E}\left|\xi_{i}\right| I\left(\left|\xi_{i}\right|>c\right) \rightarrow 0, \quad c \rightarrow \infty
$$
for any fixed $n$, since for each $i \leqslant n$ the relation
$$
\left\{\left|\xi_{i}\right|>c\right\} \downarrow \v... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,754 |
27. (La Vallée Poussin's Theorem.) Prove that $\left\{\xi_{n}\right\}_{n \geqslant 1}$ is uniformly integrable if and only if
$$
\sup _{n} E G\left(\left|\xi_{n}\right|\right)<\infty
$$
for some non-negative increasing function $G$, such that
$$
\frac{G(x)}{x} \rightarrow \infty, \quad x \rightarrow \infty
$$
Also ... | Solution. If $\sup _{n} \mathrm{E} G\left(\left|\xi_{n}\right|\right)<\infty$ for some non-decreasing, convex function $G: [0, \infty) \rightarrow [0, \infty)$ with $G(0)=0$ and $G(x) \rightarrow \infty$ as $x \rightarrow \infty$, then for any $c>0$ we have
$$
\sup _{n} \mathrm{E}\left|\xi_{n}\right| I\left(\left|\xi_... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,755 |
28. Let $\left\{\xi_{n}\right\}_{n \geqslant 1}$ be a uniformly integrable family of random variables. Show that the families
$$
\begin{gathered}
\left\{\zeta:|\zeta| \preccurlyeq\left|\xi_{n}\right| \text { for some } n\right\}, \\
\left\{\sum_{n} \lambda_{n} \xi_{n}: \lambda_{n} \in \mathbb{R}, \sum_{n}\left|\lambda... | Solution. We have
$$
\begin{aligned}
& \mathrm{E}|\zeta| I(|\zeta|>c)=\int_{c}^{\infty} \mathrm{P}(|\zeta|>x) d x+c \mathrm{P}(|\zeta|>c) \leqslant \\
& \leqslant \int_{c}^{\infty} \mathrm{P}\left(\left|\xi_{n}\right|>x\right) d x+c \mathrm{P}\left(\left|\xi_{n}\right|>c\right)=\mathrm{E}\left|\xi_{n}\right| I\left(\l... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,756 |
29. Let $\xi_{1}, \xi_{2}, \ldots$ be arbitrary random variables. Show that the family $\left\{n^{-1} \sum_{i=1}^{n} \xi_{i}\right\}_{n \geqslant 1}$ is uniformly integrable if at least one of the families $\left\{\xi_{n}\right\}_{n \geqslant 1}$ or $\left\{\sum_{i=1}^{n} \xi_{i} / i\right\}_{n \geqslant 1}$ is uniform... | Solution. If the family $\left\{\xi_{n}\right\}_{n \geqslant 1}$ is uniformly integrable, then the same is true for $\left\{n^{-1} \sum_{i=1}^{n} \xi_{i}\right\}_{n \geqslant 1}$ (see problem II.6.28). Let now the family $\left\{X_{n}\right\}_{n \geqslant 1}, X_{n}=\sum_{i=1}^{n} \xi_{i} / i$, be uniformly integrable. ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,757 |
30. Let $0 \leqslant \xi_{n} \xrightarrow{d} \xi$ and $\mathrm{E} \xi_{n}<\infty$. Prove that the following conditions are equivalent:
1) $\mathrm{E} \xi_{n} \rightarrow \mathrm{E} \xi<\infty$,
2) $\lim _{n} \mathrm{E} \xi_{n} \leqslant \mathrm{E} \xi<\infty$,
3) the family $\left\{\xi_{n}\right\}_{n \geqslant 1}$ is u... | Solution. By the definition of convergence $\xrightarrow{d}$, for any point of continuity $x$ of the function $f(x)=\mathrm{P}(\xi>x)$, we have $f_{n}(x)=\mathrm{P}\left(\xi_{n}>x\right) \rightarrow f(x)$. Therefore, $f_{n} \rightarrow f$ almost everywhere with respect to the Lebesgue measure (the measure of the set of... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,758 |
31. (Vitali's Criterion.) Let $p>0, \mathrm{E}\left|\xi_{n}\right|^{p}<\infty, \xi_{n} \xrightarrow{p} \xi$. Prove that the following conditions are equivalent:
1) $\mathrm{E}\left|\xi_{n}\right|^{p} \rightarrow \mathrm{E}|\xi|^{p}<\infty$
2) $\lim _{n} \mathrm{E}\left|\xi_{n}\right|^{p} \leqslant \mathrm{E}|\xi|^{p}<\... | Solution. From the convergence $\xi_{n} \xrightarrow{p} \xi$, due to the continuity of the function $\varphi(x)=|x|^{p}$, it follows that $\left|\xi_{n}\right|^{p} \xrightarrow{p}|\xi|^{p}$. Thus, by problem II.6.30, conditions 1, 2, and 3 are equivalent ( $\xi_{n} \xrightarrow{d} \xi$, since $\xi_{n} \xrightarrow{p} \... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,759 |
32. (Pratt's Lemma, [93].) Let $\xi, \eta, \zeta$ and $\xi_{n}, \eta_{n}, \zeta_{n}, n \geqslant 1$, be random variables such that $\eta_{n} \leqslant \xi_{n} \leqslant \zeta_{n}$ for all $n$,
$$
\xi_{n} \stackrel{p}{\rightarrow} \xi, \quad \eta_{n} \xrightarrow{p} \eta, \quad \zeta_{n} \stackrel{p}{\rightarrow} \zeta... | Solution. Since $0 \leqslant \xi_{n}-\eta_{n} \leqslant \zeta_{n}-\eta_{n}$ and $\mathrm{E}\left(\zeta_{n}-\eta_{n}\right) \rightarrow \mathrm{E}(\zeta-\eta)$, the family $\left\{\zeta_{n}-\eta_{n}\right\}_{n \geqslant 1}$ is uniformly integrable (see problem II.6.31) and, therefore, the family $\left\{\xi_{n}-\eta_{n}... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,760 |
34. Prove the following theorem of Beppo Levi.
Let random variables $\xi_{1}, \xi_{2}, \ldots$ be integrable, i.e., $\mathrm{E}\left|\xi_{n}\right|<$ $<\infty$ for all $n \geqslant 1$. If in addition $\mathrm{E} \xi_{n}<\infty$ and $\xi_{n} \uparrow \xi$, then the random variable $\xi$ is integrable and $\mathrm{E} \x... | Solution. By the condition, for all $n \geqslant 1$ we have the equality
$$
\xi_{n}=\xi_{1}+\zeta_{2}+\ldots+\zeta_{n}, \quad \zeta_{n}=\xi_{n}-\xi_{n-1} \geqslant 0
$$
Moreover, due to the boundedness of $\mathrm{E} \xi_{n}$, we have
$$
\sum_{n=1}^{\infty} E \zeta_{n}<\infty
$$
From this, it follows that $\xi_{n}$... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,762 |
35. Prove the following variation of Fatou's lemma.
If $0 \leqslant \xi_{n} \rightarrow \xi$ a.s. and suр $\mathrm{E} \xi_{n}<\infty$, then the random variable $\xi$ is integrable and $\mathrm{E} \xi \leqslant \sup ^{n} \mathrm{E} \xi_{n}$. | Solution. We will assume that the quantity $\xi$ is continuous; otherwise, we replace the quantities $\xi, \xi_{1}, \ldots$ with $\xi+U, \xi_{1}+U, \ldots$, where $U$ is uniformly distributed on $[0,1]$ and independent of $\xi, \xi_{1}, \ldots$. Let us denote
$$
\begin{aligned}
\zeta_{m n} & =\sum_{k=1}^{m^{2}} \frac{... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,763 |
36. Let $\xi_{n} \downarrow \xi$ a.s., $\mathrm{E}\left|\xi_{n}\right|<\infty$. Show that $\mathrm{E}\left|\xi_{n}-\xi\right| \rightarrow 0$. | Solution. We have $\mathrm{E}\left|\xi_{n}-\xi\right| \leqslant \mathrm{E}\left|\xi_{n}^{+}-\xi^{+}\right|+\mathrm{E}\left|\xi_{n}^{-}-\xi^{-}\right|$. Here $0 \leqslant$ $\leqslant \xi_{n}^{+} \downarrow \xi^{+}, 0 \leqslant \xi_{n}^{-} \uparrow \xi^{-}$ and $\sup \mathrm{E} \xi_{n}^{-}<\infty$. Since $\mathrm{P}\left... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,764 |
37. Let the random variables $\xi_{n}, n \geqslant 1$, be such that $\mathrm{D} \xi_{n}=$ $=o\left(\left(\mathrm{E} \xi_{n}\right)^{2}\right)$ and
$$
\mathrm{E} \xi_{n} \rightarrow \infty, \quad n \rightarrow \infty
$$
Prove that $\xi_{n} \xrightarrow{p} \infty$. | Solution. By Chebyshev's inequality
$$
\mathrm{P}\left(\xi_{n}\mathrm{E} \xi_{n} / 2\right) \leqslant \frac{4 \mathrm{D} \xi_{n}}{\left(\mathrm{E} \xi_{n}\right)^{2}} \rightarrow 0
$$
and, consequently, for any $c>0$ we have
$$
\mathrm{P}\left(\xi_{n}>c\right) \rightarrow 1, \quad n \rightarrow \infty
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,765 |
38. Provide an example of random variables $\xi_{n}, n \geqslant 1$, such that
$$
\mathrm{E} \sum_{n=1}^{\infty} \xi_{n} \neq \sum_{n=1}^{\infty} \mathrm{E} \xi_{n}
$$ | Solution. Define $\xi_{n}, n \geqslant 1$, by
$$
\xi_{n}=n I(n U \leqslant 1)-(n-1) I((n-1) U \leqslant 1),
$$
where $U$ is a random variable with a uniform distribution on $[0,1]$. Then
$$
\sum_{n=1}^{\infty} \xi_{n}=\lim _{n} \sum_{i=1}^{n} \xi_{i}=\lim _{n} n I(n U \leqslant 1)=0
$$
with probability one. Thus
$... | 0\neq1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,766 |
39. For any finite number of independent integrable random variables $\xi_{1}, \ldots, \xi_{n}$, the equality
$$
\mathrm{E} \prod_{i=1}^{n} \xi_{i}=\prod_{i=1}^{n} \mathrm{E} \xi_{i} \cdot
$$
holds. Show that if $\xi_{1}, \xi_{2}, \ldots$ is a sequence of independent integrable random variables, then, in general,
$$... | Solution. Let $\xi_{i}, i=1,2, \ldots$, be Bernoulli random variables,
$$
\mathrm{P}\left(\xi_{i}=0\right)=\mathrm{P}\left(\xi_{i}=2\right)=\frac{1}{2}
$$
Then $\mathrm{E} \xi_{i}=1$ and $\prod_{i=1}^{\infty} \mathrm{E} \xi_{i}=1$. On the other hand,
$$
\mathrm{P}\left(\prod_{i=1}^{\infty} \xi_{i}>0\right)=\lim _{n}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,767 |
40. Let the random variable $X$ be such that
$$
\varlimsup_{n} \frac{\mathrm{P}(|X|>2 n)}{\mathrm{P}(|X|>n)}<\frac{1}{2}
$$
Prove that then $\mathbf{E}|X|<\infty$. | Solution. Since
$$
\mathrm{E}|X|=\int_{\mathbb{R}_{+}} \mathrm{P}(|X|>x) d x
$$
the finiteness of $\mathrm{E}|X|$ is equivalent to the convergence of the integral
$$
\int_{1}^{\infty} \mathrm{P}(|X|>x) d x = \sum_{n=0}^{\infty} \int_{2^{n}}^{2^{n+1}} \mathrm{P}(|X|>x) d x \leqslant \sum_{n=0}^{\infty} 2^{n+1} \mathr... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,768 |
41. Let $X$ be a non-negative random variable with a continuous distribution function $F=F(x)$. Provide an example of some non-negative function $G=G(x)$, depending on $F$, such that $\mathrm{E} G(X)<\infty$ and at the same time, the finiteness of $\mathrm{E} G(\alpha X)^{2}$ for some $0<\alpha<1$ ensures the existence... | Solution. We can take $G=(1-F)^{-1/2}$. Then
$$
\mathrm{E} G(X)=\int_{0}^{\infty}[1-F(x)]^{-1/2} d F(x)=\int_{0}^{1} \frac{d y}{\sqrt{y}}=2
$$
If at the same time
$$
\mathrm{E} G(\alpha X)^{2}=\int_{0}^{\infty} \frac{d F(x)}{1-F(\alpha x)}=\frac{1-F(z)}{1-F(\alpha z)} \leqslant \int_{z}^{\infty} \frac{d F(x)}{1-F(\a... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,769 |
42. We will call a polynomial $P=P(x, y)$ of degree $n$ admissible if it has non-zero coefficients for $x^n$ and $y^n$. Let $X$ and $Y$ be such independent random variables that some admissible polynomials $P(X, Y)$ and $Q(X, Y)$ of degrees $n$ and $m$ respectively define independent variables.
It is somewhat surprisi... | Solution. (a) First, note that one can always choose a sufficiently large $a>0$ such that for all $x, y$ in $\mathbb{R}$, the inequalities
$$
|P(x, y)|0$ there always exists a $c=c(b)>0$ such that for all $|y| \leqslant b$, $b^{2} \leqslant|x|$, the inequalities
$$
(c|x|)^{n}0$, we see that $\mathrm{E}|X|^{2 p}0$ and... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,770 |
43. Let $X$ be a random variable taking values $n=$ $=0,1,2, \ldots$ with probabilities $p_{n}$. The function
$$
G(s)=\mathrm{E} s^{X}=p_{0}+p_{1} s+p_{2} s^{2}+\ldots, \quad|s| \leqslant 1
$$
is called the generating function of the random variable $X$ (see § 3 of the appendix). Establish the following formulas:
(a... | Solution. By direct verification, we obtain
$$
\begin{gathered}
\sum_{n=0}^{\infty} e^{-\lambda} \frac{\lambda^{n}}{n!} s^{n}=e^{-\lambda} \sum_{n=0}^{\infty} \frac{(\lambda s)^{n}}{n!}=e^{\lambda s-\lambda} \\
\sum_{k=0}^{\infty} p(1-p)^{n} s^{n}=p \sum_{n=0}^{\infty}[(1-p) s]^{n}=\frac{p}{1-(1-p) s} \\
\mathrm{E}s^{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,771 |
44. Let $X$ be a random variable taking values in the set $\mathbb{Z}_{+}$. Its factorial moments $m_{(n)}$ are defined by the formula
$$
m_{(n)}=\mathrm{E} X(X-1) \ldots(X-n+1)=\mathrm{E}(X)_{n}, \quad n \in \mathbb{N}
$$
Show that for any $n \geqslant 1$ the following statements are true:
(a) if $\mathrm{E} X^{n} ... | Solution. By the Lebesgue dominated convergence theorem, for each $0 < s < 1$, $G^{(n)}(s)$ exists and
$$
G^{(n)}(s)=\mathrm{E}(X)_{n} s^{X} .
$$
Moreover, $\mathrm{E}(X)_{n} \leqslant \mathrm{E}^{n}$ and $\mathrm{E} X^{n} I(X>2 n) \leqslant 2^{n} \mathrm{E}(X)_{n} I(X>2 n)$. Therefore,
$$
\mathrm{E} X^{n}<\infty \L... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,772 |
45. Let $X$ be a random variable taking values $0,1, \ldots, n$, with binomial moments $b_{0}, b_{1}, \ldots, b_{n}$, defined by the formula
$$
b_{k}=\mathrm{E} C_{X}^{k}=\frac{\mathrm{E}(X)_{k}}{k!} \equiv \frac{1}{k!} \mathrm{E} X(X-1) \ldots(X-k+1)
$$
Show that the generating function has the form
$$
G(s)=\mathrm... | Solution. Since $\mathrm{E} C_{X}^{k}=\mathrm{E} C_{X}^{k} I(X \geqslant k)$, by the binomial formula of Newton
$$
\begin{array}{rl}
\mathrm{E} s^{X}=\mathrm{E}(s-1+1)^{X}=\mathrm{E} \sum_{k=0}^{X} C_{X}^{k}(s-1)^{k}=\sum_{k=0}^{n}(s-1)^{k} & \mathrm{E} C_{X}^{k} I(X \geqslant k)= \\
& =\sum_{k=0}^{n} b_{k}(s-1)^{k}
\... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,773 |
47. Let $A_{1}, \ldots, A_{n}$ be arbitrary events and $\Sigma_{n}=I_{A_{1}}+\ldots+I_{A_{n}}$. Show that the generating function $G_{\Sigma_{n}}(s)=\mathrm{E} s^{\Sigma_{n}}$ is given by the following formula:
$$
G_{\Sigma_{n}}(s)=\sum_{m=0}^{n} S_{m}(s-1)^{m}
$$
where
$$
S_{m}=\sum_{1 \leqslant i_{1}<\ldots<i_{m} ... | Solution. The representation for $G_{\Sigma_{n}}$ follows from the following relations:
$$
\begin{aligned}
& G_{\Sigma_{n}}(s)=\mathrm{E} \prod_{i=1}^{n}\left(1+I_{A_{i}}(s-1)\right)= \\
& =\mathrm{E}\left[1+\sum_{i=1}^{m} I_{A_{i}}(s-1)+\sum_{1 \leqslant i_{1}<i_{2} \leqslant n} I_{A_{i_{1}}} I_{A_{i_{2}}}(s-1)^{2}+\... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,775 |
48. Let $G(s)=p_{0}+p_{1} s+p_{2} s^{2}+\ldots$ be the generating function of a discrete random variable $X, \mathrm{P}(X=n)=p_{n}, n=0,1,2, \ldots$ (see problem II.6.43). Let
$$
q_{n}=\mathrm{P}(X>n), \quad r_{n}=\mathrm{P}(X \leqslant n), \quad n=0,1,2, \ldots
$$
Show that the generating functions of the sequences ... | Solution. Since $q_{n}+r_{n}=1$ for all $n$, when $|s|n) s^{n} \uparrow \mathrm{E} X, \quad s \uparrow 1
$$
follows from the formula $\mathrm{E} X=\sum_{n \geqslant 0} \mathrm{P}(X>n)$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,776 |
49. Along with the generating function $G(s)$, in many cases the concept of the moment generating function is also useful:
$$
M(s)=\mathrm{E} e^{s X}
$$
(assuming that $s$ are such that $\mathrm{E} e^{s X} < \infty$). If the moment generating function $M(s)$ is finite in a neighborhood of zero, i.e., if there exists ... | Solution. (a) Let us assume for definiteness that for some $a>0$ the relation holds
$$
\left(\mathrm{E} e^{a|X|} \leqslant\right) \mathrm{E} e^{a X}+\mathrm{E} e^{-a X}0} x^{n} e^{-(a-|s|) x} \cdot \mathrm{E} e^{a|X|}0, p+q+r=1$. Let us define the distribution of ( $X, Y$ ) using the table:
| $X \backslash Y$ | 0 | 1... | M_{X+Y}()=M_{X}()M_{Y}() | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,777 |
50. Let $M(s)=\mathrm{E} e^{s X}$ be the moment generating function of a non-negative random variable $X$. Show that for all $p, s>0$ the following inequality holds:
$$
\mathrm{E} X^{p} \leqslant\left(\frac{p}{e s}\right)^{p} M(s)
$$ | Solution. The statement follows from the inequality
$$
x^{p} \leqslant\left(\frac{p}{e s}\right)^{p} e^{s x}, \quad x>0
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,778 |
51. Let $\xi$ be a random variable with distribution function $F$ and $E|\xi|^{-1}<\infty$. Prove that the function $F$ is differentiable at zero and $F^{\prime}(0)=0$. | Solution. Note that $\mathrm{P}(\xi=0)=0$, if $\mathrm{E}|\xi|^{-1}<\infty$. Consequently, by the property of absolute continuity of the Lebesgue integral
$$
\frac{F(x)-F(0)}{x} \leqslant E \frac{I(|\xi| \leqslant|x|)}{|x|} \leqslant E \frac{I(|\xi| \leqslant|x|)}{|\xi|} \rightarrow 0, \quad x \rightarrow 0
$$ | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,779 |
52. Show that if $\xi$ is an integrable random variable, then
$$
\mathrm{E} \xi=\int_{0}^{\infty} \mathrm{P}(\xi>x) d x-\int_{-\infty}^{0} \mathrm{P}(\xi>x) d x
$$
and if $\xi \geqslant 0$, then
$$
\mathrm{E} \xi=\int_{0}^{\infty} \mathrm{P}(\xi>x) d x
$$
and for $a>0$ the equality holds
$$
\mathrm{E}[\xi I(\xi>a)... | Solution. All the given relations follow from the formula
$$
\mathrm{E} \zeta=\int_{\mathbb{R}_{+}} \mathrm{P}(\zeta>x) d x
$$
which is valid for non-negative quantities $\zeta$. For deriving the first equality, the decomposition $\mathrm{E} \xi=\mathrm{E} \xi^{+}-\mathrm{E} \xi^{-}$ is also used. | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,780 |
53. Let $F=F(x)$ be the distribution function of a random variable $\xi$. Show that
$$
\begin{aligned}
& \mathrm{E}|\xi|>0 .
\end{aligned}
$$ | Solution. The first statement follows from the formulas
$$
\begin{aligned}
& \mathrm{E} \xi^{-}=\int_{0}^{\infty} \mathrm{P}\left(\xi^{-} \geqslant x\right) d x=\int_{0}^{\infty} F(-x) d x \\
& \mathrm{E} \xi^{+}=\int_{0}^{\infty} \mathrm{P}\left(\xi^{+}>x\right) d x=\int_{0}^{\infty}[1-F(x)] d x
\end{aligned}
$$
Mor... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,781 |
54. Let $X$ be a random variable, and $p>0$ be any positive number.
(a) Show that if
$$
n^{p-1} \mathrm{P}(|X|>n)=\frac{o(1)}{n}, \quad n \rightarrow \infty
$$
then $\mathrm{E}|X|^{r} < \infty
$$ | Solution. For all $r>0$, by Fubini's theorem
$$
\mathrm{E}|X|^{r}=\mathrm{E} \int_{0}^{\infty} r x^{r-1} I(|X|>x) d x=r \int_{0}^{\infty} x^{r-1} \mathrm{P}(|X|>x) d x
$$
Similarly,
$$
\begin{aligned}
\mathrm{E}|X| \ln (|X| \vee 1) & =\mathrm{E} \int_{0}^{\infty} I(|X|>y) d y \int_{1}^{\infty} I(|X|>x) \frac{d x}{x}... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,782 |
57. Let $X$ be a non-degenerate non-negative random variable. Show that the equality
$$
\mathrm{E} X^{a} \mathrm{E} X^{b}=\mathrm{E} X^{c} \mathrm{E} X^{d},
$$
where $a+b=c+d$ and $a, b, c, d \in \mathbb{R}$, can only hold if the sets $\{a, b\}$ and $\{c, d\}$ coincide. (It is assumed that the considered mathematical... | Solution. Let for definiteness $a=\min \{a, b, c, d\}$. Then the considered equality is equivalent to
$$
\mathrm{E} Z^{b-a}=\mathrm{E} Z^{c-a} \mathrm{E} Z^{d-a}
$$
where $b-a=c-a+d-a$ and the quantity $Z$ has a distribution function of the form
$$
\mathrm{P}(Z \leqslant z)=\frac{\mathrm{E} X^{a} I(X \leqslant z)}{\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,785 |
58. (See [86].) Let $X$ be a random variable taking a finite number of values $x_{1}<\ldots<x_{n}, n \geqslant 2$. Show that for all $a, b \in \mathbb{R}$, the following equality holds:
$$
\mathrm{D} X=\mathrm{E}(X-a)(X-b)+(\mathrm{E} X-a)(b-\mathrm{E} X),
$$
and derive from this that
$$
\left(\mathrm{E} X-x_{i}\rig... | Solution. The first equality is established by expanding the brackets. From it, the specified inequalities can easily be derived by substituting $(a, b)=\left(x_{1}, x_{n}\right)$ and $(a, b)=\left(x_{i}, x_{i+1}\right)$, and also noting that with probability one
$$
\left(X-x_{1}\right)\left(X-x_{n}\right) \leqslant 0... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,786 |
59. Let $\mathrm{P}(X \in[a, b])=1$ for some $-\infty < a < b < \infty$ and $b - a > 0$. | Solution. The function $f(x)=(x-\mathrm{E} X)^{2}$ is convex, therefore
$$
(X-\mathrm{E} X)^{2} \leqslant \frac{X-a}{b-a}(b-\mathrm{E} X)^{2}+\frac{b-X}{b-a}(a-\mathrm{E} X)^{2}
$$
so that
$\mathrm{D} X \leqslant \sup _{c}\left\{\frac{c-a}{b-a}(b-c)^{2}+\frac{b-c}{b-a}(a-c)^{2}\right\}=\sup _{c}(c-a)(b-c)=\frac{(b-a... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,787 |
60. (See [34].) Let $\mathrm{P}(|X| \leqslant b)=1, \mathrm{E} X=0$ and $\mathrm{D} X=\sigma^{2}$ for some $\sigma, b>0$. What is the maximum possible value of $\mathrm{E} e^{X}$? | Solution. We will show that the maximum possible value of $\mathrm{E} e^{X}$ is achieved on a Bernoulli random variable $X_{*}$, one of whose values is $b$ (and, consequently, the other value is $-\sigma^{2} / b$, since $\mathrm{E} X_{*}=0$ and $\left.\mathrm{E} X_{*}^{2}=\mathrm{D} X_{*}=\sigma^{2}\right)$. In other w... | \frac{e^{b}\sigma^{2}+e^{-\sigma^{2}/b}b^{2}}{\sigma^{2}+b^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,788 |
61. (See [71].) Let $\mathrm{P}(X \in[a, b])=1$ for some $a<0<b$. Prove that
$$
\mathrm{E} e^{X-\mathrm{E} X} \leqslant e^{\frac{(b-a)^{2}}{4}}
$$
Refine the last inequality by showing that
$$
\mathrm{E} e^{X-\mathrm{E} X} \leqslant e^{\frac{(b-a)^{2}}{8}}
$$ | Solution. By Jensen's inequality
$$
\mathrm{E} e^{X-\mathrm{E} X}=e^{-\mathrm{E} X} \mathrm{E} e^{X} \leqslant \mathrm{E} e^{-X} \mathrm{E} e^{X}
$$
Now, according to problem I. 6.59, we have
$$
\sup _{X} \mathrm{E} e^{X} \mathrm{E} e^{-X}=\frac{\left(e^{a}+e^{b}\right)^{2}}{4 e^{a+b}}=\operatorname{ch}^{2} \frac{d}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,789 |
62. Let $\xi$ be a random variable, $\mathrm{P}(\xi>0)>0$ and $\mathrm{E}|\xi|^{p}<\infty$ for some $p>1$. Show that
$$
\mathrm{P}(\xi>0) \geqslant \frac{(\mathrm{E} \xi I(\xi>0))^{q}}{\|\xi I(\xi>0)\|_{p}^{q}}
$$
where $q=p /(p-1)$. From this, derive that if $\mathrm{E} \xi \geqslant 0$, then for any $\varepsilon \i... | Solution. By Hölder's inequality
$$
\mathrm{E} \xi I(\xi>0) \leqslant\|\xi I(\xi>0)\|_{p}(\mathrm{P}(\xi>0))^{1 / q}
$$
which is equivalent to the first estimate. To derive the remaining estimates, it is sufficient to replace $\xi$ with $\xi-\varepsilon \mathrm{E} \xi$ and note the following:
$$
\begin{aligned}
& \m... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,790 |
63. Let $\xi$ be a random variable, $\mathrm{E} \xi^{2} < \infty$
$$
\mathrm{P}(|\xi - \mathrm{E} \xi| \geq \varepsilon) \leq \frac{\mathrm{D} \xi}{\varepsilon^2}
$$
Construct examples demonstrating the sharpness of the given estimate. | Solution. We can assume that $\mathrm{E} \xi=0$. By the Cauchy-Bunyakovsky inequality
$$
\begin{gathered}
\varepsilon=\mathrm{E}(\varepsilon-\xi) \leqslant \mathrm{E}(\varepsilon-\xi) I(\xi<\varepsilon) \leqslant\left(\mathrm{E}(\xi-\varepsilon)^{2}\right)^{1 / 2} \mathrm{P}(\xi<\varepsilon)^{1 / 2}= \\
=\left(\mathrm... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,791 |
64. Assuming that $\mathrm{E} \xi^{2}<\infty$, prove the inequality
$$
\mathrm{P}(\xi=0) \leqslant \frac{\mathrm{D} \xi}{\mathrm{E} \xi^{2}}
$$ | Solution. By the Cauchy-Bunyakovsky inequality
$$
(\mathrm{E} \xi)^{2}=(\mathrm{E} \xi I(|\xi|>0))^{2} \leqslant \mathrm{E} \xi^{2} \mathrm{P}(|\xi|>0)=\mathrm{E} \xi^{2}(1-\mathrm{P}(\xi=0))
$$
The last expression is equivalent to the desired estimate. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,792 |
65. Let $\xi$ be a non-negative random variable. Prove that if for a given $\varepsilon>0$ the probability $\mathrm{P}(\xi>\varepsilon)$ does not exceed $a$ for some $a \geqslant 0$, then
$$
\mathrm{E} \xi \leqslant \frac{\varepsilon}{1-\frac{\left(a \mathrm{E} \xi^{2}\right)^{1 / 2}}{\mathrm{E} \xi}}
$$
(assuming th... | Solution. By the Cauchy-Bunyakovsky inequality
$$
\mathrm{E} \xi I(\xi>\varepsilon) \leqslant\left(\mathrm{P}(\xi>\varepsilon) \mathrm{E} \xi^{2}\right)^{1 / 2} \leqslant\left(a \mathrm{E} \xi^{2}\right)^{1 / 2}
$$
and, therefore,
$$
\mathrm{E} \xi \leqslant \frac{\varepsilon}{\mathrm{E} \xi I(\xi \leqslant \varepsi... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,793 |
66. Let $\xi$ be a non-negative random variable. Show that for all $p>1$ the following equality holds:
$$
\int_{0}^{\infty} \frac{\mathrm{E}\left(\xi^{p} \wedge x^{p}\right)}{x^{p}} d x=\frac{p}{p-1} \mathrm{E} \xi
$$
In particular,
$$
\int_{0}^{\infty} \frac{\mathrm{E}\left(\xi^{2} \wedge x^{2}\right)}{x^{2}} d x=2... | Solution. By Fubini's theorem
$$
\int_{0}^{\infty} \frac{\mathrm{E}\left(\xi^{p} \wedge x^{p}\right)}{x^{p}} d x=\mathrm{E} \int_{0}^{\xi} 1 d x+\mathrm{E} \int_{\xi}^{\infty} \frac{\xi^{p}}{x^{p}} d x=\mathrm{E} \xi+\mathrm{E} \xi \cdot \int_{1}^{\infty} \frac{d x}{x^{p}}=\frac{p}{p-1} \mathrm{E} \xi
$$ | \frac{p}{p-1}\mathrm{E}\xi | Calculus | proof | Yes | Yes | olympiads | false | 33,794 |
67. Let the random variable $X$ be such that $\mathrm{E} X=0$ and $\mathrm{P}(X \leqslant 1)=1$. Prove that
$$
\frac{(\mathrm{E}|X|)^{2}}{2} \leqslant-\mathrm{E} \ln (1-X)
$$ | Solution. For all $0<\alpha<2$ we have
$$
\frac{(\mathrm{E}|X|)^{2}}{2} \leqslant \frac{\mathrm{E}(2-\alpha X)}{2} \mathrm{E} \frac{X^{2}}{2-\alpha X}=\mathrm{E}\left[X+\frac{X^{2}}{2-\alpha X}\right]
$$
Thus, the desired estimate will be proven if we choose $\alpha$ such that for all $x \leqslant 1$ the inequality
... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,795 |
68. Let $(\Omega, \mathscr{F})$ be an arbitrary measurable space. Define a set function $\mu=\mu(B), B \in \mathscr{F}$ on this space by
$$
\mu(B)= \begin{cases}|B|, & \text { if } B \text { is finite, } \\ \infty & \text { in other cases }\end{cases}
$$
where $|B|$ is the cardinality of the set $B$. Show that the se... | Solution. Obviously, $\mu(B) \geqslant 0$ for each $B \in \mathscr{F}$. Furthermore, if $B=\bigcup_{n} B_{n}$ for pairwise disjoint sets $B_{n} \in \mathscr{F}, n \geqslant 1$, then
$$
\mu(B)=|B|=\sum_{n}\left|B_{n}\right|=\sum_{n} \mu\left(B_{n}\right) .
$$ | proof | Other | proof | Yes | Yes | olympiads | false | 33,796 |
69. (On the Radon-Nikodym Theorem.) Consider on the measurable space $(\Omega, \mathscr{F})=([0,1], \mathscr{B}([0,1]))$ two measures: the Lebesgue measure $\lambda$ and the counting measure $\mu$ (see problem II.6.68). Show that $\lambda \ll \mu$, but at the same time, the statement of the Radon-Nikodym theorem about ... | Solution. By definition, $\mu(B)=0$ if and only if $B=\varnothing$ and, therefore, $\lambda(B)=0$. Thus, $\lambda \ll \mu$. Suppose there exists $f=\frac{d \lambda}{d \mu}$. Then for any $\omega \in[0,1]$, the following equalities hold
$$
0=\lambda(\{\omega\})=\int_{\{\omega\}} f d \mu=f(\omega),
$$
i.e., $f$ is iden... | proof | Other | proof | Yes | Yes | olympiads | false | 33,797 |
70. Let $\lambda$ and $\mu$ be two $\sigma$-finite measures on $(\Omega, \mathscr{F})$ and $f=\frac{d \lambda}{d \mu}$. Show that if $\mu(f=0)=0$, then the density $\frac{d \mu}{d \lambda}$ exists and can be taken as the function
$$
\varphi= \begin{cases}1 / f & \text { on the set }\{f \neq 0\} \\ c & \text { on the s... | Solution. If $\mu(f=0)=0$, then $\lambda(f=0)=0$ and, therefore, for all $B \in \mathscr{F}$ the following equalities hold
$$
\mu(B)=\int_{B} d \mu=\int_{B \backslash\{f=0\}} d \mu=\int_{B \backslash\{f=0\}} \varphi f d \mu=\int_{B \backslash\{f=0\}} \varphi d \lambda=\int_{B} \varphi d \lambda .
$$ | proof | Other | proof | Yes | Yes | olympiads | false | 33,798 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.