Find an error in the convolution of 2 pdfs

Question

$X\sim \exp(2)$ and $Y\sim \exp(3)$ I was asked to find $\mathbb{P}(X>Y)$ but this is equivalent to $\mathbb{P}(X-Y>0)$. Pdf's of $X$ and $-Y$ are $f_X(x)=2e^{-2x}$ and $f_{-Y}(y)=-3e^{-3y}$ respectively. So let $Z=X-Y$ then

$ \begin{align} f_Z(t)=f_{X-Y}(t)=f_X*f_{-Y}(t)&=\int_\mathbb{R} -3e^{-(3t-x)}\mathbb{1}_{[0,\infty)}(t-x)2e^{-2x}\mathbb{1}_{[0,\infty)}(x) \\&= \int_\mathbb{R} -3e^{-3x}\mathbb{1}_{(-\infty,t)}(x)2e^{-2x}\mathbb{1}_{[0,\infty)}(x) \\&=\int_\mathbb{R}-6e^{-5x}\mathbb{1}_{[-\infty,t]\cap[0,\infty)}(x) \\&=\int_0^t-6e^{-5x} \\&= \frac{6}{5} (-1 + e^{-5 t})\end{align}$

So $$F(t)=\int_0^t\frac{6}{5} (-1 + e^{-5 t})=\frac{6}{5} (-1 + e^{-5 t}) t$$

but then as $t\to\infty$ one gets that $F(t)$ diverges $$\lim_{t\to \infty}F(t)=\lim_{t\to\infty}\frac{6}{5} (-1 + e^{-5 t}) t =\infty\neq 1$$

So $F(t)$ is not a CDF and in consequence $f_Z(t)$ not PDF.

What did I do wrong?

Answer 1

$f_{X}\left(x\right)=2e^{-2x}\mathbf{1}_{\left(0,\infty\right)}\left(x\right)$ and $f_{-Y}\left(y\right)=f_{Y}\left(-y\right)=3e^{3y}\mathbf{1}_{\left(0,\infty\right)}\left(-y\right)$ leading to:

$$\left(f_{X}\ast f_{Y}\right)\left(t\right)=\int2e^{-2z}\mathbf{1}_{\left(0,\infty\right)}\left(z\right)3e^{3t-3z}\mathbf{1}_{\left(0,\infty\right)}\left(3z-3t\right)dz=6e^{3t}\int_{\max\left\{ 0,t\right\} }^{\infty}e^{-5z}dz$$

Then we discern the cases $t>0$ and $t\leq0$.

• If $t>0$ then $\left(f_{X}\ast f_{Y}\right)\left(t\right)=6e^{3t}\int_{t}^{\infty}e^{-5z}dz=6e^{3t}\times\frac{1}{5}e^{-5t}=\frac{6}{5}e^{-2t}$.
• If $t\leq0$ then $\left(f_{X}\ast f_{Y}\right)\left(t\right)=6e^{3t}\int_{0}^{\infty}e^{-5z}dz=\frac{6}{5}e^{3t}$

Then: $$P(X>Y)=P(X-Y>0)=\int\frac{6}{5}e^{-2t}dt=\frac35$$

It is more handsome here to go for:$$P(X>Y)=\int_0^{\infty} P(X>Y\mid Y=y)f_Y(y)dy=$$$$\int_0^{\infty} P(X>y)f_Y(y)dy=\int_0^{\infty}e^{-2y}3e^{-3y}dy=\frac35$$