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 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 accepted Difference between $P(Y \ge y)$ and $P(Y > y)$. 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 $y \ge 0$. May30 revised Difference between $P(Y \ge y)$ and $P(Y > y)$. typo May30 asked Difference between $P(Y \ge y)$ and $P(Y > y)$. May3 awarded Announcer Apr14 awarded Popular Question 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. Mar25 accepted What does it mean to say “the random variable $X$ conditioned on $X$ being non-negative”? Mar25 asked What does it mean to say “the random variable $X$ conditioned on $X$ being non-negative”? Mar5 awarded Benefactor Mar5 accepted Martingale and bounded stopping time Mar1 asked Lower bound on building heap. Feb27 comment Martingale and bounded stopping time @ByronSchmuland Yes. Feb27 comment Martingale and bounded stopping time @ByronSchmuland Probability: Theory and Examples, 4th edition Feb27 awarded Promoter Feb26 comment Martingale and bounded stopping time I gave my method another try and it didn't work. So I decided to offer a bounty.