Recent content by Moo1

A monkey that can probably speak a better English than yours, but that wouldn't bother looking for the translation of an Italian sentence no one cares about. And I did mean wording, not working. Sincerely yours, Monkey cow.

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I'm afraid your wording is as wrong as ever. You're always putting some weird notations and it's quite obvious that you're not used to using common things in probability. Hence the probability that you understood wrongly is superior to the probability of the question being wrongly worded.

(Surprised)

Well it'd be nice if you shared it with us, for some people may need to have this kind of proof at hand :) (although I personally don't need it (Evilgrin))

Hello, A bit late, but I can answer. Don't condition by X_1, but by X_{n-1}. Then note that S_n=\sum_{i=1}^{X_{n-1}} Z_i^{(n)} to write the expectation of a product : E\left[\prod_{i=1}^{X_{n-1}} s^{Z_i^{(n)}}\bigg|X_{n-1}\right] and finish it off.

In France, there's the Poincaré's institute, but you've got to be really good :D I think it's the same for Princeton's IAS

It's like if you ask why when you mix white with black, you obtain something nearer to black than white (in equal quantities). By the way, there isn't only one gene determining eye or hair color.

That's almost perfect. For the latter one, it's $E[e^{B_1+B_2}]=E[e^{B_1}]E[e^{B_2}]=e^{1/2}e^{2/2}=e^{3/2}$ Maybe you ought to put more details for the step, especially for (e). See, not that difficult ? :p

Yep, that's a good start ! :)

I'm not scarce on details because it could be an assignment, but trust me you just need the basic properties of a brownian motion to solve these questions. They can be found here : http://en.wikipedia.org/wiki/Brownian_motion#Mathematics (it can be B or W, most of the time, people don't make any...

Hello, Yes, B indeed stands for a brownian motion, so you have to keep the basic properties in mind. B0 = 0 Bt-B0 follows a normal distribution with mean 0 and variance (not standard deviation) t. So in order to solve most of these questions, you need to know the moments of a normal...

Well if you want to compute it without mentioning gamma function, it's possible, but you'd have to do successive integrations by parts. But this is indeed the solution. And you'd recognize a geometric distribution because : $\displaystyle \frac{\mu}{(1+\mu)^{k+1}}=\frac{\mu}{1+\mu}\cdot...

That's exactly it for the formula ! But your computation of the integral isn't correct. Recall that k is a positive integer and that $\displaystyle k!=\Gamma(k+1)=\int_0^\infty e^{-t} t^k ~dt$ Make the proper substitution to get $e^{-b(1+\mu)}$ instead of $e^{-t}$ and it'll be all good ! :) I...

An exponential distribution is continuous. Its pdf is $\mu e^{-\mu b}$ So the expectation you're looking for is just $\displaystyle \int_0^\infty \mu e^{-\mu b}\cdot e^{-b}\cdot\frac{b^k}{k!} ~db$, where k is a constant. If you didn't know this formula, I'll explain it later, I have to go sleep