Inequality involving independent random variables

Question

We have independent random variables $X$ and $Y$. Why does the following equality hold? $$P(X>Y)=\int P(X>Y\mid Y=z)f_Y(z) \, dz,$$ where $f_Y$ is the density of $Y$. Any ideas?

Answer 1

It holds because of the law of total probability. I think here it tells us that, it computes the probability of $X$ when bigger than $Y$ for all values of $Y$. But since, both events are independent,

\begin{equation} P(A \cap B) = \frac{P(A\mid B)}{P(B)} = \frac{P(A) \cdot P(B)}{P(B)} = P(A) \end{equation}

it can be reduced as,

\begin{equation} f(X > Y) = \int P(X > Y = z) f_{Y} (Y = z) \, dz \end{equation}

Assuming that $P(X)$ is a discrete function and $f_Y(y)$ a continuous one, the law of total probability can be expressed as,

\begin{equation} P(X=x) = \int P (X=x\mid Y=y) f_Y(Y=y) \, dy \end{equation}