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seen Dec 13 at 14:59

Student.


May
31
comment Difference between $P(Y \ge y)$ and $P(Y > y)$.
Thanks! Somehow I believed the sum should starts from $0$ without much thinking.
May
30
accepted Difference between $P(Y \ge y)$ and $P(Y > y)$.
May
30
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.
May
30
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.
May
30
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.
May
30
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 $y \ge 0$.
May
30
revised Difference between $P(Y \ge y)$ and $P(Y > y)$.
typo
May
30
asked Difference between $P(Y \ge y)$ and $P(Y > y)$.
May
3
awarded  Announcer
Apr
14
awarded  Popular Question
Mar
25
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.
Mar
25
accepted What does it mean to say “the random variable $X$ conditioned on $X$ being non-negative”?
Mar
25
asked What does it mean to say “the random variable $X$ conditioned on $X$ being non-negative”?
Mar
5
awarded  Benefactor
Mar
5
accepted Martingale and bounded stopping time
Mar
1
asked Lower bound on building heap.
Feb
27
comment Martingale and bounded stopping time
@ByronSchmuland Yes.
Feb
27
comment Martingale and bounded stopping time
@ByronSchmuland Probability: Theory and Examples, 4th edition
Feb
27
awarded  Promoter
Feb
26
comment Martingale and bounded stopping time
I gave my method another try and it didn't work. So I decided to offer a bounty.