Asymptotic probability of difference of two positive random variables

Question

Assume I have two positive, iid random variables, $X$ and $Y$. I need to compute $P\{X-Y > u\}$. I was thinking of doing $P\{X > Y + u \} = P\{X > v\}$, where $v = Y+u$. (Reason: Since $Y$ and $u$ are positive, their sum will also be positive and hence it should not matter if I replace the right hand side by $v$.)

Is this the right way to approach the problem? If yes, is my argument technically sound? If not, what am I missing?

edit: $X$ and $Y$ have stable distributions

Answer 1

It is always stable with the same exponent. Let me sum up the comments.

Let $X$ has characteristic function $f(t)$, then

$Ee^{it(X-Y)} = Ee^{itX}Ee^{-itY} = f(t)f(-t)$ by independence.

Now looking at the form of form of the characteristic function on http://en.wikipedia.org/wiki/Stable_distribution

we get $f(t)f(-t) = \exp[-2|ct|^{\alpha}]$

This means this is a stable distribution with these parameters.

stability parameter $\alpha$, skewness is 0, scale parameter is $2^{\frac{1}{\alpha}}c$ and location parameter 0.

Answer 2

An alternative way of trying this is to let $F,f$ be the cdf,pdf of $X$ (and hence of $Y$) and write (assuming continuity) $$P(X-Y>u) = P(Y<X+u) = \int F(x+u) f(x) dx.$$ So if you know the distribution of $X$ then you're done.