This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

AND
OR
NO

Found problems: 5

Let $X_1, X_2$ be independent, identically distributed random variables such that $X_i\geq 0$ for all $i$. Let $\mathrm EX_i=m$, $\mathrm{Var} (X_i)=\sigma ^2<\infty$. Show that, for all $0<\alpha\leq 1$ $$\lim_{n\to\infty} n\,\mathrm{Var} \left( \left[ \frac{X_1+\ldots +X_n}{n}\right] ^\alpha\right)=\frac{\alpha ^ 2 \sigma ^ 2}{m^{2(1-\alpha)}}$$ [Gy. Michaletzki]
Let $(\Omega, \mathcal A, P)$ be a probability space, and let $(X_n, \mathcal F_n)$ be an adapted sequence in $(\Omega, \mathcal A, P)$ (that is, for the $\sigma$-algebras $\mathcal F_n$, we have $\mathcal F_1\subseteq \mathcal F_2\subseteq \dots \subseteq \mathcal A$, and for all $n$, $X_n$ is an $\mathcal F_n$-measurable and integrable random variable). Assume that $$\mathrm E (X_{n+1} \mid \mathcal F_n )=\frac12 X_n+\frac12 X_{n-1}\,\,\,\,\, (n=2, 3, \ldots )$$ Prove that $\mathrm{sup}_n \mathrm{E}|X_n|<\infty$ implies that $X_n$ converges with probability one as $n\to\infty$. [I. Fazekas]
For problem 11 , i couldn’t find the correct translation , so i just posted the hungarian version . If anyone could translate it ,i would be very thankful . [tip=see hungarian]Az $X$ ́es$ Y$ valo ́s ́ert ́eku ̋ v ́eletlen v ́altoz ́ok maxim ́alkorrel ́acio ́ja az $f(X)$ ́es $g(Y )$ v ́altoz ́ok korrela ́cio ́j ́anak szupr ́emuma az olyan $f$ ́es $g$ Borel m ́erheto ̋, $\mathbb{R} \to \mathbb{R}$ fu ̈ggv ́enyeken, amelyekre $f(X)$ ́es $g(Y)$ v ́eges sz ́ora ́su ́. Legyen U a $[0,2\pi]$ interval- lumon egyenletes eloszl ́asu ́ val ́osz ́ınu ̋s ́egi v ́altozo ́, valamint n ́es m pozit ́ıv eg ́eszek. Sz ́am ́ıtsuk ki $\sin(nU)$ ́es $\sin(mU)$ maxim ́alkorrela ́ci ́oja ́t. [/tip] Edit: [hide=Translation thanks to @tintarn] The maximal correlation of two random variables $X$ and $Y$ is defined to be the supremum of the correlations of $f(X)$ and $g(Y)$ where $f,g:\mathbb{R} \to \mathbb{R}$ are measurable functions such that $f(X)$ and $g(Y)$ is (almost surely?) finite. Let $U$ be the uniformly distributed random variable on $[0,2\pi]$ and let $m,n$ be positive integers. Compute the maximal correlation of $\sin(nU)$ and $\sin(mU)$. (Remark: It seems that to make sense we should require that $E[f(X)]$ and $E[g(Y)]$ as well as $E[f(X)^2]$ and $E[g(Y)^2]$ are finite. In fact, we may then w.l.o.g. assume that $E[f(X)]=E[g(Y)]=0$ and $E[f(Y)^2]=E[g(Y)^2]=1$.)[/hide]
Let $\xi_{(k_1, k_2)}, k_1, k_2 \in\mathbb N$ be random variables uniformly bounded. Let $c_l, l\in\mathbb N$ be a positive real strictly increasing infinite sequence such that $c_{l+1}/ c_l$ is bounded. Let $d_l=\log \left(c_{l+1}/c_l\right), l\in\mathbb N$ and suppose that $D_n=\sum_{l=1}^n d_l\uparrow \infty$ when $n\to\infty$ Suppose there exist $C>0$ and $\varepsilon>0$ such that $$\left| \mathbb E \left\{ \xi_{(k_1,k_2)}\xi_{(l_1,l_2)}\right\}\right| \leq C\prod_{i=1}^2 \left\{ \log_+\log_+\left( \frac{c_{\max\{ k_i, l_i\}}}{c_{\min\{ k_i, l_i\}}}\right)\right\}^{-(1+\varepsilon)}$$ for each $(k_1, k_2), (l_1,l_2)\in\mathbb N^2$ ($\log_+$ is the positive part of the natural logarithm). Show that $$\lim_{\substack{n_1\to\infty \\ n_2\to\infty}} \frac{1}{D_{n_1}D_{n_2}}\sum_{k_1=1}^{n_1} \sum_{k_2=1}^{n_2} d_{k_1}d_{k_2}\xi_{(k_1,k_2)}=0$$ almost surely. (translated by j___d)
Let $(\Omega, \mathcal{A},\mathbb{P})$ be a standard probability space, and $\mathcal{X}$ be the set of all bounded random variables. For $t>0$, defined the mapping $R_t$ by \[R_t(X)=t\log \mathbb{E}[\exp(X/t)], \quad X \in \mathcal{X},\] and for $t \in (0,1)$ define the mapping $Q_t$ by \[Q_t(X)=\inf\{x \in \mathbb{R}: \mathbb{P}(X>x) \le t\}, \quad X \in \mathcal{X}.\] For two mappings $f,g:\mathcal{X} \to [-\infty,\infty)$, defined the operator $\square$ by \[f\square g(X)=\inf\{f(Y)+g(X-Y): Y \in \mathcal{X}\}, \quad X \in \mathcal{X}.\] (a) Show that, for $t,s>0$, \[R_t \square R_s=R_{t+s}.\] (b) Show that, for $t,s>0$ with $t+s<1$, \[Q_t \square Q_s=Q_{t+s}.\]