If $X\mid Y$ is exponentially distributed, and $Y$ is exponentially distributed, find the distribution of $X$

Question

Say $X \mid Y \sim \mathrm{Exp}(\lambda)$, can we use the memoryless property and say $X \sim \mathrm{Exp}(\lambda)$?

Answer 1

No. Use the Law of Total Probability.

If $Y\sim\mathcal {Exp}(\lambda)$ and $X\mid Y\sim\mathcal{Exp}(Y)$, then

$$\mathsf P(X>a) ~{=\mathbf 1_{a<0}+\mathbf 1_{a\geq 0}\cdot\int_\Bbb R \mathsf P(Y=y)\,\mathsf P(X>a\mid Y=y)~\mathsf d y\\=\mathbf 1_{a<0}+\mathbf 1_{a\geq 0}\cdot\int_0^\infty \lambda e^{-\lambda y}\cdot e^{-ya}~\mathsf d y}$$

Which simplifies to an interesting expression.