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18. Consider sampling without replacement. $n$ balls are chosen from an urn containing $M_{1}$ white and $M_{2}$ black balls, $M_{1}+M_{2}=M$. The event $B_{n_{1}, n_{2}}$ consists of $n_{1}$ of the selected balls being white, $n_{2}-$
black, $n_{1}+n_{2}=n$. Show that if $M \rightarrow \infty$ and $M_{1} \rightarrow \... | Solution. By analogy with the reasoning from the solution of problem I.2.21 (a), we obtain that $\mathrm{P}\left(B_{n_{1}, n_{2}}\right)=\frac{C_{M_{1}}^{n_{1}} C_{M_{2}}^{n_{2}}}{C_{M}^{n}}$. Then
$$
\begin{aligned}
\mathrm{P}\left(B_{n_{1}, n_{2}}\right)=\frac{C_{M_{1}}^{n_{1}} C_{M_{2}}^{n_{2}}}{C_{M}^{n}}=C_{n}^{n... | C_{n}^{n_{1}}p^{n_{1}}(1-p)^{n_{2}} | Combinatorics | proof | Yes | Yes | olympiads | false | 33,474 |
19. Prove that the point $m=\left(m_{1}, \ldots, m_{k}\right)$, at which the maximum value of the probability of a multinomial distribution is achieved, satisfies the inequalities $(n+1) p_{i}-1 \leqslant m_{i} \leqslant(n+k-1) p_{i}$, $1 \leqslant i \leqslant k$. | Solution. Let
$$
p\left(n_{1}, \ldots, n_{k}\right)=\frac{n!}{n_{1}!\ldots n_{k}!} p_{1}^{n_{1}} \ldots p_{k}^{n_{k}}
$$
If the maximum of $p\left(n_{1}, \ldots, n_{k}\right)$ is achieved at point $m$, then local "movements" of $m$ can only decrease $p$. For example,
$$
\frac{p\left(m_{1}, m_{2}, \ldots\right)}{p\le... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,475 |
20. One-dimensional Ising model. Let there be $n$ particles located at points $1, \ldots, n$.
Suppose each particle belongs to one of two types, with $n_{1}$ particles of the first type and $-n_{2}\left(n_{1}+n_{2}=n\right)$ of the second type. We will assume that all $n!$ arrangements of the particles are equally lik... | Solution. Each sequence of particle types of length $n$ can be uniquely associated with a sequence of letters $A, B, a, b$ of length $n-1$. The letter $A$ at the $i$-th position means that the particle at the $i$-th position is of the first type, and the particle at the $(i+1)$-th position is of the second type, and vi... | \frac{C_{n_{1}-1}^{m_{22}}C_{n_{2}-1}^{m_{11}}}{C_{n}^{n_{1}}} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,476 |
21. Let $N$ be the size of some population that needs to be estimated with "minimal means" without simply counting all elements of this set. This kind of question is of interest, for example, when estimating the number of residents in a particular country, city, etc.
In 1786, Laplace proposed the following method to e... | Solution. (a) Assuming all people are distinguishable, we consider an unordered sample of size $n$ without replacement. The total number of elementary outcomes is $C_{N}^{n}$, and we consider the outcomes to be equally probable.
Choosing $m$ elements from $M$ marked ones can be done in $C_{M}^{m}$ ways, and choosing $... | \widehat{N}=[\frac{Mn}{}] | Combinatorics | proof | Yes | Yes | olympiads | false | 33,477 |
22. In (elementary) combinatorics, binomial coefficients $C_{M}^{n} \equiv \frac{(M)_{n}}{n!}=\frac{M!}{n!(M-n)!}$ (also denoted as $\binom{M}{n}$) and the number of arrangements $(M)_{n} \equiv M(M-1) \ldots(M-n+1)$ are usually defined only for integers $n, M \in N=\{1,2, \ldots\}$. In many questions of analysis, it i... | Solution. (a) For $n \geq 0$, both sides of the equality are polynomials of degree $n$ in $x$, the values of which, according to problem I.2.2(b), coincide for $x \in \mathbb{N}, x \geqslant n$. Therefore, the polynomials are identically equal.
Identical reasoning proves points (b) and (c) using problems I.2.2 (b $^{*... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,478 |
23. From an urn containing $N$ balls, of which $n$ are white and $N-n$ are black, a selection without replacement is made ( $n \geqslant 2$ ). Consider ordered samples of size $M$. What is the probability of events $A_{i, j}$ and $A_{i, j, k}$, where $A_{i, j}$ is the event that white balls are in the $i$-th and $j$-th... | Solution. The number of elementary outcomes $(N)_{\text {m }}$. Let's calculate the number of elementary outcomes in $A_{i, j}$: on the $i$-th place, we can place $n$ white balls, and on the $j$-th place, the remaining $n-1$ balls. The other $M-2$ places are filled in $(N-2)_{M-2}$ ways. Therefore,
$$
\mathrm{P}\left(... | \frac{n(n-1)(n-2)}{N(N-1)(N-2)} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,479 |
24. Provide the formula for the probability $P_{n}$ that among thirteen cards drawn from a full deck of 52 cards, $n$ cards will be of the spades suit. | Solution. In total, there are $C_{52}^{13}$ unordered sets that can be drawn. Among them, we are interested in sets where $n$ cards are of the spades suit (which can be chosen in $C_{13}^{n}$ ways) and $13-n$ cards are of another suit (which can be chosen in $C_{39}^{13-n}$ ways). The desired probability is
$$
P_{n}=\... | P_{n}=\frac{C_{13}^{n}C_{39}^{13-n}}{C_{52}^{13}} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,480 |
25. On a circle, there are $n \geqslant 3$ points. A pair of them is chosen at random. Find the probability that the chosen points were "neighbors". | Solution. Elementary outcomes - pairs of numbers $(m, k), m \neq k$, $1 \leqslant m, k \leqslant n$. The total number of elementary outcomes is $(n)_2$, suitable ones are $-n \cdot 2$ (the first point is chosen arbitrarily, then there are only two options for the second). The desired probability is $P_{n}=2 n /(n)_2=2 ... | \frac{2}{n-1} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,481 |
26. («The Problem of Married Couples».) In how many ways can $n$ married couples ( $n \geqslant 3$ ) be seated at a round table so that men and women alternate, but spouses do not sit next to each other? | Solution. Let us assign numbers $1, \ldots, 2n$ (clockwise) to the seats around the table, and let a woman sit at seat 1 for definiteness. Let $A_k$ be the event that seats $k$ and $k+1$ are occupied by a certain married couple ($1 \leqslant k \leqslant 2n$ and seat $2n+1$ is identified with seat 1). Then the required ... | \frac{1}{n!}\sum_{k=0}^{n}(-1)^{k}(n-k)!\frac{2n}{2n-k}C_{2n-k}^{k} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,482 |
27. (Latin squares.) Consider the numbers $1,2, \ldots, n$, and from them form $n \times n$ matrices (‘squares’) with the property that each digit appears only once in each row and each column. For $n=2$, examples are the squares
| 1 | 2 |
| :--- | :--- |
| 2 | 1 |, $\quad$ | 2 | 1 |
| :--- | :--- |
| 1 | 2 |.
For $n... | Solution. Sequentially according to the number, we fill the rows of the matrix, moving from left to right. When filling the cell with number $(i, j)$, there are at least $n-i-j+2$ options (all numbers except the $(i-1)$-th number in the $j$-th column and the $(j-1)$-th number in the $i$-th row are suitable). By multipl... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,483 |
28. From a box containing $n$ white and $m$ black balls $(n \geqslant m)$, all balls are drawn one by one at random. Find the probability of the following events:
(a) there will be a moment when the number of white and black balls in the sequence of drawn balls will be equal
(b) at any moment, the number of white bal... | Solution. (a) The number of all sequences is $C_{n+m}^{n}$. If a sequence starts with a black ball, it is favorable. There are $C_{n+m-1}^{m-1}$ such sequences. Between the favorable sequences $y$, the first ball of which is white, and the sequences $x$, starting with a black ball, there is a bijection $\varphi$. It is... | \frac{n-+1}{n+1} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,484 |
1. Provide examples showing that, in general, the equalities
$$
\mathrm{P}(B \mid A)+\mathrm{P}(B \mid \bar{A})=1, \quad \mathrm{P}(B \mid A)+\mathrm{P}(\bar{B} \mid \bar{A})=1
$$
are false. | Solution. If $B=\varnothing$, then
$$
\mathrm{P}(B \mid A)+\mathrm{P}(B \mid \bar{A})=\mathrm{P}(\varnothing \mid A)+\mathrm{P}(\varnothing \mid \bar{A})=0 \neq 1
$$
If $B=A$, then
$$
\mathrm{P}(B \mid A)+\mathrm{P}(\bar{B} \mid \bar{A})=\mathrm{P}(A \mid A)+\mathrm{P}(\bar{A} \mid \bar{A})=2 \neq 1
$$ | proof | Other | math-word-problem | Yes | Yes | olympiads | false | 33,485 |
2. An urn contains $M$ balls, of which $M_{1}$ are white. A sample of size $n$ is considered. Let $B_{j}$ be the event that the ball drawn on the $j$-th step is white, and $A_{k}$ be the event that there are exactly $k$ white balls in the sample of size $n$. Show that in the classical situation, whether for sampling wi... | Solution. Let's number the balls: the first $M_{1}$ are white, the rest are black. Consider the space of elementary outcomes $\Omega=\{\omega\}$, where $\omega=\left(a_{1}, \ldots, a_{n}\right)$ is an ordered set of $n$ numbers from 1 to $M$, indicating the number of the ball drawn on the $i$-th step.
In the case of s... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,486 |
3. Let $A_{1}, \ldots, A_{n}$ be independent events, $\mathrm{P}\left(A_{i}\right)=p_{i}$.
(a) Show that the events $\widetilde{A}_{1}, \ldots, \widetilde{A}_{n}$, where $\widetilde{A}_{i}$ denotes either $A_{i}$ or the complement of $A_{i}$, are also independent.
(b) Show that
$$
\mathrm{P}\left(\bigcup_{i=1}^{n} A_... | Solution. (a) Let's prove the statement for two events. Indeed, if $A$ and $B$ are independent, then
$$
\begin{aligned}
& \mathrm{P}(\bar{A} B)=\mathrm{P}(B \backslash A B)=\mathrm{P}(B)-\mathrm{P}(A B)=\mathrm{P}(B)-\mathrm{P}(A) \mathrm{P}(B)= \\
&= \mathrm{P}(B)(1-\mathrm{P}(A))=\mathrm{P}(B) \mathrm{P}(\bar{A})
\e... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,487 |
4. Let $A$ and $B$ be independent events. In terms of $\mathrm{P}(A)$ and $\mathrm{P}(B)$, express the probabilities of the events consisting in exactly $k$, at least $k$, and at most $k$ of the events $A$ and $B$, $k=0,1,2$. | Solution. For brevity, let $a:=\mathrm{P}(A), b:=\mathrm{P}(B)$, $\xi$ - the number of events that occurred. Then
$$
\mathrm{P}(\xi \leqslant 2)=\mathrm{P}(\xi \geqslant 0)=1 \quad(\text { since for any outcome } 0 \leqslant \xi \leqslant 2) \text {. }
$$
Further,
$$
\begin{aligned}
& p_{0}:=\mathrm{P}(\xi=0)=\mathr... | p_{0}=1--b+,\;p_{1}=+b-2ab,\;p_{2}=,\;\mathrm{P}(\xi\geqslant1)=+b-,\;\mathrm{P}(\xi\leqslant1)=1- | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,488 |
5. Let event $A$ be such that it is independent of itself, i.e., $A$ and $A$ are independent. Show that then $\mathrm{P}(A)$ is equal to 0 or 1.
Show also that if events $A$ and $B$ are independent and $A \subseteq B$, then either $\mathrm{P}(A)=0$, or $\mathrm{P}(B)=1$. | Solution. Let $A$ be independent of $A$ and $x:=\mathrm{P}(A)$, then
$$
x=\mathrm{P}(A)=\mathrm{P}(A \cap A)=\mathrm{P}(A) \mathrm{P}(A)=x^{2}
$$
This equation has exactly 2 roots: 0 and 1.
Let $A$ and $B$ be independent and $A \subseteq B$, then $\mathrm{P}(A)=\mathrm{P}(A B)=\mathrm{P}(A) \mathrm{P}(B)$, i.e., $\m... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,489 |
6. Let event $A$ be such that $\mathrm{P}(A)$ is 0 or 1. Show that $A$ and any event $B$ are independent. | Solution. If $\mathrm{P}(A)=0$, then by the additivity and non-negativity of probability, $\mathrm{P}(A B) \leqslant \mathrm{P}(A)=0$, hence, $\mathrm{P}(A B)=0$, i.e.,
$$
0=\mathrm{P}(A) \mathrm{P}(B)=\mathrm{P}(A B)=0
$$
and, therefore, $A$ and $B$ are independent.
If $\mathrm{P}(A)=1$, then $\mathrm{P}(\bar{A})=0... | proof | Other | proof | Yes | Yes | olympiads | false | 33,490 |
7. Consider the electrical circuit shown in the figure. Each of the relays $A, B, C, D$ and $E$, operating independently, is either in the open (i.e., non-working) state (and thus does not allow the electrical signal to pass), or in the closed (working) state (and then the signal is passed) with probabilities $p$ and $... | Solution. Let events $A, B, C, D$ and $E$ denote the open (non-working) state of the corresponding relay, and let event $S$ consist of the signal, input at the "input," being received at the "output." Using the inclusion-exclusion principle and the independence of the relays' operation, we get
$$
\begin{gathered}
\mat... | \frac{(2q^{2}-q^{4})p}{q(1-p^{2})^{2}+pq^{2}(2-q^{2})} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,491 |
8. Let $\mathrm{P}(A+B)>0$. Show that
$$
\mathrm{P}(A \mid A+B)=\frac{\mathrm{P}(A)}{\mathrm{P}(A)+\mathrm{P}(B)}
$$ | Solution. The statement of the problem follows from the formula of conditional probability and the fact that $A \subseteq A+B$. | proof | Other | proof | Yes | Yes | olympiads | false | 33,492 |
9. Let event $A$ be independent of events $B_{n}, n \geqslant 1$, and $B_{i} \cap B_{j}=\varnothing, i \neq j$. Verify that events $A$ and $\bigcup_{n=1}^{\infty} B_{n}$ are independent. | Solution. If $B_{i} \cap B_{j}=\varnothing, i \neq j$, then $\left(B_{i} \cap A\right) \cap\left(B_{j} \cap A\right)=\varnothing, i \neq j$. Then
$$
\begin{aligned}
\mathrm{P}\left(A \cap \bigcup_{n=1}^{\infty} B_{n}\right) & =\mathrm{P}\left(\bigcup_{n=1}^{\infty}\left(A \cap B_{n}\right)\right)=\sum_{n=1}^{\infty} \... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,493 |
10. Show that if $\mathrm{P}(A \mid C)>\mathrm{P}(B \mid C)$ and $\mathrm{P}(A \mid \bar{C})>\mathrm{P}(B \mid \bar{C})$, then $\mathrm{P}(A)>\mathrm{P}(B)$. | Solution. Multiplying the inequality $\mathrm{P}(A \mid C)>\mathrm{P}(B \mid C)$ by $\mathrm{P}(C)$, we get $\mathrm{P}(A C)>\mathrm{P}(B C)$. Similarly, $\mathrm{P}(A \bar{C})>\mathrm{P}(B \bar{C})$. Adding the last two inequalities, we obtain the required statement. | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,494 |
11. Show that if $\mathrm{P}(B)>0$, then
$$
\mathrm{P}(A \mid B)=\mathrm{P}(A \mid B C) \mathrm{P}(C \mid B)+\mathrm{P}(A \mid B \bar{C}) \mathrm{P}(\bar{C} \mid B)
$$ | Solution. By the definition of conditional probability
$$
\begin{aligned}
\mathrm{P}(A \mid B)= & \frac{\mathrm{P}(A B)}{\mathrm{P}(B)}=\frac{\mathrm{P}(A B C)+\mathrm{P}(A B \bar{C})}{\mathrm{P}(B)}=\frac{\mathrm{P}(A B C)}{\mathrm{P}(B C)} \frac{\mathrm{P}(B C)}{\mathrm{P}(B)}+ \\
& +\frac{\mathrm{P}(A B \bar{C})}{\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,495 |
12. Let $X$ and $Y$ be independent binomial random variables with parameters $n$ and $p$. Show that
$$
\mathrm{P}(X=k \mid X+Y=m)=\frac{C_{n}^{k} C_{n}^{m-k}}{C_{2 n}^{m}}, \quad k=0,1, \ldots, \min (m, n)
$$ | Solution. Since $X+Y$ has a binomial distribution with parameters $2n$ and $p$, we obtain that
$$
\begin{aligned}
\mathrm{P}\{X=k \mid X+Y=m\} & =\frac{\mathrm{P}\{X=k ; Y=m-k\}}{\mathrm{P}\{X+Y=m\}}= \\
& =\frac{C_{n}^{k} p^{k} q^{n-k} \cdot C_{n}^{m-k} p^{m-k} q^{n-(m-k)}}{C_{2 n}^{m} p^{m} q^{2 n-m}}=\frac{C_{n}^{k... | \frac{C_{n}^{k}C_{n}^{-k}}{C_{2n}^{}} | Combinatorics | proof | Yes | Yes | olympiads | false | 33,496 |
13. Let $A, B, C$ be pairwise independent and equally probable events, and $A \cap B \cap C=\varnothing$. Find the maximum possible value for the probability $\mathrm{P}(A)$. | Solution. Let $\mathrm{P}(A)=p$. By the condition, $\mathrm{P}(B)=\mathrm{P}(C)=p$, $\mathrm{P}(A B)=\mathrm{P}(B C)=\mathrm{P}(A C)=p^{2}$.
Consider the probability
$$
\begin{aligned}
& \mathrm{P}(A \backslash(B \cup C))=\mathrm{P}(A \backslash(A(B \cup C)))= \\
& =\mathrm{P}(A)-\mathrm{P}(A(B \cup C))=\mathrm{P}(A)... | \frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,497 |
14. A ball was added to an urn containing one white ball - either white or black (with equal probabilities of selection). After this, one ball was randomly drawn from the urn. It turned out to be white. What is the conditional probability that the remaining ball in the urn is also white? | Solution. Let
$$
\begin{aligned}
A & =\{\text { a white ball was added to the urn }\}, \\
C & =\{\text { a white ball was drawn }\}, \\
D & =\{\text { a white ball remains }\} .
\end{aligned}
$$
Note that $A=C D$. Then
$$
\begin{aligned}
\mathrm{P}(D \mid C)=\frac{\mathrm{P}(D C)}{\mathrm{P}(C)}=\frac{\mathrm{P}(A)}... | \frac{2}{3} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,498 |
15. If events $A$ and $B$ are independent, then by definition $\mathrm{P}(A B)=\mathrm{P}(A) \mathrm{P}(B)$. Describe the cases for arbitrary $A$ and $B$ where the inequalities $\mathrm{P}(A B) \leqslant \mathrm{P}(A) \mathrm{P}(B)$ and $\mathrm{P}(A B) \geqslant \mathrm{P}(A) \mathrm{P}(B)$ hold. | Solution. If $\mathrm{P}(A)=0$ or $\mathrm{P}(B)=0$, then by problem I.3.6, events $A$ and $B$ are independent, and therefore both inequalities are satisfied.
Let $\mathrm{P}(A) \neq 0, \mathrm{P}(B) \neq 0$, then
$$
\mathrm{P}(A B) \leqslant \mathrm{P}(A) \mathrm{P}(B) \Leftrightarrow \frac{\mathrm{P}(A B)}{\mathrm{... | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 33,499 |
2. Prove the validity of the following inclusion-exclusion formula for indicators of three events $A, B$, and $C$:
$$
I_{A \cup B \cup C}=I_{A}+I_{B}+I_{C}-\left(I_{A \cap B}+I_{A \cap C}+I_{B \cap C}\right)+I_{A \cap B \cap C} .
$$
Represent the indicator $I_{A_{1} \cup . . . \cup A_{n}}$ of the union of events $A_... | Solution. We have
$$
\begin{aligned}
I_{A \cup B \cup C} & =I_{A \cup B}+I_{C}-I_{(A \cup B) \cap C}=I_{A}+I_{B}-I_{A \cap B}+I_{C}-I_{(A \cap C) \cup(B \cap C)}= \\
& =I_{A}+I_{B}+I_{C}-I_{A \cap B}-\left(I_{A \cap C}+I_{B \cap C}-I_{(A \cap C) \cap(B \cap C)}\right)= \\
& =I_{A}+I_{B}+I_{C}-\left(I_{A \cap B}+I_{A \... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,501 |
3. Let $\xi_{1}, \ldots, \xi_{n}$ be independent Bernoulli random variables,
$$
\mathrm{P}\left\{\xi_{i}=0\right\}=1-\lambda_{i} \Delta, \quad \mathrm{P}\left\{\xi_{i}=1\right\}=\lambda_{i} \Delta
$$
where $\Delta$ is a small number, $\Delta>0, \lambda_{i}>0$. Show that
$$
\begin{aligned}
& \mathrm{P}\left\{\xi_{1}+... | Solution. Using the independence of random variables $\xi_{1}, \ldots, \xi_{n}$, we obtain
$$
\begin{aligned}
\mathrm{P}\left\{\xi_{1}+\ldots+\xi_{n}=1\right\} & =\sum_{i=1}^{n} \mathrm{P}\left\{\xi_{1}=0, \ldots, \xi_{i}=1, \ldots, \xi_{n}=0\right\}= \\
& =\sum_{i=1}^{n} \mathrm{P}\left\{\xi_{1}=0\right\} \ldots \mat... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,502 |
4. Show that $\inf _{-\infty<a<\infty} \mathrm{E}(\xi-a)^{2}$ is achieved at $a=\mathrm{E} \xi$ and, consequently,
$$
\inf _{-\infty<a<\infty} \mathrm{E}(\xi-a)^{2}=\mathrm{D} \xi
$$ | Solution. Consider the difference
$$
\begin{aligned}
\mathrm{E}(\xi-a)^{2}-\mathrm{D} \xi=\left(\mathrm{E} \xi^{2}-2 a \mathrm{E} \xi+a^{2}\right) & -\left(\mathrm{E} \xi^{2}-(\mathrm{E} \xi)^{2}\right)= \\
& =(\mathrm{E} \xi)^{2}-2 a \mathrm{E} \xi+a^{2}=(\mathrm{E} \xi-a)^{2} \geqslant 0
\end{aligned}
$$
Equality i... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,503 |
5. Let $\xi$ be a random variable with distribution function $F_{\xi}(x) = \mathrm{P}\{\xi \leqslant x\}$ and $\mu=\mu(\xi)$ (or $\mu=\mu\left(F_{\xi}\right)$) - the median of the random variable $F_{\xi}(x)$, i.e., such a point that
$$
F_{\xi}(\mu-) \leqslant \frac{1}{2} \leqslant F_{\xi}(\mu)
$$
(For other ways to ... | Solution. Assuming that $\mu=0$ (otherwise, we make the substitution $\widetilde{\xi}=\xi-\mu)$, we find that for $a \geqslant 0$ the equality holds
$$
\mathrm{E}|\xi-a|=\mathrm{E}|\xi|+\mathrm{E} f(\xi)
$$
where
$$
f(x)= \begin{cases}a, & x \leqslant 0 \\ a-2 x, & 0 < x
\end{cases}
$$
Since $\mathrm{E} f(\xi) I_{\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,504 |
6. Let $P_{\xi}(x)=\mathrm{P}\{\xi=x\}$ and $F_{\xi}(x)=\mathrm{P}\{\xi \leqslant x\}$. Show that for $a>0$ and $-\infty<b<\infty$ the following equalities hold:
$$
\begin{aligned}
& P_{a \xi+b}(x)=P_{\xi}\left(\frac{x-b}{a}\right), \\
& F_{a \xi+b}(x)=F_{\xi}\left(\frac{x-b}{a}\right) .
\end{aligned}
$$
Show also th... | Solution. If $a>0$ and $-\infty<b<+\infty$, then
$$
\begin{aligned}
& P_{a \xi+b}(x)=\mathrm{P}\{a \xi+b=x\}=\mathrm{P}\left\{\xi=\frac{x-b}{a}\right\}=P_{\xi}\left(\frac{x-b}{a}\right), \\
& F_{a \xi+b}(x)=\mathrm{P}\{a \xi+b \leqslant x\}=\mathrm{P}\left\{\xi \leqslant \frac{x-b}{a}\right\}=F_{\xi}\left(\frac{x-b}{a... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,505 |
7. Let $\xi$ and $\eta$ be two random variables, $\mathrm{D} \xi>0, \mathrm{D} \eta>0$ and $\rho=$ $\rho(\xi, \eta)$ be their correlation coefficient. Show that $|\rho| \leqslant 1$. Moreover, if $|\rho|=1$, then there exist constants $a$ and $b$ such that $\eta=a \xi+b$. Furthermore, if $\rho=1$, then
$$
\frac{\eta-\... | Solution. Let
$$
\widetilde{\xi}=\frac{\xi-\mathrm{E} \xi}{\sqrt{D \xi}}, \quad \widetilde{\eta}=\frac{\eta-\mathrm{E} \eta}{\sqrt{\mathrm{D} \eta}}
$$
Then $\mathbf{E} \widetilde{\xi}=\mathbf{E} \widetilde{\eta}=0, \mathbf{D} \widetilde{\xi}=\mathbf{E} \widetilde{\xi}^{2}=\mathbf{D} \widetilde{\eta}=\mathbf{E} \wide... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,506 |
8. Let $\xi$ and $\eta$ be two random variables, $\mathrm{E} \xi=\mathrm{E} \eta=0, \mathrm{D} \xi=\mathrm{D} \eta=1$ with correlation coefficient $\rho=\rho(\xi, \eta)$. Show that
$$
\mathrm{E} \max \left(\xi^{2}, \eta^{2}\right) \leqslant 1+\sqrt{1-\rho^{2}}
$$ | Solution. Note that for any values of $\xi$ and $\eta$, the equality holds:
$$
\max \left(\xi^{2}, \eta^{2}\right)=\frac{1}{2}\left(\xi^{2}+\eta^{2}+\left|\xi^{2}-\eta^{2}\right|\right)
$$
Since $\mathrm{E} \xi^{2}=\mathrm{E} \eta^{2}=1$, to solve the problem, it remains to show that
$$
\mathrm{E}\left|\xi^{2}-\eta^... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,507 |
9. Let $L(x, y)$ be a loss function, i.e., $L(x, y) \geqslant 0$, and $L(x, y)=0$ only when $x=y$, and $\xi$ be a random variable with distribution function $F$. Suppose the minimum of the function $\mathrm{E} L(\xi, a)$ is achieved at the point $a^{*}$. Show that
(a) $a^{*}=\mathrm{E} \xi$, if $L(x, y)=(x-y)^{2}$;
(... | Solution. (a) The statement follows from the equality $\mathrm{E}(\xi-a)^{2}=\mathrm{D} \xi+$ $+(\mathrm{E} \xi-a)^{2}$.
(b) The statement follows from part (c).
(c) Let, for definiteness, the $q$-th quantile be 0, and $a>0$. We have
$$
q \mathrm{E}(\xi-a)^{+}+(1-q) \mathrm{E}(\xi-a)^{-}=q \mathrm{E} \xi^{+}+(1-q) \... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,508 |
10. Prove the inequality
$$
|\mu-E \xi| \leqslant \sqrt{D \xi}
$$
where $\mu$ is the median of the random variable $\xi$. In which cases is equality achieved? | Solution. By Lyapunov's inequality, we have
$$
|\mathrm{E} \xi-\mu| \leqslant \mathrm{E}|\xi-\mu| \leqslant \mathrm{E}|\xi-\mathrm{E} \xi| \leqslant \sqrt{\mathrm{D} \xi}
$$
Equality is achieved if and only if $\xi$ is either a constant or a Bernoulli random variable with $p=q=1 / 2$, and $\mu$ is its minimum value. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,509 |
11. Prove the inequality
$$
|\mu-a|^{p} \leqslant 2 \mathrm{E}|\xi-a|^{p}
$$
where $a \in \mathbb{R}, p \geqslant 1$ such that $\mathrm{E}|\xi|^{p}<\infty$, and $\mu$ is the median of the random variable $\xi$. | Solution. Let $p=1$. Then, according to problem I.4.9(b), we have
$$
|\mu-a| \leqslant \mathrm{E}|\xi-\mu|+\mathrm{E}|\xi-a| \leqslant 2 \mathrm{E}|\xi-a|
$$
Now let $p>1$. For $\mu \geqslant a$, by Hölder's inequality, we have
$$
|\mu-a| \mathbf{P}(\xi \geqslant \mu) \leqslant \mathrm{E}|\xi-a| I(\xi \geqslant \mu)... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,510 |
12. Using the property
$$
l_{U_{i=1}^{n} A_{i}}=1-\prod_{i=1}^{n}\left(1-I_{A_{i}}\right)
$$
of indicators from problem I.4.1, prove the inclusion-exclusion formula for the probability of the union of events:
$$
\begin{aligned}
& \mathrm{P}\left(A_{1} \cup \ldots \cup A_{n}\right)=\sum_{1 \leqslant i_{1} \leqslant n... | Solution. Let's take the expectation of both sides of the equality $\sum_{i=1}^{n} I_{A_{i}}=1-\prod_{i=1}^{n}\left(1-I_{A_{i}}\right)$. Then the left side of the equality is
$$
\mathrm{E}_{\bigcup_{i=1}^{n} A_{i}}=\mathrm{P}\left(\bigcup_{i=1}^{n} A_{i}\right)
$$
and the right side, due to the linearity of expectati... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,511 |
13. If $\xi_{1}, \ldots, \xi_{n}$ are independent random variables, $\varphi_{1}=\varphi_{1}\left(\xi_{1}, \ldots\right.$ $\left.\ldots, \xi_{k}\right)$ and $\varphi_{2}=\varphi_{2}\left(\xi_{k+1}, \ldots, \xi_{n}\right)$ are two functions of $\xi_{1}, \ldots, \xi_{k}$ and $\xi_{k+1}, \ldots, \xi_{n}$ respectively, the... | Solution. Let $Y_{i}$ be the set of values that the random variable $\xi_{i}$ takes with positive probabilities, $i=1, \ldots, n$. Fix arbitrary $z_{1}$ and $z_{2}$. Let
$$
\begin{aligned}
& J_{z_{1}}^{1}=\left\{\left(y_{1}, \ldots, y_{k}\right): \varphi_{1}\left(y_{1}, \ldots, y_{k}\right)=z_{1}\right\} \\
& J_{z_{2}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,512 |
14. Show that random variables $\xi_{1}, \ldots, \xi_{n}$ are independent if and only if for all $x_{1}, \ldots, x_{n}$ the equality
$$
F_{\xi_{1}, \ldots, \xi_{n}}\left(x_{1}, \ldots, x_{n}\right)=F_{\xi_{1}}\left(x_{1}\right) \ldots F_{\xi_{n}}\left(x_{n}\right)
$$
holds, where
$$
F_{\xi_{1}, \ldots, \xi_{n}}\left... | Solution. The necessity is evident from the definition of independent random variables. Let's prove the sufficiency. Fix arbitrary $y_{1}, \ldots, y_{n}$. Introduce the difference operator $\Delta_{y_{i}}: \mathbb{R}^{n} \rightarrow \mathbb{R}$, acting by the formula
$$
\begin{aligned}
& \Delta_{y_{i}} F_{\xi_{1}, \ld... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,513 |
15. Show that a random variable $\xi$ is independent of itself (i.e., $\xi$ and $\xi$ are independent) if and only if $\xi(\omega)=\mathrm{const}$ almost surely ${ }^{2}$ for $\omega \in \Omega$. | Solution. Sufficiency. Let $\xi(\omega)=c$ a.s., $\omega \in \Omega$, where $c$ is a constant. For any $c_{1}$ and $c_{2}$, both $\mathrm{P}\left\{\xi=c_{1}, \xi=c_{2}\right\}$ and $\mathrm{P}\left\{\xi=c_{1}\right\} \mathrm{P}\left\{\xi=c_{2}\right\}$ are equal to 1 when $c=c_{1}=c_{2}$ and 0 in all other cases. There... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,514 |
16. Under what conditions on $\xi$ are the random variables $\xi$ and $\sin \xi$ independent? | Solution. Let $y_{1}, \ldots, y_{n}$ be all the values ${ }^{3}$ that $\xi$ takes with positive probability. Then for any $i$ and $j, 1 \leqslant i0
\end{aligned}
$$
This means, $\sin y_{i}=\sin y_{j}$. Due to the arbitrariness of $i$ and $j$, we get $\sin y_{1}=\ldots=\sin y_{n}$, i.e.,
$$
\sin \xi=c \text { a.s., }... | \sin\xi= | Other | math-word-problem | Yes | Yes | olympiads | false | 33,515 |
17. Let $\xi$ and $\eta$ be independent random variables and $\eta \neq 0$. Express the probabilities $\mathrm{P}\{\xi \eta \leqslant z\}$ and $\mathrm{P}\left\{\frac{\xi}{\eta} \leqslant z\right\}$ in terms of the functions $F_{\xi}(x)$ and $F_{\eta}(y)$. | Solution. Let the random variable $\eta$ take values $x_{1}, \ldots, \ldots, x_{n}$, where $x_{1}, \ldots, x_{p}$ are negative, and $x_{p+1}, \ldots, x_{n}$ are positive. Then
$$
\begin{aligned}
F_{\xi \eta}(z)= & \mathrm{P}\{\xi \eta \leqslant z\}=\sum_{k=1}^{n} \mathrm{P}\left\{\eta=x_{k}, \xi x_{k} \leqslant z\righ... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,516 |
18. Let $\xi, \eta, \zeta$ be random variables such that $|\xi| \leqslant 1,|\eta| \leqslant 1$, $|\zeta| \leqslant 1$. Prove the validity of Bell's inequality: $|\mathrm{E} \zeta \zeta - \mathrm{E} \eta \zeta| \leqslant$ $\leqslant 1 - \mathbf{E} \xi \eta$. | Solution. Due to the constraints on the random variables, the inequality $\xi(1+\eta) \leqslant 1+\eta$ holds, or equivalently, $\xi-\eta \leqslant 1-\xi \eta$. Similarly, we obtain $\eta-\xi \leqslant 1-\xi \eta$, i.e., $|\xi-\eta| \leqslant 1-\xi \eta$. Since $|\zeta| \leqslant 1$, we have $|\xi \zeta-\eta \zeta| \le... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,517 |
19. Let $\xi, \eta, X, Y$ be random variables bounded in absolute value by one. Prove the Clauser-Horne-Shimony-Holt inequality
$$
|\mathrm{E} \xi X+\mathrm{E} \xi Y+\mathrm{E} \eta X-\mathrm{E} \eta Y| \leqslant 2 .
$$
Under what condition is equality achieved? | Solution. We have
$$
\begin{aligned}
&|\mathrm{E} \xi X+\mathrm{E} \xi Y+\mathrm{E} \eta X-\mathrm{E} \eta Y| \leqslant \\
& \leqslant|\mathrm{E} \xi(X+Y)|+|\mathrm{E} \eta(X-Y)| \leqslant \mathrm{E}|X+Y|+\mathrm{E}|X-Y|
\end{aligned}
$$
and equality is achieved if (with probability one)
$$
\xi=\left\{\begin{array}{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,518 |
20. In $n$ urns, $k$ balls are thrown independently. (For each ball, the probability of it landing in any specific urn is $1 / n$.) Find the expected number of non-empty urns. | Solution. Let $\xi$ be the number of non-empty urns. Then $\xi=\sum_{i=1}^{n} I_{A_{i}}$, where $A_{i}=\{i-$ th urn is non-empty $\}$. Note that
$$
\mathrm{P}\left(\bar{A}_{i}\right)=\mathrm{P}\{\text { no ball falls into the } i \text {-th urn }\}=\left(\frac{n-1}{n}\right)^{k} .
$$
Then
$$
\mathrm{E} \xi=\sum_{i=1... | n(1-(\frac{n-1}{n})^{k}) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,519 |
21. (Reformulation of problem I.3.12.) Let $\xi_{1}, \ldots, \xi_{n}$ be i.i.d. ${ }^{4}$ random variables, $\mathrm{P}\left\{\xi_{1}=1\right\}=p, \mathrm{P}\left\{\xi_{1}=0\right\}=1-p$, where $0<p<1$. Let $S_{k}=\xi_{1}+\ldots+\xi_{k}, k \leqslant n$. Show that for $1 \leqslant m \leqslant n$ the following equality h... | Solution. Note that $S_{n}, S_{m}$, and $S_{n}-S_{m}$ have a binomial distribution with parameters $p$ and respectively $n, m$, and $n-m$. Moreover, $S_{m}$ and $S_{n}-S_{m}$ are independent. Using this, we obtain
$$
\begin{aligned}
\mathrm{P}\left\{S_{m}=k \mid S_{n}=l\right\} & =\frac{\mathrm{P}\left\{S_{m}=k ; S_{n... | \frac{C_{}^{k}C_{n-}^{-k}}{C_{n}^{}} | Combinatorics | proof | Yes | Yes | olympiads | false | 33,520 |
22. Let $\xi_{1}, \ldots, \xi_{n}$ be independent random variables and
$$
\xi_{\min }=\min \left(\xi_{1}, \ldots, \xi_{n}\right), \quad \xi_{\max }=\max \left(\xi_{1}, \ldots, \xi_{n}\right)
$$[^3]
Show that
$$
\mathrm{P}\left\{\xi_{\text {min }} \geqslant x\right\}=\prod_{i=1}^{n} \mathrm{P}\left\{\xi_{i} \geqslant... | Solution. For any set of random variables $\xi_{1}, \ldots, \xi_{n}$, the following equalities hold:
$$
\left\{\xi_{\min } \geqslant x\right\}=\bigcap_{i=1}^{n}\{\xi \geqslant x\} \quad \text { and } \quad\left\{\xi_{\max }<x\right\}=\bigcap_{i=1}^{n}\left\{\xi_{i}<x\right\} .
$$
The statement of the problem now dire... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,521 |
23. Let $\xi_{1}, \ldots, \xi_{n}$ be i.i.d. random variables, $\mathrm{P}\left\{\xi_{1}=1\right\}=\mathrm{P}\left\{\xi_{1}=\right.$ $=-1\}=1 / 2$. Let $S_{2 n}=\xi_{1}+\ldots+\xi_{2 n}, M_{2 n}=\max \left(S_{1}, \ldots, S_{2 n}\right)$. Show that for $k \leqslant n$ the following equality holds:
$$
\mathrm{P}\left\{M... | Solution. Let $A_{i}=\left\{S_{i}=k, S_{j}<k, 1 \leqslant j<i\right\}$. Then
$$
\begin{aligned}
& \mathrm{P}\left\{M_{2 n} \geqslant k, S_{2 n}=0\right\}= \\
& \quad=\sum_{i=1}^{n} \mathrm{P}\left(A_{i} \cap\left\{S_{2 n}=0\right\}\right)=\sum_{i=1}^{n} \mathrm{P}\left(A_{i} \cap\left\{\xi_{i+1}+\ldots+\xi_{n}=-k\righ... | \frac{1}{2}[\frac{1}{\mathrm{P}{S_{2n}=0}}-1] | Combinatorics | proof | Yes | Yes | olympiads | false | 33,522 |
24. Provide an example of two random variables $\xi$ and $\eta$, having the same distribution function ( $F_{\xi}=F_{\eta}$ ), but such that $\mathrm{P}\{\xi \neq \eta\}>0$. | Solution. Consider the sample space $\Omega=$ $=\left\{\omega_{1}, \omega_{2}\right\}, \mathrm{P}\left(\omega_{1}\right)=\mathrm{P}\left(\omega_{2}\right)=1 / 2$. Define the random variables pointwise: $\xi\left(\omega_{1}\right)=\eta\left(\omega_{2}\right)=1, \xi\left(\omega_{2}\right)=\eta\left(\omega_{1}\right)=-1$.... | |\xi-\eta|\equiv2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,523 |
25. Let $\xi, \eta$ and $\zeta$ be random variables, and the distribution functions of $\xi$ and $\eta$ coincide. Is it true that then the distribution functions of $\xi \zeta$ and $\eta \zeta$ coincide? | Solution. Answer: no. Consider the example given in problem I.4.24. Let $\zeta:=\xi$, then $\xi \zeta=\xi^{2} \equiv 1, \eta \zeta \equiv-1$, and the distribution functions of these random variables do not coincide. | notfound | Algebra | proof | Yes | Yes | olympiads | false | 33,524 |
26. Provide an example of two dependent random variables $\xi$ and $\eta$ such that $\xi^{2}$ and $\eta^{2}$ are independent. | Solution. Consider the example from problem I.4.24. We have $\xi^{2}=\eta^{2} \equiv 1$. By problem I.4.15, the quantity $\xi^{2}$ is independent of $\eta^{2}$. At the same time,
$$
\mathrm{P}(\xi=1, \eta=-1)=1 / 2 \neq 1 / 4=\mathrm{P}(\xi=1) \mathrm{P}(\eta=-1)
$$ | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,525 |
27. Let $\xi$ be a discrete random variable. Consider the following three ways of defining the median $\mu=\mu(\xi)$ of this random variable (cf. I.4.5):
(a) $\max (\mathrm{P}\{\xi>\mu\}, \mathrm{P}\{\xi<\mu\}) \leqslant 1 / 2$;
(b) $\mathrm{P}\{\xi<\mu\} \leqslant 1 / 2 \leqslant \mathrm{P}\{\xi \leqslant \mu\}$;
(... | Solution. We will show that $M_{a}=M_{b}$. Indeed, $\mu \in M_{a}$ if and only if
$$
\left\{\begin{array} { l }
{ \mathrm { P } ( \xi \mu ) \leqslant 1 / 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\mathrm{P}(\xi<\mu) \leqslant 1 / 2, \\
\mathrm{P}(\xi \leqslant \mu) \leqslant 1 / 2
\end{array} \Leftrigh... | proof | Other | math-word-problem | Yes | Yes | olympiads | false | 33,526 |
28. Let an urn contain $N$ balls, of which $a$ are white, $b$ are black, and $c$ are red, with $a+b+c=N$. Suppose $n$ balls are drawn, and let $\xi$ be the number of white balls and $\eta$ be the number of black balls among them. Show that if the selection is "with replacement," then
$$
\operatorname{cov}(\xi, \eta)=-... | Solution. (a) Sampling with replacement. Introduce random variables $\xi_{1}, \ldots, \xi_{n}$, where $\xi_{i}$ takes the value 1 if a white ball is drawn on the $i$-th step, and 0 otherwise, and $\eta_{1}, \ldots, \eta_{n}$, where $\eta_{i}$ takes the value 1 if a black ball is drawn on the $i$-th step, and 0 otherwis... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,527 |
1. Let $\xi$ and $\eta$ be random variables with correlation coefficient $\rho$. Show the validity of the following two-dimensional analogue of Chebyshev's inequality:
$$
\mathrm{P}\{|\xi-\mathrm{E} \xi| \geqslant \varepsilon \sqrt{\mathrm{D} \xi} \text { or }|\eta-\mathrm{E} \eta| \geqslant \varepsilon \sqrt{\mathrm{... | Solution. Consider the random variables $\widetilde{\xi}:=\frac{\xi-\mathrm{E} \xi}{\sqrt{D \xi}}$ and $\widetilde{\eta}:=\frac{\eta-E \eta}{\sqrt{D \eta}}$. Then $\mathbf{E} \widetilde{\xi}=\mathbf{E} \widetilde{\eta}=0, \mathbf{D} \widetilde{\xi}=\mathbf{D} \widetilde{\eta}=1$ and
$$
\begin{aligned}
& \mathrm{P}\{|\... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,528 |
2. Let $\xi=\left(\xi_{1}, \ldots, \xi_{n}\right)$ be a random vector in $\mathbb{R}^{n}$ with a non-degenerate covariance matrix $C=\mathrm{E}(\xi-\mathrm{E} \xi)(\xi-\mathrm{E} \xi)^{\top}$. Prove the following analogue of Chebyshev's inequality:
$$
\mathrm{P}\left((\xi-\mathrm{E} \xi)^{\top} C^{-1}(\xi-\mathrm{E} \... | Solution. Let $\mathrm{E} \xi=0$ for simplicity. Since the covariance matrix is non-negatively defined, $\xi^{\top} C^{-1} \xi \geqslant 0$ and Chebyshev's inequality can be applied:
$$
\begin{aligned}
\mathrm{P}\left(\xi^{\top} C^{-1} \xi>\varepsilon\right) & \leqslant \varepsilon^{-1} \mathrm{E} \xi^{\top} C^{-1} \x... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,529 |
3. Let $f=f(x)$ be a non-negative even function, non-decreasing for positive values of $x$. Show that for a random variable $\xi=\xi(\omega)$ with $|\xi(\omega)| \leqslant C$, where $C>0$, the following lower bound estimate holds:
$$
\mathrm{P}\{|\xi| \geqslant \varepsilon\} \geqslant \frac{\mathrm{E} f(\xi)-f(\vareps... | Solution. Due to the evenness of the function $f=f(x)$, we have $\mathrm{E} f(\xi)=\mathrm{E} f(|\xi|)$. Let $A=\{\omega:|\xi(\omega)| \geqslant \varepsilon\}$. By the properties of mathematical expectation and the properties of the function $f$, we have
$$
\mathrm{E} f(|\xi|)=\mathrm{E} f(|\xi|) I_{A}+\mathrm{E} f(|\... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,530 |
4. Let $\xi_{1}, \ldots, \xi_{n}$ be a sequence of independent random variables, $\mathrm{D} \xi_{i} \leqslant C$. Show that then
$$
\mathrm{P}\left\{\left|\frac{\xi_{1}+\ldots+\xi_{n}}{n}-\frac{\mathrm{E}\left(\xi_{1}+\ldots+\xi_{n}\right)}{n}\right| \geqslant \varepsilon\right\} \leqslant \frac{C}{n \varepsilon^{2}}... | Solution. Let $S_{n}=\xi_{1}+\ldots+\xi_{n}$. By the corollary of Chebyshev's inequality (see formula (4) in V1.I.5) and properties of variance,
$$
\mathrm{P}\left\{\left|\frac{S_{n}-\mathrm{E} S_{n}}{n}\right| \geqslant \varepsilon\right\} \leqslant \frac{\mathrm{D} \frac{S_{n}-\mathrm{E} S_{n}}{n}}{\varepsilon^{2}}=... | \frac{C}{n\varepsilon^{2}} | Inequalities | proof | Yes | Yes | olympiads | false | 33,531 |
5. Let $\xi_{1}, \ldots, \xi_{n}$ be independent Bernoulli random variables, $\mathrm{P}\left\{\xi_{i}=1\right\}=p>0, \mathrm{P}\left\{\xi_{i}=-1\right\}=1-p$. Show that the following Bernstein's inequality holds: there exists $a>0$ such that
$$
\mathrm{P}\left\{\left|\frac{S_{n}}{n}-(2 p-1)\right| \geqslant \varepsil... | Solution. Let's transform the probability of interest:
$$
\mathrm{P}\left\{\left|\frac{S_{n}}{n}-(2 p-1)\right| \geqslant \varepsilon\right\}=\mathrm{P}\left\{\left|\frac{S_{n}+n}{2 n}-p\right| \geqslant \frac{\varepsilon}{2}\right\}
$$
Notice that $\left(S_{n}+n\right) / 2=\widetilde{S}_{n}=\widetilde{\xi}_{1}+\ldot... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,532 |
6. Let $\xi$ be a non-negative random variable and $a>0$. Find the maximum possible value for the probability $\mathrm{P}\{\xi \geqslant a\}$ in each of the following cases, when it is known that
(a) $\mathrm{E} \xi=m$;
(b) $\mathrm{E} \xi=m, \mathrm{D} \xi=\sigma^{2}$. | Solution. (a) In the case $a \leqslant m$, the maximum value of $\mathrm{P}\{\xi \geqslant a\}$ is 1, when $\mathrm{P}\{\xi=m\}=1$. In the case $a>m$, we use Chebyshev's inequality $\mathrm{P}\{\xi \geqslant a\} \leqslant \mathrm{E} \xi / a$. Equality is achieved when $\mathrm{P}\{\xi=a\}=\frac{m}{a}$ and $\mathrm{P}\{... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 33,533 |
7. Let $S_{0}=0, S_{n}=\xi_{1}+\ldots+\xi_{n}, n \leqslant N$, where $\xi_{1}, \ldots, \xi_{n}$ are independent Bernoulli random variables, $\mathrm{P}\left\{\xi_{n}=1\right\}=p>0$, $\mathrm{P}\left\{\xi_{n}=0\right\}=q, n \leqslant N$. Denote $P_{n}(k)=\mathrm{P}\left\{S_{n}=k\right\}$. Show that for $n<N$ and $k \geq... | Solution. By the formula of total probability
$$
\begin{aligned}
& P_{n+1}(k)=\mathrm{P}\left(S_{n+1}=k \mid \xi_{n+1}=1\right) \mathrm{P}\left\{\xi_{n+1}=1\right\}+ \\
& \quad+\mathrm{P}\left\{S_{n+1}=k \mid \xi_{n+1}=0\right\} \mathrm{P}\left\{\xi_{n+1}=0\right\}= \\
& =\mathrm{P}\left\{S_{n}=k-1\right\} \mathrm{P}\... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,534 |
8. Let $\xi_{1}, \ldots, \xi_{N}$ be independent Bernoulli random variables, $\mathrm{P}\left\{\xi_{i}=1\right\}=\mathrm{P}\left\{\xi_{i}=-1\right\}=1 / 2, i=1, \ldots, N$. Let $S_{m}=\xi_{1}+\ldots$ $\ldots+\xi_{m}$. Show that for $2 m \leqslant N$ the following equality holds:
$$
\mathrm{P}\left\{S_{1} \ldots S_{2 m... | Solution. The value $S_{i}$ will never turn to zero for $i=1, \ldots, n$, if they are all either positive or negative, i.e.,
$$
\left\{S_{1} \cdot \ldots \cdot S_{2 m} \neq 0\right\}=\left\{\min _{1 \leqslant i \leqslant 2 m} S_{i} \geqslant 1\right\} \cup\left\{\max _{1 \leqslant i \leqslant 2 m} S_{i} \leqslant-1\ri... | 2^{-2}C_{2}^{} | Combinatorics | proof | Yes | Yes | olympiads | false | 33,535 |
9. Consider $M$ cells, numbered $1, \ldots, M$. Suppose that in the cell with number $n$ there is one white ball and $n$ black ones. A "random" selection of balls is made from each of the $M$ cells. Let
$$
\xi_{n}= \begin{cases}1, & \text { if a white ball is drawn from the cell with number } n, \\ 0, & \text { if a b... | Solution. By Chebyshev's inequality
$$
\mathrm{P}\left\{\left|\frac{S_{M}}{\ln M}-1\right| \geqslant \varepsilon\right\} \leqslant \frac{\mathrm{E}\left(\frac{S_{M}}{\ln M}-1\right)^{2}}{\varepsilon^{2}}
$$
To prove the statement of the problem, it is sufficient to show that
$$
\mathrm{E}\left(S_{M}-\ln M\right)^{2}... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,536 |
10. Let $\xi_{1}, \ldots, \xi_{n}$ be independent Bernoulli random variables such that $\mathrm{P}\left\{\xi_{k}=1\right\}=p_{k}, \mathrm{P}\left\{\xi_{k}=0\right\}=1-p_{k}, 1 \leqslant k \leqslant n$, and $\frac{1}{n} \sum_{k=1}^{n} p_{k}=a$. Show that for a given $a, 0<a<1$, the variance $\mathrm{D} S_{n}$ of the qua... | Solution. Note that
$$
\mathrm{D} S_{n}=\sum_{k=1}^{n} \mathrm{D} \xi_{k}=\sum_{k=1}^{n} p_{k}-p_{k}^{2}=n a-\sum_{k=1}^{n} p_{k}^{2}
$$
To solve the problem, it is necessary to show that $\sum_{k=1}^{n} p_{k}^{2}$ reaches its minimum value when $p_{1}=p_{2}=\ldots=p_{n}=a$.
For any concave function $f(\cdot)$, the ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,537 |
11. Let $\xi_{1}, \ldots, \xi_{n}$ be independent Bernoulli random variables,
$$
\mathbf{P}\left\{\xi_{k}=1\right\}=p, \quad \mathbf{P}\left\{\xi_{k}=0\right\}=1-p, \quad 1 \leqslant k \leqslant n
$$
Find the conditional probability that the first one («success») appears on the $m$-th step, given that the «success» o... | Solution. Let $S_{n}:=\xi_{1}+\ldots+\xi_{n}$. The quantity $S_{n}$ has a binomial distribution with parameters $n$ and $p$. By the definition of conditional probability,
$$
\begin{aligned}
& \mathrm{P}\left\{\xi_{m}=1 \mid S_{n}=1\right\}= \\
& =\frac{\mathrm{P}\left\{\xi_{1}=\ldots=\xi_{m-1}=\xi_{m+1}=\ldots=\xi_{n}... | \frac{1}{n} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,538 |
12. Let $\left(p_{1}, \ldots, p_{r}\right)$ and $\left(q_{1}, \ldots, q_{r}\right)$ be two probability distributions. Show that the following Gibbs' inequality holds:
$$
-\sum_{i=1}^{r} p_{i} \ln p_{i} \leqslant-\sum_{i=1}^{r} p_{i} \ln q_{i}
$$ | Solution. Note that by the properties of the logarithmic function, the inequality to be proved is equivalent to the inequality
$$
\sum_{i=1}^{r} p_{i} \ln \frac{q_{i}}{p_{i}} \leqslant 0
$$
We use the inequality $\ln x \leqslant x-1$, which holds for any positive $x$. Then
$$
\sum_{i=1}^{r} p_{i} \ln \frac{q_{i}}{p_... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,539 |
1. Consider the random variables from problem I.5.11. Let $p=1 / 2$ and $Z_{n}=2 S_{n}-n$ (the number of excess units over zeros in $n$ trials). Show that
$$
\sup _{j}\left|\sqrt{\pi n} \mathrm{P}\left\{Z_{2 n}=j\right\}-e^{-j^{2} /(4 n)}\right| \rightarrow 0, \quad n \rightarrow \infty
$$ | Solution. Let $2 n=m$ and $k=j / 2+n$. Then the proof of the required statement reduces to proving that
$$
\sup _{k}\left|\sqrt{\frac{\pi m}{2}} \mathrm{P}\left\{S_{m}=k\right\}-e^{-\frac{(k-m p)^{2}}{2 m p q}}\right|\left(\equiv \sup _{k} \varepsilon(k, m)\right) \rightarrow 0, \quad m \rightarrow \infty
$$
With thi... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,540 |
2. Prove that in the Poisson theorem (with $p=\lambda / n$) the following estimate holds:
$$
\sup _{k}\left|P_{n}(k)-\frac{\lambda^{k} e^{-\lambda}}{k!}\right| \leqslant \frac{\lambda^{2}}{n}
$$ | Solution. Let $\eta_{1}, \ldots, \eta_{n}$ and $\zeta_{1}, \ldots, \zeta_{n}$ be independent random variables, having respectively Poisson distribution with parameter $\lambda / n$ and Bernoulli distribution with
$$
\mathrm{P}\left\{\zeta_{i}=0\right\}=e^{\lambda / n}(1-\lambda / n), \quad \mathrm{P}\left\{\zeta_{i}=1... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,541 |
3. Let $\xi_{1}, \ldots, \xi_{n}$ be independent and identically distributed random variables, $\mathrm{P}\left\{\xi_{k}=1\right\}=$ $\mathrm{P}\left\{\xi_{k}=-1\right\}=1 / 2$ (symmetric Bernoulli scheme), $S_{n}=\xi_{1}+\ldots+\xi_{n}$ and $P_{n}(k)=\mathrm{P}\left\{S_{n}=k\right\}$, where $k \in E_{n}=\{0, \pm 1, \l... | Solution. By the formula of total probability
$$
\begin{gathered}
P_{n+1}(k)=\mathrm{P}\left\{S_{n+1}=k\right\}=\mathrm{P}\left\{S_{n+1}=k \mid \xi_{n+1}=-1\right\} \mathrm{P}\left\{\xi_{n+1}=-1\right\}- \\
-\mathrm{P}\left\{S_{n+1}=k \mid \xi_{n+1}=1\right\} \mathrm{P}\left\{\xi_{n+1}=1\right\}= \\
=\frac{1}{2} \math... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,542 |
4. (Continuation of problem I.6.3.) The sequence of values $S_{0}=0$, $S_{1}=\xi_{1}, S_{2}=\xi_{1}+\xi_{2}, \ldots, S_{n}=\xi_{1}+\ldots+\xi_{n}$ can be considered as the trajectory of a particle's random walk, starting from zero and moving up or down by one unit at integer moments in time.
Now, let's assume that the... | Solution. (a) By the local limit theorem
$$
P_{t}(x)=\lim _{\Delta \rightarrow 0} \frac{P_{n \Delta}(k \sqrt{\Delta})}{\sqrt{\Delta}}=\lim _{\Delta \rightarrow 0} \frac{1}{\sqrt{2 \pi n \Delta}} e^{-\frac{k^{2}}{2 n}}=\frac{\sqrt{2}}{\sqrt{\pi t}} e^{-\frac{x^{2}}{2 t}}
$$
(b) By direct calculation of the partial der... | \frac{\partialP_{}(x)}{\partial}=\frac{1}{2}\frac{\partial^{2}P_{}(x)}{\partialx^{2}} | Calculus | proof | Yes | Yes | olympiads | false | 33,543 |
5. Generalizing the results of the previous problem, assume that a wandering particle moves up by $\Delta x$ with probability $p(\Delta)=\frac{1}{2}+\Delta x$ and moves down by $\Delta x$ with probability $q(\Delta)=\frac{1}{2}-\Delta x$. Again, let $\Delta x=\sqrt{\Delta}$ and $n \Delta \rightarrow t, k \Delta x \righ... | Solution. By the local limit theorem
$$
\begin{aligned}
P_{t}(x)=\lim _{\Delta \rightarrow 0} \frac{P_{n \Delta}(k \sqrt{\Delta})}{\sqrt{\Delta}}=\lim _{\Delta \rightarrow 0} \frac{\sqrt{2}}{\sqrt{\pi n \Delta\left(1 / 4-\Delta x^{2}\right)}} e^{-\frac{(k / 2-n \Delta x)^{2}}{2 n\left(1 / 4-\Delta x^{2}\right)}}= \\
=... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,544 |
6. What needs to be changed in the limiting transition made in the two previous problems so that the limiting function satisfies the equation
$$
\frac{\partial P_{t}(x)}{\partial t}=-\mu \frac{\partial P_{t}(x)}{\partial x}+\frac{1}{2} \sigma^{2} \frac{\partial^{2} P_{t}(x)}{\partial x^{2}}
$$
(called the Fokker-Plan... | Solution. Suppose that a wandering particle moves up by $\Delta x$ with probability $p(\Delta)=\frac{1}{2}+\Delta x \frac{\mu}{2 \sigma^{2}}$ and moves down by $\Delta x$ with probability $q(\Delta)=\frac{1}{2}-\Delta x \frac{\mu}{2 \sigma^{2}}$. Let $\Delta x=\sigma \sqrt{\Delta}$ and $n \Delta \rightarrow t$, $k \Del... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,545 |
7. Let $F_{n}=F_{n}(t), t \in[0,1], n \geqslant 1$, be a non-decreasing function satisfying the condition $F_{n}(t) \rightarrow t$ for all rational $t \in Q \cap[0,1]$. Show that this convergence will be uniform:
$$
\sup _{t \in[0,1]}\left|F_{n}(t)-t\right| \rightarrow 0, \quad n \rightarrow \infty
$$ | Solution. Fix $\varepsilon>0$ and take $m>1 / \varepsilon$. For sufficiently large $n$ we have $\left|F_{n}(i / m)-i / m\right|<\varepsilon$ for all $i=0, \ldots, m$. Thus, due to the non-decreasing nature of $F_{n}(t)$ for any $t \in\left[\frac{i}{m} ; \frac{i+1}{m}\right]$ we get
$$
F_{n}(t)-t \leqslant F_{n}\left(\... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,546 |
8. Show that for $x>0$ the inequalities hold
$$
\frac{x}{1+x^{2}} \varphi(x)<1-\Phi(x)<\frac{\varphi(x)}{x}
$$
where $\varphi(x)=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2}$ and $\Phi(x)=\int_{-\infty}^{x} \varphi(y) d y$. | Solution. Consider the functions
$$
f(x):=\frac{\varphi(x)}{x}+\Phi(x)-1 \text { and } g(x):=\frac{x}{1+x^{2}} \varphi(x)+\Phi(x)-1
$$
Since $\varphi^{\prime}(x)=-x \varphi(x)$ and $\Phi^{\prime}(x)=\varphi(x)$, we obtain that
$$
\begin{aligned}
& f^{\prime}(x)=\frac{-x^{2} \varphi(x)-\varphi(x)}{x^{2}}+\varphi(x)=-... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 33,547 |
9. Show that for the Poisson distribution, the following local theorem holds: for each $k=0,1,2, \ldots$ and $\lambda \rightarrow \infty$ the condition
$$
\sqrt{\lambda}\left|\frac{\lambda^{k}}{k!} e^{-\lambda}-\frac{1}{\sqrt{2 \pi \lambda}} \exp \left\{-\frac{1}{2 \lambda}(k-\lambda)^{2}\right\}\right| \rightarrow 0
... | Solution. If $\sqrt{\lambda}$ is brought under the modulus sign, the minuend will have the form $\frac{1}{k!} \lambda^{k+\frac{1}{2}} e^{-\lambda}$. Since $e^{\lambda}$ grows faster than any power function as $\lambda \rightarrow+\infty$, $\frac{1}{k!} \lambda^{k+\frac{1}{2}} e^{-\lambda} \rightarrow 0$.
Similarly, th... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,548 |
1. Suppose it is a priori known that the parameter $\theta$ takes values in the set $\Theta_{0} \subseteq[0,1]$. Determine when there exists an unbiased estimate for the parameter $\theta$ that takes values only in this set $\Theta_{0}$. | Solution. Let $\Theta_{0}$ be a one-point set ( $\Theta_{0}=\left\{\theta_{0}\right\}$ ). Then $T_{n} \equiv \theta_{0}$ is an unbiased estimate for $\theta_{0}$.
Let $\Theta_{0}$ consist of at least two points. We will show that for the existence of an unbiased estimate for $\theta$, taking values from $\Theta_{0}$, ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,549 |
2. The $\chi^{2}$ (chi-squared) goodness-of-fit criterion. Let $\xi_{1}, \ldots, \xi_{n}$ be independent Bernoulli random variables, $\mathrm{P}\left\{\xi_{i}=1\right\}=p, \mathrm{P}\left\{\xi_{i}=0\right\}=1-p, 1 \leqslant i \leqslant n$. In this problem, we consider the question of testing the hypothesis $H_{0}: p=p_... | Solution. In solving the problem, results and formulas in parentheses from B1.I.6 will be used. We have
$$
\begin{aligned}
\mathrm{P}\left(\chi^{2}(\xi)\right. & \leqslant x)=\mathrm{P}\left(\sqrt{x} \leqslant \frac{S_{n}(\xi)-\mathrm{E} S_{n}(\xi)}{\sqrt{\mathrm{D} S_{n}}} \leqslant \sqrt{x}\right)= \\
& =\mathrm{P}\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,550 |
3. Let $\xi$ be a random variable with a binomial distribution
$$
\mathrm{P}_{\theta}\{\xi=k\}=C_{n}^{k} \theta^{k}(1-\theta)^{n-k}, \quad 0 \leqslant k \leqslant n,
$$
where $n$ is a given number and $\theta$ is an "unknown parameter" that needs to be estimated based on (a single) observation of this random variable... | Solution. The distribution of the random variable $\xi$ can be considered as the distribution of the sum of independent identically distributed variables $\xi_{1}, \ldots, \xi_{n}$, where $\mathrm{P}_{\theta}\left\{\xi_{1}=1\right\}=\theta, \mathrm{P}_{\theta}\left\{\xi_{1}=0\right\}=1-\theta$.
As shown in V1.I.7,
$$... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,551 |
4. Two proofreaders A and B, after reading the proof, found $a$ and $b$ typos, respectively, of which $c$ turned out to be common. Assuming that the proofreaders worked independently of each other, give a "reasonable" estimate of the number of remaining undetected typos. | Solution. Consider the following statistical model. $2 n$ independent Bernoulli trials $\xi_{1}, \ldots, \xi_{2 n}$ are conducted. The event $\left\{\xi_{2 k}=1\right\}$ means that proofreader $A$ noticed the $k$-th typo, $\left\{\xi_{2 k-1}=1\right\}$ means that proofreader $B$ noticed the $k$-th typo, $k=1, \ldots, n... | \frac{}{}- | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,552 |
1. Provide an example of two random variables $\xi$ and $\eta$ that are not independent, but for which
$$
\mathrm{E}(\xi \mid \eta)=\mathrm{E} \xi
$$ | Solution. Consider the sample space $\Omega=\left\{\omega_{1}, \omega_{2}, \omega_{3}\right\}, \mathrm{P}\left(\omega_{1}\right)=\mathrm{P}\left(\omega_{2}\right)=1 / 4, \mathrm{P}\left(\omega_{3}\right)=1 / 2$. Random variables $\xi$ and $\eta$ are defined pointwise as follows:
$\xi\left(\omega_{1}\right)=1, \quad \x... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,553 |
2. The conditional variance of a random variable $\xi$ with respect to a partition $\mathscr{D}$ is the random variable
$$
\mathrm{D}(\xi \mid \mathscr{D})=\mathrm{E}\left[(\xi-\mathrm{E}(\xi \mid \mathscr{D}))^{2} \mid \mathscr{D}\right] .
$$
Show that
$$
\mathrm{D} \xi=\mathrm{ED}(\xi \mid \mathscr{D})+\mathrm{DE}... | Solution. By expanding the brackets and taking $\mathscr{D}$-measurable random variables out from under the expectation sign, we obtain
$$
\begin{aligned}
& \mathrm{D}(\xi \mid \mathscr{D})=\mathrm{E}\left(\xi^{2}-2 \xi \mathrm{E}(\xi \mid \mathscr{D})+[\mathrm{E}(\xi \mid \mathscr{D})]^{2} \mid \mathscr{D}\right)= \\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,554 |
3. Prove that for any function $f=f(\eta)$, the conditional expectation $\mathrm{E}(\xi \mid \eta)$ has the following property:
$$
\mathrm{E}[f(\eta) \mathrm{E}(\xi \mid \eta)]=\mathrm{E}[\xi f(\eta)]
$$ | Solution. Since the function $f(\eta)$ is $\mathscr{D}_{\eta}$-measurable,
$$
f(\eta) \cdot \mathrm{E}(\xi \mid \eta)=\mathrm{E}(\xi \cdot f(\eta) \mid \eta)
$$
If we calculate the expectation of both sides of the equality, we obtain the statement of the problem. | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,555 |
4. Let $\xi$ and $\eta$ be random variables. Show that $\inf _{f} \mathrm{E}(\eta-f(\xi))^{2}$ is achieved at the function $f^{*}(\xi)=\mathrm{E}(\eta \mid \xi)$. (Thus, the optimal mean-square estimate of $\eta$ based on $\xi$ is the conditional expectation $\mathbf{E}(\eta \mid \xi)$.) | Solution. For an arbitrary function $f=f(x)$, we obtain
$$
\begin{aligned}
& \mathrm{E}(\eta-f(\xi))^{2}=\mathrm{E}\left[\left(\eta-f^{*}(\xi)\right)+\left(f^{*}(\xi)-f(\xi)\right)\right]^{2}= \\
& \quad=\mathrm{E}\left(\eta-f^{*}(\xi)\right)^{2}+2 \mathrm{E}\left[\left(\eta-f^{*}(\xi)\right)\left(f^{*}(\xi)-f(\xi)\ri... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,556 |
5. Let $\xi_{1}, \ldots, \xi_{n}, \tau-$ be independent random variables, where $\xi_{1}, \ldots, \xi_{n}$ are identically distributed, and $\tau$ takes values $1, \ldots, n$. Show that if $S_{\tau}=\xi_{1}+\ldots+\xi_{\tau}-$ is the sum of a random number of random variables, then
$$
\mathrm{E}\left(S_{\tau} \mid \ta... | Solution. Since
$$
1=\sum_{k=1}^{n} I(\tau=k) \quad \text { and } \quad \mathrm{E}\left(\xi_{i} \mid \tau\right)=\mathrm{E} \xi_{1}, \quad i=1, \ldots, n,
$$
we obtain that
$$
\begin{aligned}
\mathrm{E}\left(S_{\tau} \mid \tau\right)=\mathrm{E}\left(\sum_{k=1}^{n} I(\tau=k) S_{k} \mid \tau\right)=\sum_{k=1}^{n} I(\t... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,557 |
6. Let the random variable $\xi$ and the partition $\mathscr{D}$ be independent (i.e., for any $D \in \mathscr{D}$, the random variables $\xi$ and $I_{D}$ are independent). Prove that then $\mathrm{E}(\xi \mid \mathscr{D})=\mathrm{E} \xi$. | Solution. By definition
$$
\begin{aligned}
& \mathrm{E}(\xi \mid \mathscr{D})(\omega)=\sum_{i=1}^{k} \mathrm{E}\left(\xi \mid D_{i}\right) I_{D_{i}}(\omega)=\sum_{i=1}^{k} \frac{\mathrm{E}\left(\xi I_{D_{i}}\right)}{\mathrm{P}\left(D_{i}\right)} I_{D_{i}}(\omega)= \\
&=\sum_{i=1}^{k} I_{D_{i}}(\omega) \mathrm{E} \xi=\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,558 |
7. Let $(\Omega, \mathscr{F}, \mathrm{P})$ be some discrete probability space and $\xi=\xi(\omega)$ be a random variable taking values $x_{1}, \ldots, x_{k}$ with probabilities $\mathrm{P}\left\{\xi=x_{i}\right\}=p_{i}$. The entropy of the random variable $\xi$ (or the experiment $\mathscr{E}_{\xi}$, consisting of obse... | Solution. For convenience of notation, let $p_{i}:=\mathrm{P}\left(\xi=x_{i}\right)$, $i=1, \ldots, k ; q_{j}:=\mathrm{P}\left(\eta=y_{j}\right), j=1, \ldots, l$.
By independence,
$$
\begin{aligned}
& H(\xi, \eta)=-\sum_{i=1}^{k} \sum_{j=1}^{l} p_{i j} \log _{2} p_{i j}= \\
& =-\sum_{i=1}^{k} \sum_{j=1}^{l} p_{i} q_{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,559 |
8. Let $(\xi, \eta)$ be a pair of random variables taking values $\left(x_{i}, y_{j}\right), i=1, \ldots, k, j=1, \ldots, l$.
The conditional entropy of the random variable $\eta$ given the event $\left\{\xi=x_{i}\right\}$ is defined as
$$
H_{x_{i}}(\eta)=-\sum_{j=1}^{l} \mathrm{P}\left\{\eta=y_{j} \mid \xi=x_{i}\rig... | Solution. (a) We have
$$
\begin{aligned}
H_{\xi}(\eta)= & \sum_{i=1}^{k} \mathrm{P}\left\{\xi=x_{i}\right\} H_{x_{i}}(\eta)= \\
= & -\sum_{i=1}^{k} \mathrm{P}\left\{\xi=x_{i}\right\} \sum_{j=1}^{l} \mathrm{P}\left\{\eta=y_{j} \mid \xi=x_{i}\right\} \log _{2} \mathrm{P}\left\{\eta=y_{j} \mid \xi=x_{i}\right\}= \\
= & -... | proof | Other | proof | Yes | Yes | olympiads | false | 33,560 |
9. Let $(\xi, \eta)$ be a pair of random variables. The quantity
$$
I_{\xi}(\eta)=H(\eta)-H_{\xi}(\eta)
$$
is called the amount of information about $\eta$ contained in $\xi$. The terminology is justified by the fact that the difference $H(\eta)-H_{\xi}(\eta)$ shows how much knowledge of the random variable $\xi$ can... | Solution. (a) The statement follows from Problem I.8.8(a).
(b) We have $I_{\xi}(\eta)=H(\eta) \Leftrightarrow H_{\xi}(\eta)=0 \Leftrightarrow H_{x_{i}}(\eta)=0, i=1, \ldots, k \Leftrightarrow$ $\Leftrightarrow \mathrm{P}\left\{\eta=y_{j} \mid \xi=x_{i}\right\}=0$ or 1 for any $i=1, \ldots, k, j=1, \ldots, l \Leftright... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,561 |
10. Let $\xi_{1}, \ldots, \xi_{n}$ be independent and identically distributed (i.i.d.) Bernoulli random variables, $\mathrm{P}\left\{\xi_{i}=1\right\}=p$, $\mathrm{P}\left\{\xi_{i}=0\right\}=1-p$, and let $S_{n}=\xi_{1}+\ldots+\xi_{n}$. Show that
(a) $\mathrm{P}\left(\xi_{1}=x_{1}, \ldots, \xi_{n}=x_{n} \mid S_{n}=k\r... | Solution. (a) We have
$$
\begin{array}{r}
\mathrm{P}\left(\xi_{1}=x_{1}, \ldots, \xi_{n}=x_{n} \mid S_{n}=k\right)=\frac{\mathrm{P}\left(\xi_{1}=x_{1}, \ldots, \xi_{n}=x_{n}, S_{n}=k\right)}{\mathrm{P}\left(S_{n}=k\right)}= \\
=\frac{I_{x}(k) p^{k} q^{n-k}}{C_{n}^{k} p^{k} q^{n-k}}=\frac{I_{\{x\}}(k)}{C_{n}^{x}} .
\en... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,562 |
1. Show that the following formulas are valid:
$$
\mathrm{E} S_{\tau_{n}^{x}}^{x}=x+(p-q) \mathrm{E} \tau_{n}^{x}, \quad \mathrm{E}\left[S_{\tau_{n}^{x}}^{x}-\tau_{n}^{x} \mathrm{E} \xi_{1}\right]^{2}=\mathrm{D} \xi_{1} \cdot \mathrm{E} \tau_{n}^{x}+x^{2}
$$ | Solution. By definition (1) we have
$$
\begin{aligned}
\tau_{n}^{x}=\min \left\{0 \leqslant l \leqslant n: S_{l}^{x}=\right. & A \text { or } B\}= \\
& =\min \left\{0 \leqslant l \leqslant n: S_{l}=A-x \text { or } B-x\right\}
\end{aligned}
$$
Let the last expression be denoted by $\tau_{n}^{\prime}$; if
$$
A-x<S_{l... | proof | Algebra | proof | Yes | Yes | olympiads | false | 33,563 |
3. Let in the Bernoulli scheme $p=q=1 / 2$. Show that
$$
\mathrm{E}\left|S_{n}\right| \sim \sqrt{\frac{2}{\pi} n}, \quad n \rightarrow \infty
$$ | Solution. Let's ensure that for $n \geqslant 1$ the following formula (a "discrete version of Tanaka's formula") holds:
$$
\left|S_{n}\right|=\sum_{k=1}^{n} \operatorname{sign}\left(S_{k-1}\right) \Delta S_{k}+N_{n}
$$
where $S_{0}=0, S_{k}=\xi_{1}+\ldots+\xi_{k}, \Delta S_{k}=\xi_{k}$,
$$
\operatorname{sign} x= \be... | \sqrt{\frac{2}{\pi}n} | Calculus | proof | Yes | Yes | olympiads | false | 33,565 |
4. Two players independently toss (each their own) fair coins. Show that the probability that they will have the same number of heads after $n$ tosses is $2^{-2 n} \sum_{k=0}^{n}\left(C_{n}^{k}\right)^{2}$.
Derive from this the equality
$$
\sum_{k=0}^{n}\left(C_{n}^{k}\right)^{2}=C_{2 n}^{n}
$$
Let $\sigma_{n}$ be th... | Solution. Let $\xi_{i}^{(k)}=1$ (or -1), if the player with number $k=1,2$ gets heads (or tails) on the $i$-th step. Then
$\mathrm{P}\left\{\begin{array}{c}\text { the number of heads } \\ \text { after } n \text { tosses is the same for both players }\end{array}\right\}=\mathrm{P}\left\{\sum_{i=1}^{n} \xi_{i}^{(1)}=\... | notfound | Combinatorics | proof | Yes | Yes | olympiads | false | 33,566 |
5. Let $\xi_{0}, \xi_{1}, \ldots, \xi_{N}$ be independent Bernoulli random variables, $\mathrm{P}\left\{\xi_{i}=1\right\}=\mathrm{P}\left\{\xi_{i}=-1\right\}=1 / 2, S_{n}=\xi_{1}+\ldots+\xi_{n}$ and $X_{n}=\xi_{0}(-1)^{S_{n}}-$
a discrete telegraph signal, $1 \leqslant n \leqslant N$. Find the mean and variance of the ... | Solution. Since the random variables $\xi_{0}$ and $(-1)^{S_{n}}$ are independent and $\mathrm{E} \xi_{0}=0$, we obtain
$$
\mathrm{E} X_{n}=\mathrm{E}\left(\xi_{0}(-1)^{S_{n}}\right)=\mathrm{E} \xi_{0} \mathrm{E}(-1)^{S_{n}}=0
$$
Since $X_{n}^{2} \equiv 1$, we have
$$
\mathrm{D} X_{n}=\mathrm{E} X_{n}^{2}-\left(\mat... | \begin{aligned}&\mathrm{E}X_{n}=0,\\&\mathrm{D}X_{n}=1,\\&\mathrm{P}{X_{n}=1\mid\xi_{0}=1}=\begin{cases}1,&neven,\\0,&nodd,\end{cases}\\ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,567 |
6. Let $\xi_{1}, \ldots, \xi_{N}$ be independent Bernoulli random variables, $\mathrm{P}\left\{\xi_{i}=1\right\}=p, \mathrm{P}\left\{\xi_{i}=-1\right\}=1-p, S_{i}=\xi_{1}+\ldots+\xi_{i}, 1 \leqslant i \leqslant N$, $S_{0}=0$. Let $\mathscr{R}_{N}$ be the range, i.e., the number of distinct points visited by the random ... | Solution. Let's represent $\mathscr{R}_{N}$ in the following form:
$$
\begin{aligned}
\mathscr{R}_{N}=1+I\left(S_{1} \neq 0\right)+I\left(S_{2} \neq\right. & \left.0, S_{2} \neq S_{1}\right)+\ldots \\
& \ldots+I\left(S_{N} \neq 0, S_{N} \neq S_{1}, \ldots, S_{N} \neq S_{N-1}\right)
\end{aligned}
$$
Let $V_{n}^{N}=\su... | \frac{\mathscr{R}_{N}}{N}arrow\mathrm{P}{S_{n}\neq0,n\geqslant0} | Other | math-word-problem | Yes | Yes | olympiads | false | 33,568 |
7. Let $\xi_{1}, \ldots, \xi_{N}$ be identically distributed random variables (not necessarily Bernoulli), $S_{0}=0, S_{i}=\xi_{1}+\ldots+\xi_{i}, 1 \leqslant i \leqslant N$. Let
$$
N_{n}=\sum_{k=1}^{n} I\left(S_{k}>0\right)
$$
- the number of positive terms in the sequence $S_{0}, S_{1}, \ldots, S_{n}$. Show that th... | Solution. Let us denote
$$
B=\left\{s \in \mathbb{R}^{n}: s_{i}>0, \quad i \leqslant k, \text { and } s_{i} \leqslant s_{k}, \quad k0\right)=b_{i}, i \leqslant n\right\}$, and the summation is over all vectors $b$ consisting of $n-k$ zeros and $k$ ones. Note that if $S \in A_{b}$, then $S\left[\pi_{b}\right] \in B$, w... | proof | Other | proof | Yes | Yes | olympiads | false | 33,569 |
1. At what speed does $\mathrm{E} \min \left(\sigma_{2 n}, 2 n\right)$ tend to $\infty$ as $n \rightarrow \infty$? (Here $\sigma_{2 n}=\min \left\{1 \leqslant k \leqslant 2 n: S_{k}=0\right\}$ and we assume $\sigma_{2 n}=\infty$ (or $\left.\sigma_{2 n}=2 n\right)$, if $S_{k} \neq 0$ for all $1 \leqslant k \leqslant 2 n... | Solution. According to item 1 of V1.I. 10, we have
$\mathrm{E} \min \left(\sigma_{2 n}, 2 n\right)=\sum_{k=1}^{n} u_{2(k-1)}+2 n u_{2 n}=$
$$
=\sum_{k=0}^{n-1} 2^{-2 k} C_{2 k}^{k}+2 n \cdot 2^{-2 n} C_{2 n}^{n}
$$
By Stirling's formula, $2 n \cdot 2^{-2 n} C_{2 n}^{n} \sim 2 \sqrt{n \pi^{-1}}$, and by problem I.9.3... | 4\sqrt{n\pi^{-1}} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33,570 |
3. Based on the ideas and methods presented in V1.I.10, show that for a symmetric ($p=q=1/2$) Bernoulli random walk $\left\{S_{k}, k \leqslant n\right\}, S_{0}=0, S_{k}=\xi_{1}+\ldots+\xi_{k}$, the following formulas hold (where $N$ is a natural number):
(a) $\mathrm{P}\left\{\max _{1 \leqslant k \leqslant n} S_{k} \g... | Solution. (a) Let $\mathscr{G}_{1}$ be the set of paths such that $\left\{\max _{1 \leqslant k \leqslant n} S_{k} \geqslant N, S_{n} < N\right\}$. Let $\mathscr{G}_{2}$ be the set of paths such that $\left\{\max _{1 \leqslant k \leqslant n} S_{k} \geqslant N, S_{n} \geqslant N\right\}$. We will construct a one-to-one c... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,572 |
4. Let $\xi_{1}, \ldots, \xi_{2 n}$ be independent Bernoulli random variables, $\mathrm{P}\left\{\xi_{k}=1\right\}=\mathrm{P}\left\{\xi_{k}=-1\right\}=1 / 2, k \leqslant 2 n$. Set $S_{0}=0$, $S_{k}=\xi_{1}+\ldots+\xi_{k}$, and let
$$
g_{2 n}=\max \left\{0<2 k \leqslant 2 n: S_{2 k}=0\right\}
$$
- the moment of the la... | Solution. For $k=0$, the formula follows from formula (9); for $k=n-$, it follows from formula (1). Let $1<k<n$. Then
\[
\begin{aligned}
\mathrm{P}\left\{g_{2 n}=2 k\right\} & =\mathrm{P}\left\{S_{2 k}=0, S_{2 k+1} \neq 0, \ldots, S_{2 n} \neq 0\right\}= \\
& =\mathrm{P}\left\{S_{2 k}=0, \xi_{2 k+1} \neq 0, \ldots, \x... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,573 |
5. Under the conditions of the previous problem, let $\theta_{2 n}$ denote the position of the first maximum of the sequence $S_{0}, S_{1}, \ldots, S_{2 n}$, i.e., $\theta_{2 n}=k$ if $S_{0}1$ does not exist, then we set $\theta_{2 n}=0$.
Show that
(a) $\mathrm{P}\left\{\theta_{2 n}=0\right\}=u_{2 n}$,
(b) $\mathrm{P}... | Solution. (a) We have
$$
\mathrm{P}\left\{\theta_{2 n}=0\right\}=\mathrm{P}\left\{S_{1} \leqslant 0, \ldots, S_{2 n} \leqslant 0\right\}
$$
By symmetry, the last probability is equal to $\mathrm{P}\left\{S_{1} \geqslant 0, \ldots, S_{2 n} \geqslant 0\right\}$, which by formula (11) is equal to $u_{2 n}$.
(b) Let the... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,574 |
6. Let $S_{k}=\xi_{1}+\ldots+\xi_{k}, k \leqslant 2 n$, where $\xi_{1}, \ldots, \xi_{2 n}$ are independent and identically distributed (i.i.d.) random variables, $\mathrm{P}\left\{\xi_{1}=1\right\}=\mathrm{P}\left\{\xi_{1}=-1\right\}=1 / 2$. Show that
(a) for $r= \pm 1, \ldots, \pm n$ the following equality holds:
$$... | Solution. Due to symmetry and the property of binomial coefficients $C_{2 n}^{n-r}=C_{2 n}^{n+r}$, it is sufficient to prove both formulas only for positive $r$.
(a) We have
$$
\begin{aligned}
\mathrm{P}\left\{S_{1} \neq 0, \ldots, S_{2 n-1} \neq 0, S_{2 n}=2 r\right\} & = \\
& =\mathrm{P}\left\{S_{1}>0, \ldots, S_{2... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 33,575 |
7. Let $\left\{S_{k}, k \leqslant n\right\}, S_{0}=0, S_{k}=\xi_{1}+\ldots+\xi_{k}$, be a symmetric Bernoulli random walk (with independent and identically distributed random variables $\xi_{1}, \ldots, \xi_{n}$ such that $\left.\mathrm{P}\left\{\xi_{1}=1\right\}=\mathrm{P}\left\{\xi_{1}=-1\right\}=1 / 2\right)$. Let
... | Solution. Let $p, r, t$ be arbitrary values of random variables. We will show that between the trajectories of type $\mathscr{G}_{1}$, which belong to the set $\left\{m_{n}=-p, M_{n}=r, S_{n}=t\right\}$, and the trajectories of type $\mathscr{G}_{2}$, which belong to the set $\left\{M_{n}=p, m_{n}=-r, S_{n}=-t\right\}$... | proof | Other | proof | Yes | Yes | olympiads | false | 33,576 |
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