Why is if $S_n \sim H$ and $P(J_n \le x)=G(-\infty, x/r]$ are independent then $P(S_n-J_n \in A)=\int_0^\infty G(A/r+y/r)H(dy)$?

Question

Let $S_n$ and $J_n$ be independent random variables and $S_n, J_n \ge 0$. Assume that $S_n\sim H$ and $P(J_n \le x)=G(-\infty, x/r]$ for some $r>0$. Then why do we have $$P(S_n-J_n \in A)=\int_0^\infty G(A/r+y/r)H(dy)$$by the convolution form? I know that $P(S_n + J_n \in A)= \int_0^\infty G(A/r-y/r)H(dy)$ by the formula but I don't know how to apply the convolution form for $-J_n$ to get the above identity.

Answer 1

Should it be $$P(J_n - S_n)?$$

There is an intuitive and a formal approach.

Intuitively, how you obtain the second formula? You assume that $S_n$ is in $[y,y+dy)$, which happens with the probability denoted as $H(dy)$, and then consider the conditional on this probability that $S_n + J_n$ is in $A$, i.e. $J_n$ is in $A - y - O(dy)$, or $J_n / r$ is in $A/r - y/r - O(dy)$, which happens with the probability $G(A/r - y/r) + O(dy)$, because $$P(J_n / r <= x) = G(-\infty, x].$$ Ignoring the second order term, we obtain the integral formula.

This way, for the first formula, arguing similar, in the end we have $$P(J_n / r \in A/r + y/r + O(dy)) = G(A/r + y/r) + O(dy),$$ and, hence, we have this instead in the integral.

Formally, you may consider $T_n = -S_n$, and $K(B)=H(-B)$, then $$\begin{align} P(J_n - S_n \in A) &= P(T_n + J_n \in A) \\ &= \int_{-\infty}^0 G(A/r - y/r)K(dy) \\ &= \int_0^\infty G(A/r + y/r)H(dy).\end{align}$$