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Aug
24
revised Summing the binomial pmf over $n$, part 2
Simplified the expression for clarity.
Aug
23
asked Summing the binomial pmf over $n$, part 2
Aug
23
comment Summing the binomial pmf over $n$
Looking at this a second time, here's a follow-up: what if I wanted to add an extra factor of $n^{-\alpha}$ into my original sum, where $\alpha > 1$? So essentially I need to compute: $$ \sum_{n \geq k} {n \choose k} q^{n-k} n^{-\alpha} $$ Would going down the road of generating functions still be fruitful here? (Just looking for a simple yes/no to provide direction.)
Aug
22
comment Summing the binomial pmf over $n$
I absolutely appreciate it. Thanks so much.
Aug
22
comment Summing the binomial pmf over $n$
This is one of those times where I really, really wish I could accept more than one answer. Love it. Thanks.
Aug
22
accepted Summing the binomial pmf over $n$
Aug
22
comment Summing the binomial pmf over $n$
Generally speaking, the dumber an answer makes me feel (all in hindsight, granted), the better it is. This one is right there at the top.
Aug
22
revised Summing the binomial pmf over $n$
edited body
Aug
22
asked Summing the binomial pmf over $n$
Aug
21
comment Transforming a sequence to distinguish a limit
@DavidC.Ullrich Hmm. I see, too bad. I've added an additional detail...
Aug
21
revised Transforming a sequence to distinguish a limit
added 287 characters in body
Aug
17
asked Transforming a sequence to distinguish a limit
Aug
11
revised Expected value of a piecewise function in two variables
changed to \cases
Aug
11
suggested approved edit on Expected value of a piecewise function in two variables
Aug
11
asked Central limit theorem: where is the martingale in this proof?
Aug
11
accepted Martingale CLT: “without loss of generality”?
Aug
11
comment Martingale CLT: “without loss of generality”?
Ah, OK. I see. I forgot that $t \in [0,1]$ so that I was wondering how this could be true with the variance converging to an increasing sequence. I see now. Thanks.
Aug
11
comment Stopped Brownian motion proof
@saz: No, you're right. Thanks for that. I see your comment, but I guess I'm just asking about simple arithmetic where he substitutes the last difference $t_M$ rather than just $T_M$, in: $\sum_{m=1}^{\infty} E(t_m; M \geq m)$ to $\sum_{m=1}^{\infty} E(E(t_M | \mathcal{B}(T_{m-1} )); M \geq m)$.
Aug
10
asked Martingale CLT: “without loss of generality”?
Aug
10
comment Stopped Brownian motion proof
@saz: I don't quite follow what equation you're referring to. I see where Durrett works with this later in the proof but I'm not sure how this relates to the identity in my post. Sorry if I'm just being slow.