Thread: Bound probability of sum of two variables

If x1 and x2 are independent, Pr(a<x1<b)=p, Pr(c<x2<d)=p, and y=x1+x2. can we prove that Pr(a+c<y<b+d)>= p? I have proved it if x1 and x2 are normally distributed. Is this also true for general distributions?

Hello,

Consider and

We have that (by independence).


We also know that

Then recall that if we have an implication A => B, then

This lets us write that Pr(a+c<y<b+d)>= p²

Pr(a+c<y<b+d)>= p² is too conservative because p²<p. It can be shown that if x1 and x2 are independently and normally distributed, Pr(a+c<y<b+d)>= p. I am wondering if this is also true for other distributions.

Originally Posted by g322

Pr(a+c<y<b+d)>= p² is too conservative because p²<p. It can be shown that if x1 and x2 are independently and normally distributed, Pr(a+c<y<b+d)>= p. I am wondering if this is also true for other distributions.

False in general. Take to be independent exponential rvs and a = c = 0, b = d = 1.

Moo's work suggests to me that his bound is the best one can expect without more assumptions.

Originally Posted by Guy

Moo's work suggests to me that his bound is the best one can expect without more assumptions.

There is no doubt about it ! It's my work your talking about, constantly skimming perfection and exactness.

(hey, I'm just kidding)