Assume X and Y are independent exponential random variables with parameters $\lambda_1$ and $\lambda_2$ respectively. Compute $P(X<Y)$.
I have alot of trouble solving questions like these. Is there a general way to tackle problems like these? Here's my approach
$$ \begin{align}
P(X<Y) &= \int_{-\infty}^{\infty} P(X<Y|Y=y) ~\ f_Y(y) dy\\
&= \int_{-\infty}^{\infty} P(X<y|Y=y) ~\ f_Y(y) dy\\
&= \int_{-\infty}^{\infty} P(X<y) ~\ f_Y(y) dy ~~~~~~~~~~~~\ \text{by independence}\\
&= \int_{-\infty}^{\infty} F_X(y) \cdot f_Y(y) dy ~~~~~~~~~~~~\ \text{not sure what to do after this}\\
&= \int_{-\infty}^{\infty} \int_\infty^y f_X(x) \cdot f_Y(y) ~\ dxdy ~~~~~~~~~~~~\ \text{yes?}\\
&= \int_{-\infty}^{\infty} \int_\infty^y \lambda_1 e^{-\lambda_1 y} \cdot \lambda_2 e^{-\lambda_2 y} ~\ dxdy \\
&= \int_{-\infty}^{\infty} \int_\infty^y \lambda_1 \lambda_2 e^{-(\lambda_1 + \lambda_2) y} ~\ dxdy \\
…
\end{align}
$$
Would like any feedback + if u know any good links, let me know!
Best Answer
You're on the right track, but it should be $f_X(x)$ rather than $f_X(y)$ in the third-to-last line. Also, the lower bound for the integrals should be $0$ since an exponential random variable is always positive.
It might be simpler to just use the fact that $$ F_X(y)=1-e^{-\lambda_1y}$$ if $y>0$, then compute the $y$ integral.