Let $X_{1}, X_{2}, X_{3}, ...$ be a sequence of iid and integrable random variables with $E[X_{1}] = \mu$. Show that $$ \frac{1}{n} (X_{1}+X_{2}+X_{3}+...+X_{n}) \to \mu   $$ in $L^{1}$.  My thoughts so far: Strong law of large numbers implies almost surely convergence. So I still need to get from a.s. convergence to $L^{1}$ convergence. In my lectrure notes I have found two possible ways of doing so:  (a) If $\quad$ $X_{n} \to X \; a.s.$ $\quad$ and $\quad$ $E(|X_{n}|^{p}) \to E(|X|^{p})$ $\quad$ then $\quad$$X_{n} \to X$ in $L^{p}$.  (b) If $\quad$ $X_{n} \to X \;$  in prob.  $\quad$ and $\quad$ $\{X_{n} : n \in \mathbb{N}\}$  is UI $\quad$ then $\quad$$X_{n} \to X$ in $L^{1}$.  Which is more promising? I tried both but...it didnt really work out...  Many thanks in advance.
122607
I've seen the  stated mainly in two (or three) forms: $S_n/n$ converges in probability (weak law) and converges almost surely (strong law).  Also, there is convergence in the $L^2$-norm for uncorrelated random variables ($L^2$ weak law). However there is a backwards martingale proof of the strong law of large numbers (any graduate level probability theory book, for example Durrett, should have it).  The important thing is that $M_{-n}:=S_n/n$ is a backward martingale, and backward martingales converge both a.e. and in the $L^1$-norm.  Then, in particular, $S_n/n$ converges in the $L^1$-norm.  Does $S_n/n$ really converge in the $L^1$-norm?  If yes, why is this never mentioned? If no, what is wrong with my above proof?
110503
There is a sequence of independent random Poisson variables $Y_1, Y_2,\ldots$ with divergent parameter $Y_i \sim \text{Poisson}(\frac{1}{n})$. Does this sequence converge almost surely? In $L^2$ space? In probability?  I am really confused as I did some research on my own and almost every question I come up to has to do with the sum of a sequence; but I do not believe this has to do anything with the sum of $Y_n$ right? to prove it converges almost surely, I have to prove that $Y_n$ approaches the limit of $e^{-1}*\sum_{i=1}^{\infty} 1/i$? Any hints would be appreciated. Thanks.
3091
This is a follow-on from my  about samples from a distribution.  Suppose $X_1 \ldots X_{n-1}, X_n$ are random variables all following some fixed distribution $D$. How do I prove that $P(X_n &gt; X_1 \wedge \ldots \wedge X_n &gt; X_{n-1}) = \frac{1}{n}$ (this is true intuitively and experimentally).
46295
I have a sequence of independent random variables $X_1, X_2, \ldots$ where  $$ X_n =   \begin{cases}     1      &amp; \quad \text{with probability} \  1/n  \\     0    &amp; \quad \text{with probability} \ 1-1/n   \end{cases} $$  In my book, it claims that while $X_n \to 0$ in probability, it is not true almost surely. The book gives a very short outline of the proof, but I am not sure if it is correct. It writes that: for $0 &lt; \epsilon &lt;1$, $P\left(\bigcup_{m=n}^\infty \{|X_m - X| &gt; \delta\}\right) = 1- \lim_{n \to \infty}P(X_m = 0 \ \forall m \text{ s.t. } n \leq m)$. I am not sure why this line can show that almost sure convergence to $0$ is not possible.   The definition of almost sure converge I take is to be: $\forall \epsilon, \delta &gt;0, \exists n_0=n_0(\epsilon, \delta) \text{ s.t. } P(\bigcup_{m=n}^\infty \{|X_m -X|&gt; \delta\})\leq \epsilon$. Can someone tell me what the author's line intends to show? Thank you
121382
Can somebody provide me with a sequence of (real-valued) functions, say on $[0,1]$ with the Lebesgue measure, such that the sequence converges in probability, or maybe in $\Vert \cdot \Vert _{L^2}$, but does not converge almost surely?
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?
38305
Assume that $x_n$ is a sequence such that $x_n$ converge to a constant $x$. I would like to prove that   $\left( 1+ x_n/n \right)^n \rightarrow e^x$  Obvious if $x_n = x$ the result would be given. The extra sequence complicates things and I would like a hint or solution on how to proof this.
32443
Exercise 5.3.1 in Durrett's "Probability Theory and Examples" states  Let $X_n$, $n\ge 0$, be a submartingale with $\sup X_n &lt; \infty$. Let $\xi_n=X_n-X_{n-1}$, and suppose $E(\sup \xi_n^+)&lt;\infty$. Show that $X_n$ converges almost surely.  Here are my thoughts: It looks like we need to use theorem 5.2.8 (Martingale convergence theorem), which states that $X_n$ converges almost surely if $\sup EX_n^+&lt;\infty$. I am trying to understand why we can just do the following (or at least I'm guessing we can't):  Suppose $X_n &gt; 0$ (edit: on some set with positive measure), then $0&lt; \sup X_n &lt; \infty$, and therefore $EX_n^+ \le \sup X_n &lt; \infty$. So we're done.  Otherwise, if $X_n \le 0$ a.e., then $\sup X_n^+ = 0$ a.e. Once again, we're done.  Am I missing something? Since I didn't use all the information, I probably am.  Thanks for the help.
80741
I'm trying to understand my teacher's solution to this problem:     Let $X_1,X_2,...$ sequence of iid random variables with expected value   $\mu$. Define $Y_n = \frac{2}{n(n+1)} \sum_{j=1}^{n} jX_j$. Prove that   $Y_n \to_\mathbb{P} \mu$.   His solution:  Let $\sigma^2 = \rm{Var}(X_i)$.  $E(Y_n) = \frac{2}{n(n+1)} \sum_{j=1}^{n} jE(X_n) = \mu$.  $\displaystyle\rm{Var}(Y_n) = \frac{4}{n^2(n+1)^2} \sum_{j=1}^{n} j^2 \rm{Var}(X_n) = \frac{4\sigma^2}{n^2(n+1)^2}\bigg(\frac{n(n+1)(2n+1)}{6}\bigg) = \frac{2\sigma^2}{3}\bigg(\frac{2n+1}{n(n+1)}\bigg) \leq \frac{2\sigma^2}{3}\bigg(\frac{2}{3}\bigg) = \sigma^2 ,\forall n$  Then it follows by Tchebychev's Weak Law of Large Numbers.  My doubt is in the last step. Isn't  Tchebychev's WLLN only true for uncorrelated random variables? In this case, $\mathbb{E}(Y_n Y_m) \neq \mu^2$, since they aren't independent, so $\rm{Cov}(Y_n,Y_m) \neq 0$, right?  Thanks.  Edit: As u/clarinetist noted, there's a problem in this exercise, since the WLLN states the convergence of the arithmetic mean of random variables, whose expected value isn't $\mu$.
26574
Suppose $X_{1},X_{2},\ldots$ be i.i.d. random variables such that $E\left[X_{i}\right]=\mu$ and $Var(X_{i})=\sigma^{2}&lt;\infty$. Let $\bar{X}=\left(X_{1}+\cdots+X_{n}\right)/n$. Show that $\frac{1}{n}\overset{n}{\underset{i=1}{\sum}}\left(X_{i}-\bar{X}\right)^{2}\rightarrow\sigma^{2}$ a.s.  Solution: Let $S_{n}=\frac{1}{n}\overset{n}{\underset{i=1}{\sum}}\left(X_{i}-\bar{X}\right)^{2}$. By Chebyshev's Inequality, $\sum_{n=1}^{\infty}P\left(\left|S_{n}-E\left[S_{n}\right]\right|&gt;\varepsilon\right)\leq\sum_{n=1}^{\infty}\frac{Var(S_{n})}{\varepsilon^{2}}$. And we know that $E\left[S_{n}\right]\rightarrow\sigma^{2}$, so I would like to finish this proof by using first Borel-Cantelli Lemma if I can show the right hand side is summable. However, $Var(S_{n})=\frac{2\sigma^{4}}{n-1}$, so the sum is infinity, which means it doesn't converge almost surely. Where am I wrong?
51013
Show that if $(x_n)$ is a convergent sequence then the sequence given by the averages $$y_n = \frac{x_1 + x_2 + \cdots + x_n}{n}$$ also converges to the same limit.  Attempt at Proof.   Since $(x_n)$ converges we can say that for all $m$, such that $m \ge N \Rightarrow |x_n - L|\lt \epsilon$.   Base Case. Let n=1 and we have for all m such that  $m \ge N_0 \Rightarrow |x_1 - L|\lt \epsilon$ and also for $m \ge N_1 \Rightarrow |x_{n+1} - L|\lt \epsilon$.  Induction Hypothesis. Assume that for an appropriate choice of $N_2$ we have for all m, $m \ge N_2 \Rightarrow |y_{n} - L|\lt \epsilon$.   Choose $max\{N_1,N_2\}$ such that for all $m \ge max\{N_1,N_2\}$ implies $$|\frac{x_1 + x_2 + \cdots + x_n}{n}- L|+ |x_{n+1} - L|\lt 2\epsilon$$  $$= |\frac{x_1 + x_2 + \cdots + x_n+ n x_{n+1}}{n}-2L |\lt 2\epsilon$$  $$= |\frac{x_1 + x_2 + \cdots + x_n+ n x_{n+1}}{n+1}-L |\le |\frac{x_1 + x_2 + \cdots + x_n+ n x_{n+1}}{2n}-L |\lt \epsilon$$ and $$|\frac{x_1 + x_2 + \cdots + x_n+  x_{n+1}}{n+1}-L |\lt|\frac{x_1 + x_2 + \cdots + x_n+ n x_{n+1}}{n+1}-L|  \lt \epsilon.$$  Is this approach correct? If not, can you please provide the correct proof. I feel as though I made a mistake in the calculations. Thanks in advance.
5355
Suppose $X$ is a non-negative random variable satisfying  \begin{align*} P[0 \le X &lt; \infty ]=1. \end{align*} Show   a)  \begin{align*} \lim_{n \to \infty} n E \left( \frac{1}{X}1_{[X&gt;n]} \right) =0 \end{align*} b)  \begin{align*} \lim_{n \to \infty} n^{-1} E \left( \frac{1}{X}1_{[X&gt;n^{-1}]} \right) =0 \end{align*}  What I did  I showed part a) already here is the proof:  \begin{align*} n E \left( \frac{1}{X}1_{[X&gt;n]} \right) \le n E \left( \frac{1}{n}1_{[X&gt;n]} \right)=n\cdot\frac{1}{n}P[X&gt;n]=P[X&gt;n] \end{align*} So $\lim_{n \to \infty} n E \left( \frac{1}{X}1_{[X&gt;n]} \right) \le \lim_{n \to \infty} P[X&gt;n]=0$.  What about part b)? The same techniques no longer works.  I was thinking that  \begin{align*} \lim_{n \to \infty} n E \left( \frac{1}{X}1_{[X&gt;n]} \right) \le \lim_{n \to \infty} n E \left( \frac{1}{X} \right) \le 0 \end{align*} but I am not sure that $E \left( \frac{1}{X} \right) &lt; \infty$
112504
I've already proved that for any $p &gt; 0$ and for any $\alpha \in \mathbb{R}$, the sequence $\frac{n^\alpha}{(1 + p)^n}$ converges to $0$.  Now, I want to prove that $\lim_{n \to \infty} x^n = 0$ as long as $|x| &lt; 1$.  I've split it into two possible cases:  (1) $0 &lt; x &lt; 1$.  Let $p = \frac{1}{x} - 1$, so that $x = \frac{1}{1 + p}$ and $p &gt; 0$ (since $x &lt; 1$). Then we let $\alpha = 0$ to see that $$\lim_{n \to \infty} x^n = \lim_{n \to \infty} \frac{n^0}{(1 + p)^n} = 0$$  (2) $-1 &lt; x &lt; 0$.  Here we cannot use the sequence $\frac{n^\alpha}{(1 + p)^n}$ again, since if we let $p = \frac{1}{x} - 1$, then $p$ isn't necessarily positive (for example, if $x = - \frac{1}{2}$, then $p = -3 &lt; 0$).  How can I prove Case (2)?
18290
Problem Prove that the sequence $$x_n=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\ln n,~~~(n=1,2,\cdots)$$is convergent. One Proof This proof is based on the following inequality  $$\frac{1}{n+1}&lt;\ln \left(1+\frac{1}{n}\right)&lt;\frac{1}{n}$$  where $n=1,2,\cdots$, which will be used repeatedly. On one hand, we obtain that $$\ln 2-\ln 1&lt;1,~~\ln 3-\ln 2&lt;\frac{1}{2},~~\ln 4-\ln 3&lt;\frac{1}{3},~~\cdots,~~\ln (n+1)-\ln n&lt;\frac{1}{n}.$$ Adding up all of these，we have that $\ln(n+1)&lt;1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}.$ Hence,$$x_{n+1}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}+\frac{1}{n+1}-\ln(n+1)&gt;\frac{1}{n+1}&gt;0.$$ This shows that $x_n$ is bounded below. On the other hand,$$x_n-x_{n+1}=-\frac{1}{n+1}+\ln(n+1)-\ln n=\ln \left(1+\frac{1}{n}\right)-\frac{1}{n+1}&gt;0.$$ This shows that $x_n$ is decreasing. Combining the two aspects, according to Monotone Bounded Theorem, we can assert that $\lim\limits_{n \to \infty}x_n$ exists. Let $\gamma$ (so-called Euler–Mascheroni Constant) denote the limit, i.e. $$\gamma=\lim_{n \to \infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\ln n\right),$$which equals $0.577216 \cdots$. We may also express that as $$1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}=\gamma+\ln n+\varepsilon_n,$$where $\varepsilon_n$ represents an infinitesimal related to $n$ under the process $n \to \infty$.
115036
If I have an i.i.d. random variables $X_i$ from a uniform distribution on $[0,1]$.   How would I find scaling sequences $a_n$, $b_n$ such that $a_n(M_n − b_n)$ converges in distribution to a non-trivial limit function $G$ for  $ Y_i = X_i,$ and $M_n = \max(Y_1,...,Y_n) $
111865
Problem: Let $(x_i)$ be a sequence in (0,1) such that $\lim_{n \to \infty} (1/n)\sum_{i=1}^n x_i^k$ exists for every $k=0,1,2,...$ and $f\in C[0,1]$. Show that $\lim_{n \to \infty} (1/n)\sum_{i=1}^n f(x_i)$ exists.  I've used the fact that $B_n(f)$ converges uniformly to $f$ to get   $\lim_{n \to \infty} (1/n)\sum_{i=1}^n f(x_i) = \lim_{n \to \infty} (1/n)\sum_{i=1}^n \sum_{k=0}^n f(\frac{k}{n}) {n \choose_k }x_i^k (1-x_i)^{n-k}$  And then you can arrange to get   $\lim_{n \to \infty} (1/n) \sum_{k=0}^n \left[ f(\frac{k}{n}) {n \choose_k }\sum_{i=1}^nx_i^k (1-x_i)^{n-k}\right]$  I want to use the fact that $(1-x_i) &lt; 1$ to consolidate some of the equation but I get nowhere.
26150
Show that. for any discrete random variable X that takes on values in the range [0,1]. Var[X] $\le$  1/4.  I translate it into a inequality like this: $x_1, x_2, x_3 \cdots ,x_n$ where $0 \le x_i \le 1$, and $p_1, p_2, p_3 \cdots ,p_n$ where $p_1+ p_2+ p_3 \cdots +p_n = 1$, prove that $\sum _1^nx_i^2p_i - (\sum _1^n x_ip_i)^2 \le {1\over 4}$ , how to prove it?  At first I tried cauchy inequatlity, but I fail :(