# Two Probability Questions

1. Oct 7, 2009

### e(ho0n3

The problem statement, all variables and given/known data
Let {Xi} be a sequence of uniform random variables on [0,1]. What is P(sup {Xi} = 1) and P(Xi = Xj), i ≠ j?

The attempt at a solution
If sup{Xi} = 1, then either Xi is 1 for some i or there is an increasing sequence of the Xi's that converges to 1. The probability of the former is 0. I don't know how to calculate the probability of the latter.

For the second probability, if things were finite I would condition on the value of Xj and compute the probability that way. But in this case, I have no idea what to do.

2. Oct 7, 2009

### jbunniii

Consider $P(\sup\{x_i\} \neq 1)$. If the supremum is not 1, it must be less than one, say $\sup\{x_i\} = \alpha < 1$. If this is the case, then none of the $x_i$ can be greater than $\alpha$. What is the probability of that?

For the second question, note that $x_i = x_j$ if and only if $x_i - x_j = 0$.

3. Oct 7, 2009

### e(ho0n3

Good tip. For a particular i, $P(X_i > \alpha) = 1 - P(X_i \le \alpha) = 1 - \alpha$, so the probability that $X_i < \alpha$ for all i must be $\prod_i (1 - \alpha) = 0$. Hmm...this can't be right.

Right. I don't know why this didn't occur to me. Thanks.

4. Oct 7, 2009

### jbunniii

That's what I get. This shows that $P(\sup\{x_i\} \leq \alpha) = 0$ for all $\alpha < 1$. So $P(\sup\{x_i\} = 1)$ must be 1.

It makes sense that this must be the case. If I choose infinitely many uniform numbers in [0,1], why should ALL of these numbers avoid some interval $[\alpha,1]$ with nonzero probability? What's special about the rightmost edge of the interval? Nothing at all.

Last edited: Oct 7, 2009
5. Oct 7, 2009

### e(ho0n3

You're right. My intuition about such things is, as you can tell, not very good. Thanks.