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 May13 comment If the sum of two independent random variables is in $L^2$, is it true that both of them are in $L^1$? @Batman That's why proving $X$ and $Y$ are in $L^1$ should be an easier task, isn't it? May13 comment If the sum of two independent random variables is in $L^2$, is it true that both of them are in $L^1$? Can you give an example s.t. $\mathbb E(X+Y)^2 < \infty$ but $\mathbb E X = \infty$ and $\mathbb E Y = \infty$? May12 comment 1D biased random walk - is the event of infinte many returns a tail event? You should check Hewitt–Savage zero–one law - en.wikipedia.org/wiki/… May7 comment hat problem and probability If the colors are assigned at equal probability and independently, you will not be able do better than 1/2. Nov17 comment The limit of integer valued random variables must be integer valued? OK. BTW, does my example shows that the portmanteau theorem works only for random variables on $\mathbb R$? Nov17 comment The limit of integer valued random variables must be integer valued? It's not clear whether the author meant to define $D_n$ and $D$ in $\mathbb R$ or compactification of $\mathbb R$. That's a question that comes to me again and again -- When the author says "random variables", does he/she allow them to take $\infty$ as value or not? Nov17 comment The limit of integer valued random variables must be integer valued? Thanks. I'm actually aware of this theorem and it is how I thought I had proved it. But when I came up with the "counter example" that I got confused again. Can we say that $\infty \in \mathbb Z$? Aug31 comment Martingale with bounded increment. @Did, I will remember to do that next time! Jun19 comment Is it possible to have $\mathbb E S_n \to \infty$ but $\inf S_n = -\infty$ a.s.? @ByronSchmuland Hmm, you're right. $S_n / n \to 2p - 1 > 0$ almost surely, so $S_n \to \infty$ almost surely Jun19 comment Is it possible to have $\mathbb E S_n \to \infty$ but $\inf S_n = -\infty$ a.s.? @ByronSchmuland Yeah, you're right about case 1. It should be $S_n = 0$ for all $n$. Jun17 comment How many ways can $5$ rings be placed on $4$ fingers? Shouldn't this goes to infinity? Jun16 comment combinatorics - Distribution of Distinct Balls into Distinct Boxes By nPk do you mean $\binom n k$? Jun3 comment Scheffe’s Theorem @ChrisJanjigian I see, it is $1 \le n \le \infty$. Thanks! Jun3 comment Scheffe’s Theorem @ChrisJanjigian How do we get $\int f_\infty = 1$? Dominated convergence theorem? May31 comment Difference between $P(Y \ge y)$ and $P(Y > y)$. Thanks! Somehow I believed the sum should starts from $0$ without much thinking. May30 comment Difference between $P(Y \ge y)$ and $P(Y > y)$. As for the sum, I was thinking if $>$ and $\ge$ versions of Lemma 2.2.8 is equivalent. Then we should have $\mathbb E(Y) = \int_0^\infty P (Y \ge y) \, \mathrm dy = \sum_{y \ge 0} P(Y \ge y)$, for $Y$ takes only integer values which seems to be wrong. May30 comment Difference between $P(Y \ge y)$ and $P(Y > y)$. So actually we can replace $>$ with $\ge$ in the proof? I'm confused because latter in the book, exercise 2.2.7, a generalized version of Lemma 2.2.8 is given, with $>$ replaced by $\ge$, that's why I'm considering whether there are difference between the two. May30 comment Difference between $P(Y \ge y)$ and $P(Y > y)$. But the integral $\int_0^\infty py^{p-1} 1_{(Y > y)} \, \mathrm dy$ should not change if we replace $>$ with $\ge$, right? This is an integral with respect to Lebesgue measure, which shouldn't change when omitting a single point. May30 comment Difference between $P(Y \ge y)$ and $P(Y > y)$. As you pointed out in the link, the sum should start from $y \ge 1$, not $y \ge 0$. Mar25 comment What does it mean to say “the random variable $X$ conditioned on $X$ being non-negative”? Seems that this is the only possible explanation.