# Sequence of random variables

1. Mar 31, 2015

### Dassinia

Hi, I'm trying to solve this exercise but I really don't know how
1. The problem statement, all variables and given/known data

Let X1, X2,.. be a sequence of iid random variables following a uniform distribution on (0,1). Define the random variable N≥2 as the first point in which the sequence (X1,X2,...) stops decreasing. i.e If N=n :
X1[PLAIN]http://www.ilemaths.net/img/smb-bleu/supegal.gifX2[PLAIN]http://www.ilemaths.net/img/smb-bleu/supegal.gif...[PLAIN]http://www.ilemaths.net/img/smb-bleu/supegal.gifXn-1<Xn [Broken]

2.P(X1[PLAIN]http://www.ilemaths.net/img/smb-bleu/infegal.gift,N [Broken] pair)=1-exp(-t) use series expansion of exp(t)+exp(-t) et exp(t)-exp(-t)
3. E[N]=e

2. Relevant equations

3. The attempt at a solution
I solved the third one
For the first one, I think that we have to use induction proof, but I don't see how to do that here ?
Thanks

Last edited by a moderator: May 7, 2017
2. Mar 31, 2015

### Ray Vickson

For the first one, try out some simple cases first: do it for n = 2, n = 3, etc. You will soon see how to do the general case.

What you wrote for 1) is wrong: you say
$$P(X_1 \leq t, N=n) = tn - \frac{1}{(n-1)!} - \frac{tn}{n!}$$
when your expression is parsed using standard mathematical rules. I suspect you might have meant
$$\frac{t^{n-1}}{(n-1)!} - \frac{t^n}{n!}$$
In that case, you MUST use "^" signs and parentheses: tn means $t \times n$, but t^n means $t^n$. Similarly, t^n-1 means $t^n - 1$, but t^(n-1) means $t^{n-1}$.

Finally, I do not understand what "X1 ≤ t, N pair" means---that is, what is "N pair"?

Last edited by a moderator: May 7, 2017