I am trying to show the following: Let $\{X_i\}$ be i.i.d  with $P(X_1=(-1)^{k}k)=\frac{1}{ck^2log(k)}$ for $k\ge 2$ with $c=\sum_{k}\frac{1}{k^2log(k)}$ a normalizing constant. Show that $\mathbb{E}[|X_1|]\infty$ but there is a non-random $\mu_n$ such that $n^{-1}S_n-\mu_n\rightarrow 0$ in probability. My attempt is the following: $$\mathbb{E}[|X_1|]=\sum_{t=2}^{\infty}tP(|X_1|=t)=\sum_{t&gt;2;t=2k}^{\infty}tP(X_1=t)+\sum_{t&gt;2;t=2k+1}^{\infty}tP(X_1=-t)=\sum_{t=2}^{\infty}\frac{t}{t^2log(t)}=\infty$$ Since $X_1$ is unbounded I need to  show the $nP(|X_1|&gt;n)\rightarrow 0$ in order to guarantee  existence of $\mu_n$ that is $n=o(P(|X_1|&gt;n))$ Consider $P(|X_1|&gt;n)=\sum_{t&gt;\lfloor n\rfloor}^{\infty}P(|X_1|=t)=\sum_{t&gt;\lfloor n\rfloor}^{\infty}\frac{1}{ct^2log(t)}$ since $\frac{1}{ct^2log(t)}\rightarrow 0$ as $t\rightarrow \infty$ then $\sum_{t&gt;\lfloor n\rfloor}^{\infty}\frac{1}{ct^2log(t)}\rightarrow 0$ as $n\rightarrow \infty$. Since $\sum_{t&gt;\lfloor n\rfloor}^{\infty}\frac{1}{ct^2log(t)}$ goes to zero quicker then n goes to $\infty$ then the desired property holds and the existence is guaranteed. Is this correct or is my approach completely wrong.
123155
Let $X_n$ be iid random variables with $P(X_n=(-1)^kk)=\dfrac{C}{k^2\log k}$ for all $k\geq 2$. Here $C$ is just a constant so that the sum of probabilities is $1$. Find constant $\mu$ such that $S_n/n\to \mu$ in probability where $S_n=\sum_{k=1}^nX_k$.   Here $E(X_n)=\infty$. I'm a bit lost after this. In general how does one find constants like this?
75972
I am wondering if the following expectation can be computed in the shown manner  $$\begin{aligned} E[\log_2(1+X) \mid X&gt;a] &amp;= \int_{0}^{\infty} P( \log_2(1+X)&gt;t \mid X&gt;a) dt \\ &amp;= \int_{0}^{\infty} P(X&gt;2^t-1 \mid X&gt;a) dt \\  &amp;= \int_{0}^{\infty} \frac{P(X&gt;2^t-1 \cap X&gt;a)}{P(X&gt;a)} dt \\ &amp;= \frac{1}{P(X&gt;a)} \left(\int_{0}^{a} P(X&gt;a) dt + \int_a^{\infty} P(X&gt;2^{t}-1) dt \right)  \end{aligned}$$  Note that there is no closed form expression for the PDF of X hence I am trying to find alternative ways to compute the expectation.  Thank you
98938
Let $X$ be a non-negative integer-valued random variable with finite mean. Show that $$E(X)=\sum^\infty_{n=0}P(X&gt;n)$$  This is the hint from my lecturer.  "Start with the definition $E(X)=\sum^\infty_{x=1}xP(X=x)$. Rewrite the series as double sum."  For my opinion. I think the double sum have the form of $\sum\sum f(x)$, but how to get this form? And how to continue?
44935
I ask this because I'm trying to understand a proof of the expected value of a non negative random variable is equal to $\int_{0}^{\infty}(1-F(x))dx$ ($F$ is the distribution function of $X$) when $E[X]$ exists, and in some step I have $\lim\limits_{x\to\infty}({x·P[X&gt;x])}=0$ and I don't know why it verifies. The problem I see is the indetermination in the limit $\infty·0$.
7921
Show that $$\lim_{n \to \infty} \sup_{k \ge n} \left(\frac{1+a_{k+1}}{a_k}\right)^k \geq e$$ for any sequence $\{a_k\}$ with positive terms, and that this estimate cannot be improved. Let $$s_k = \left(\frac{1+a_{k+1}}{a_k}\right)^k $$  Then $s_k \geq 0$ as otherwise $a_{k+1} &lt; -1$. Also, $\sup_{k \geq n}\{s_k\}$ is nonincreasing on $n$. Assume $\lim\limits_{n \to \infty}\sup_{k \geq n}\{s_k\} = l &lt; e$. For $0 &lt; \varepsilon&lt; e-l $, choose $N$ such that $l \leq \sup_{k \geq n}\{s_k\} &lt; e - \varepsilon$ for all $n &gt; N$.  We know that $$\lim\limits_{n \to \infty} \left(\frac{n+1}{n}\right)^n = e$$ and that $r_n = \left(\frac{n+1}{n}\right)^n$ is increasing. Choose $N'$ such that $e- \varepsilon &lt; r_n &lt; e$ for all $n &gt; N'$.  Then for $M =\max\{N,N'\}$, $ \sup_{k \geq n}\{s_k\} &lt;r_n$ for all $n&gt;M$. Furthermore, $s_n \leq \sup_{k \geq n}\{s_k\}$, so $s_n &lt; r_n$.  Here I wanted to get some sort of contradiction but I'm not really sure how. I know that $a_n$ cannot have a limit $A$ as then $$0 \leq \lim\limits_{n \to \infty} a_{n+1} &lt; \lim\limits_{n \to \infty}\left(1 + \frac 1n \right)a_n -1,$$ which implies $A &lt; A-1$  And $a_n$ cannot be bounded by $d$ as we would have, for  $n &gt; 2d$, $a_{n+1} &lt; a_n\left(1 + \frac 1n\right) - 1 &lt; d\left(1 + \frac {1}{2d}\right) - 1 = d - \frac 12$. Continuing in this way there would be an $a_n &lt; 0$.   I was thinking of using the fact that the superior limit is the largest partial limit. What should I do next or should I use a different method?
109561
Let $X_n$ be a sequence of random variables. Can we find $c_n$ (positive constants) so that $X_nc_n \rightarrow 0$?  It is easy with Markov's equality + Borel Cantelli, with assumptions on expectations, but what if we don't know anything about expectations?
56591
This question is related to  , with the difference being that $F_X$ (the probability distribution function of random variable $X$) is an arbitrary continuous distribution function, not necessarily strictly increasing.  I think the proof is similar, but we have to take care of the possibility that $F_X$ may not be $1$-to-$1$.  I list my attempt below, and would appreciate it if someone can confirm if it's correct, and, in particular, if it can be improved.  Thanks a lot!  The goal is to show $F_Y(y)=y$ for any $y \in [0,1]$.  To do so, note that $F_Y(y)\triangleq\mathbb P(\{Y\le y\})$, and  $\{Y\le y\}=\{F_X(X)\le y\}=\{X\in F_X^{-1}([0, y])\}.$  Since $F_X$ is continuous, $F_X^{-1}([0,y])$ must be closed.  So it follows that   $\sup F_X^{-1}([0, y])=\max F_X^{-1}([0, y])=\max F_X^{-1}(\{y\}),$ which let's denote by $a$.    Therefore, $F_Y(y)=\mathbb P(\{X\le a\})=F_X(a)=y.\quad$  (Q.E.D.)
42725
I can intuitively see that this is true, but I'm having a very hard time proving it. I'm actually not even quite sure where to begin. I tried using the inequalities that define the floor function, and I'm pretty sure this is the way to go, but I'm failing to make any kind of progress.  I appreciate any help.
4583
$\def\d{\mathrm{d}}$Given that $X_1, X_2, \cdots$ are i.i.d., prove that$$ \frac{1}{n} \max_{1 \leqslant k \leqslant n} |X_k| \xrightarrow{\mathrm{a.s.}} 0 \Longleftrightarrow E(|X_1|) &lt; +\infty. $$   My try:  Because $\varliminf\limits_{n \to \infty} \dfrac{1}{n} \max\limits_{1 \leqslant k \leqslant n} |X_k|$ and $\varlimsup\limits_{n \to \infty} \dfrac{1}{n} \max\limits_{1 \leqslant k \leqslant n} |X_k|$ are both measurable with respect to the exchangeable σ-algebra of $X_1, X_2, \cdots$, they are both almost surely constants by the Hewitt-Savage 0-1 law, i.e.$$ \varliminf_{n \to \infty} \frac{1}{n} \max_{1 \leqslant k \leqslant n} |X_k| = c_1, \quad \text{a.s.} \tag{1}\\ \varlimsup_{n \to \infty} \frac{1}{n} \max_{1 \leqslant k \leqslant n} |X_k| = c_2, \quad \text{a.s.} $$ where $c_1, c_2 \in [0, +\infty]$. Also, if the distribution function of $|X_1|$ is $F(x)$, then the distribution function of $\dfrac{1}{n} \max\limits_{1 \leqslant k \leqslant n} |X_k|$ is$$ F_n(x) = P\left( \frac{1}{n} \max_{1 \leqslant k \leqslant n} |X_k| \leqslant x \right) = \prod_{k = 1}^n P(|x_k| \leqslant nx) = (F(nx))^n, $$ which implies\begin{align*} E\left( \frac{1}{n} \max_{1 \leqslant k \leqslant n} |X_k| \right) &amp;= \int_0^{+\infty} x \,\d((F(nx))^n) = \frac{1}{n} \int_0^{+\infty} y \,\d((F(y))^n)\\ &amp;= \frac{1}{n} \int_0^{+\infty} y·n (F(y))^{n - 1} \,\d F(y) = \int_0^{+\infty} y (F(y))^{n - 1} \,\d F(y)\\ &amp;\leqslant \int_0^{+\infty} y \,\d F(y) = E(|X_1|). \end{align*} Also for any $c &gt; 0$,$$ \sum_{k = 1}^\infty P(|X_1| \geqslant ck) &lt; \infty \Longleftrightarrow E(|X_1|) &lt; +\infty. \tag{2} $$ However, I failed to connect (1) and (2) to prove the proposition.  Any ideas are appreciated. Thanks in advance.
33243
Let $P(k,X)=p(1-p)^{k-1}$.  When deriving the moment generating function I start off as follows:  $E[e^{kt}X]=\sum\limits_{k=1}^{\infty}e^{kt}p(1-p)^{k-1}$.  How I end up rearranging this is as follows:  $\frac{p}{1-p}\sum\limits_{k=1}^{\infty}e^{kt}(1-p)^k=\frac{p}{1-p}\sum\limits_{k=1}^{\infty}(e^{t}(1-p))^k=\frac{p}{1-p}\frac{1}{1-e^t(1-p)}$  I'm obviously not arriving at the correct answer, but I want to know why this derivation is wrong.
16762
Lets $(X_n)_{n\in\mathbb{N}}$ be an iid sequence of standard normal random variables. Define $$M_n=\max_{1\leq i\leq n} X_i.$$ Prove that $$\lim_{n\rightarrow\infty} \frac{M_n}{\sqrt{2\log n}}=1\quad\text{a.s.}$$  I used the fact that $$\left(\frac{1}{x}-\frac{1}{x^3}\right)e^{-\frac{x^2}{2}}\leq\mathbb P(X_n&gt;x)\leq \frac{1}{x}e^{-\frac{x^2}{2}},$$  and the Borel Cantelli lemmas to prove that   $$\limsup_{n\rightarrow\infty} \frac{X_n}{\sqrt{2\log n}}=1\quad\text{a.s.}$$  I used Davide Giraudo's comment to show $$\limsup_n \frac{M_n}{\sqrt{2\log n}}=1\quad \text{a.s.}$$  I have no idea how to compute the $\liminf$. Borel-Cantelli give us tools to compute the $\limsup$ of sets, I am unsure of how to argue almost sure convergence. Any help would be appreciated.
47677
I'm searching for a really simple and beautiful proof that the sequence $(u_n)_{n \in \mathbb{N}} = \sum\nolimits_{k=1}^n \frac{1}{k} - \log(n)$ converges. At first I want to know if my answer is OK.  My try: $\lim\limits_{n\to\infty} \sum\limits_{k=1}^n \frac{1}{k} - \log (n) = \lim\limits_{n\to\infty} \sum\limits_{k=1}^n \frac{1}{k} + \sum\limits_{k=1}^{n-1} \log(k)-\log(k+1) = \lim\limits_{n\to\infty} \frac{1}{n} + \sum\limits_{k=1}^{n-1} \log(\frac{k}{k+1})+\frac{1}{k}$     $ = \sum\limits_{k=1}^{\infty} \frac{1}{k}-\log(\frac{k+1}{k})$ Now we prove that the last sum converges by the comparison test: $\frac{1}{k}-\log(\frac{k+1}{k}) &lt; \frac{1}{k^2} \Leftrightarrow k&lt;k^2\log(\frac{k+1}{k})+1$ which surely holds for $k\geqslant 1$      As $ \sum\limits_{k=1}^{\infty} \frac{1}{k^2}$ converges $ \Rightarrow \sum\limits_{k=1}^{\infty} \frac{1}{k}-\log(\frac{k+1}{k})$ converges and we name this limit $\gamma$ q.e.d
100865
Good day,  I have the problem of an old exam without solutions. It is the following:    Let $X_1,X_2,...$ be i.i.d. (independent and identically distributed) random variables on the same probability space such stat $$P\left(X_1=-\frac{1}{4}\right)=P\left(X_1=\frac{1}{4}\right)=\frac{1}{2}$$ For all $n\in \mathbb{N}$ define $M_n:=\prod_{k=1}^n (1+X_k)$.   Show that $(M_n)_{n\in\mathbb{N}} $ is a martingale w.r.t. a suitably chosen filtation (This was no problem for me) Use the law of large numbers to show that $(M_n)_{n\in\mathbb{N}}$ converges almost surely and determine the limit. (Here is the problem for me.)     This is the strong law of large numbers (SLLN) we had in the lecture:  Let $X_i$ be uncorrelated random variables, $E(X_i)=\mu, E(X_i^2)&lt;\infty$, $\exists C$ such that $\text{Var}(X_i)\leq C$ for all $i$ and define $S_n=\sum_{i=1}^n X_i$. Then $$\frac{S_n}{n} \to \mu \text{ almost surely.}$$    Therefore I have to get this $M_n$ into the form of $\frac{S_n}{n}$. I had a problems in the class where I had $$ A_n:=\left( \prod_{i=1}^n X_i \right)^{1/n}.$$ This was pretty elegant since we get the wished form via $$\log A_n = \frac{1}{n} \sum_{i=1}^n \log(X_i)=:\frac{1}{n} \sum_{i=1}^n Y_i.$$  I wanted to use this here:   $$\log M_n = \sum_{k=1}^n \log(1+X_k) = \frac{1}{n} \sum_{k} \log(1+X_k)^n $$  But now this "new" random variable $Y_{n,k}:=\log(1+X_k)^n$ is dependent of $n$. I don't know if this is a problem (now I think this is a huge problem, see below.) Let's try and use the SLLN for this: We would get $\log M_n \to E(Y_{n,k})$ if the assumptions are fulfilled but the limit should not be dependent on $n$. $$E(Y_{n,k})=E(\log(1+X_k)^n)=\frac{1}{2} \left( \log(5/4)^n +\log(3/4)^n \right) = \frac{1}{2} \log (15/16)^n \to -\infty$$  I don't know what to do. I think I could use the Martingale Convergence Theorem to get that $M_n$ converges a.s. but via this method I don't get the limit. Further the exam states that I have to use the SLLN explicitly.  I am thankful for every help/hint.
40488
Show that for a non-negative random variable $X$,   $$\mathbb E(X)=\int_0^\infty \mathbb P(X\ge x) \, dx.$$   I started with  $$\mathbb E(X)=\int_0^\infty x \, dF(x)=\int_0^\infty \int_0^x dt\,dF(x).$$ This is my try.
86181
Let $(X,\mathcal{M},\mu)$ be a measurable space, $1&lt;p_{k}&lt;\infty$ for $k=1,2,\ldots,N$ with $\sum_{k=1}^{N}\frac{1}{p_{k}}=1$ and $f_{k}\in L^{p_{k}}(\mu)$ for $k=1,\ldots,N$.  Show that $$\left\|f_{1}f_{2}\cdots f_{N}\right\|_{1}\leq \|f_{1}\|_{p_{1}}\|f_{2}\|_{p_{2}}\cdots\|f_{N}\|_{p_{N}}.$$  Remark: I tried to use the Holder inequality with $p=p_{1} $ and $q=\frac{1}{\sum_{k=2}^{\infty}\frac{1}{p_{k}}}$, but when I try to continue the process, it becomes complicated until I think that this is not the way.
80367
Let $\{\mathcal{F}_n\}_n$ be a decreasing sequence of sub-$\sigma$-fields of $\mathcal{F}$($\mathcal{F}_{n+1}\subset\mathcal{F}_n$) and let $\{X_n\}_n$ be a backward submartingale($E[|X_n|]&lt;\infty$, $X_n$ is $\mathcal{F}_n$ measurable and $E[X_n|\mathcal{F}_{n+1}]\geq X_{n+1}$).  If we know $l=\lim_{n\rightarrow\infty}E[X_n]&gt;-\infty$, how to prove that $\{X_n\}_n$ is uniformly integrable? Many thanks!  I have an idea, it is enough to show:    $\sup_nE[|X_n|]&lt;\infty$;  $\forall\epsilon&gt;0$, $\exists\delta&gt;0$ s.t. $\forall A\in\mathcal{F}$ such that $\mathbb{P}(A)&lt;\delta$, we have $\sup_nE[1_{A}|X_n|]&lt;\epsilon$.    For $|X_n|=2X_n^+-X_n$, we have  $$E[|X_n|]=2E[X_n^+]-E[X_n]\leq 2E[X_0^+]-l\leq 2E[|X_0|]-l$$  which implies $\sup_nE[|X_n|]&lt;\infty$.  Secondly, $\forall A\in\mathcal{F}$, we can suppose without loss of generality that $A\in\mathcal{F}_0$.  $$E[1_{A}|X_n|]=2E[[1_{A}X_n^+]-E[[1_{A}X_n]$$  Does someone have an idea for this term? Thanks a lot.....
97524
Let $T_n$ be a random variable with $T_n=X_1+...+X_n$ where the $X_i$'s are iid. Further we set $N(t)=max\{ n: T_n\leq n\}$ with the property $\Pr(N(t)&lt;\infty)=1$.  I want to prove that $\mathbb{E}[N(t)]&lt;\infty$.   I have done this so far. Let $F_n$ be the distribution function of $T_n$. $$\mathbb{E}[N(t)]=\sum_{n=1}^{\infty} n P(N(t)=n)=\sum_{n=1}^{\infty} n (F_n(t)-F_{n+1}(t))=\sum_{n=1}^{\infty} F_n(t)$$  Further I know that $0 \leq F_n \leq 1$. But does this implies $\mathbb{E}[N(t)]&lt;\infty$ ? I don't think so... I don't know what I can do further...   Can anybody help me how I can proof it?   Thank you for any help!!!