Ytukyg

Suppose we have two random variables $X$ and $Y$, that are independent. Suppose they have the associated distribution functions $F(x)$ and $G(y)$. Now, assume that $X$ and $Y$ are uniformly distributed. More specifically, let $X$ be uniformly distributed in $[-phi,phi]$ $Y$ be uniformly distributed in interval $[0,bar{y}]$.

Given some constants, $a,b$, how would we compute this probability?

$p(X < a, X < Y-b)$

First, I can see that this is the same as $p(X leq min{a, Y-b})$

So we would have:

$int_x int_y mathbb{1}[x leq min{a, y-b}] dG(y) dF(x)$

If we assume that $bar{y}$ is sufficiently large, how would we compute this integral?

Let $[x<min(a,y-b)]$ denote the function $mathbb R^2tomathbb R$ that takes value $1$ if $x<min(a,y-b)$ and takes value $0$ otherwise.

Then:

$$begin{aligned}P(X<a,X<Y-b) & =mathbb{E}[X<min(a,Y-b)]\
& =intint[x<min(a,y-b)]dF(x)dG(y)\
& =int F(min(a,y-b))dG(y)\
& =int_{-infty}^{a+b}F(a)dG(y)+int_{a+b}^{infty}F(y-b)dG(y)\
& =Fleft(aright)Gleft(a+bright)+int_{a+b}^{infty}F(y-b)dG(y)
end{aligned}
$$

In second equality independence is used.

In third equality it is used that $F$ is a continous CDF.

Apply this to your mentioned special case.