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
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.