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 $\displaystyle A=\{a<X_1<b~,~c<X_2<d\}$ 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
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.