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?

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- June 26th 2011, 12:56 PMg322Bound 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?

- June 26th 2011, 01:23 PMMooRe: Bound probability of sum of two variables
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)>= pē - June 26th 2011, 06:13 PMg322Re: Bound probability of sum of two variables
Pr(a+c<y<b+d)>= pē is too conservative because pē<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.

- June 27th 2011, 07:24 AMGuyRe: Bound probability of sum of two variables
- June 30th 2011, 03:08 AMMooRe: Bound probability of sum of two variables