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visits member for 3 years, 7 months
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Computer scientist student who is trying to study math and trying avoid asking stupid questions on this website.


Nov
17
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 $\mathbb R$?
Nov
17
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?
Nov
17
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 $\infty \in \mathbb Z$?
Aug
31
comment Martingale with bounded increment.
@Did, I will remember to do that next time!
Jun
19
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
Jun
19
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 $n$.
Jun
17
comment How many ways can $5$ rings be placed on $4$ fingers?
Shouldn't this goes to infinity?
Jun
16
comment combinatorics - Distribution of Distinct Balls into Distinct Boxes
By nPk do you mean $\binom n k$?
Jun
16
comment Convergence to exponential function.
I see my mistake!
Jun
3
comment Scheffe’s Theorem
@ChrisJanjigian I see, it is $1 \le n \le \infty$. Thanks!
Jun
3
comment Scheffe’s Theorem
@ChrisJanjigian How do we get $\int f_\infty = 1$? Dominated convergence theorem?
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
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 $y \ge 0$.
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.
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
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.