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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- Jun 26th 2011, 11:56 AMg322Bound 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?

- Jun 26th 2011, 12:23 PMMooRe: Bound probability of sum of two variables
Hello,

Consider $\displaystyle A=\{a<X_1<b~,~c<X_2<d\}$ and $\displaystyle B=\{a+c<Y<b+d\}$

We have that $\displaystyle P(A)=P(a<X_1<b)P(c<X_2<d)=p^2$ (by independence).

We also know that $\displaystyle \begin{cases} a<X_1<b \\ c<X_2<d \end{cases}\Rightarrow a+c<Y<b+d$

Then recall that if we have an implication A => B, then $\displaystyle P(A)\leq P(B)$

This lets us write that Pr(a+c<y<b+d)>= pē - Jun 26th 2011, 05:13 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.

- Jun 27th 2011, 06:24 AMGuyRe: Bound probability of sum of two variables
- Jun 30th 2011, 02:08 AMMooRe: Bound probability of sum of two variables