Let $Y|M = m$ be a discrete random variable with probability function $P(Y = y|M = m) = \frac{e^{-m}m^y}{y!}$ for $y \in \{0, 1, ...\}$ and $M$ a continuous random variable with density function $f_M(m) = e^{-m}$ for $m > 0$. What's the marginal probability of $Y$?
I'm aware that $Y|M = m$ ~ Poisson(m) and $M$ ~ Exponential(1), but i can't go further. Maybe i'm missing a ridiculous point.
I thought i could find $P(Y = y)$ by integrating the conditional in the range of $M$, but i was wrong. Can someone help me, please? (Sorry about my english, it's not my first language).
Don't integrate just the conditional, but the joint measure.
By the Law of Total Probability:
$$\begin{align}\mathsf P(Y{\,=\,}y) &= \int_\Bbb R \mathsf P(Y{\,=\,}y\mid M{\,=\,}m) f_M(m)\,\mathrm d m\\ &= \dfrac{1}{y!}\mathbf 1_{y\in\Bbb N}\int_0^\infty {\mathrm e^{-2m} m^y}\,\mathrm d m \\ &~~\vdots\end{align}$$