Expected value of a Geometric distribution

Question

I have a random number $X\sim Exp(\lambda)$ and a value $Y\sim Poi(X)$. I have to calculate $\mathbb{E}[Y]$.

I proved that $Y\sim Geo(\frac{\lambda}{(\lambda+1)})$, so i should have $\mathbb{E}[Y]:=\frac{\lambda+1}{\lambda}$.

Why, instead, is $\frac{1}{\lambda}$?

Text says that $\mathbb{E}[Y]=\mathbb{E}[\mathbb{E}[Y|X]]=\mathbb{E}[X]$. Trivially the first equivalence is a property of conditional expectation but what justifies the second equivalence?

Thanks in advance for any help!

Answer 1

Note that technically $Y|X\sim Poi(X)$. Then the second equivalence is using the expected value of a Poisson random variable.

Answer 2

Conditioned on $\{X=x\}$, $Y$ has Poisson distribution with mean $x$. So we can compute the distribution of $Y$ by conditioning on $X$: for any nonnegative integer $n$ we have \begin{align} \mathbb P(Y=n) &= \int_0^\infty \mathbb P(Y=n\mid X=x)f_X(x)\ \mathsf dx\\ &= \int_0^\infty e^{-x}\frac{x^n}{n!}\lambda e^{-\lambda x}\ \mathsf dx\\ &= \frac\lambda{n!} \int_0^\infty x^n e^{-(1+\lambda)x}\ \mathsf dx. \end{align} We compute the integral $\int_0^\infty x^n e^{-(1+\lambda)x}\ \mathsf dx$ by induction. For $n=0$, $$ \int_0^\infty e^{-(1+\lambda)x}\ \mathsf dx = \frac1{1+\lambda} = \frac{0!}{(1+\lambda)^1}. $$ Now suppose that $\int_0^\infty x^n e^{-(1+\lambda)x}\ \mathsf dx = \frac{n!}{(1+\lambda)^{n+1}}$ for some nonnegative integer $n$. Then using integration by parts, we have \begin{align} \int_0^\infty x^{n+1} e^{-(1+\lambda)x}\ \mathsf dx &= \left.\frac{-x^{n+1}e^{-(1+\lambda)x}}{1+\lambda}\right|_0^\infty +\frac{n+1}{1+\lambda}\int_0^\infty x^n e^{-(1+\lambda)x}\ \mathsf dx\\ &= 0 + \frac{n+1}{1+\lambda}\left(\frac{n!}{(1+\lambda)^{n+1}}\right)\\ &= \frac{(n+1)!}{(1+\lambda)^{n+2}}. \end{align} It follows that $$ \mathbb P(Y=n) = \frac\lambda{n!}\left(\frac{n!}{(1+\lambda)^{n+1}}\right) = \frac\lambda{(1+\lambda)^{n+1}}. $$ We compute the mean of $Y$ by \begin{align} \mathbb E[Y] &= \sum_{n=0}^\infty n\cdot\mathbb P(Y=n)\\ &= \sum_{n=0}^\infty \frac{n\lambda}{(1+\lambda)^{n+1}}\\ &= \frac1\lambda. \end{align}