What is the distribution for a random variable compose with another random variable?

Question

I have a a random variable $X$ with distribution $Po(M)$ where $M$ is a random variable with distribution $Exp(a)$. I want to know what is the "real" distribution of $X$. I was calculating by \begin{equation} P(X=k)=\int_0^{\infty}P(X=k|M=k)\cdot \mathcal{f}_M(x)dx=\int_0^\infty e^{-x}\frac{x^k}{k!}\cdot \frac{1}{a}e^{-\frac{x}{a}} \end{equation} Using the mean of a random variable with Gamma distribution, I have to \begin{equation} \int_0^\infty e^{-x}\frac{x^k}{k!}\cdot \frac{1}{a}e^{-\frac{x}{a}}=\left(\frac{a}{a+1}\right)^p \frac{1} {a} \end{equation} but I don't know if is this a distribution? or if I made a mistake

Answer 1

I'm assuming you meant $\left(\frac{a}{a+1}\right)^{k+1} \frac{1}{a}$. If you write this as $$\left(\frac{a}{a+1}\right)^k \frac{1}{a+1}$$ you get the PMF of the geometric distribution that counts the number of failures until the first success in a sequence of trials with success probability $1/(a+1)$.

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One other typo: it should be $P(X=k \mid M=x)$, not $P(X=k \mid M=k)$.

Also, the convention I have usually seen is that the PDF for $\text{Exponential}(a)$ is $a e^{-ax}$ rather than $\frac{1}{a} e^{-x/a}$. But if your course/textbook is using the latter convention, you can ignore this comment.