Mean of a Poisson-Lognormal Distribution (PLN)

Question

I would like to calculate the mean value of a PLN distribution, $$ f(x;\mu,\sigma)=\frac{1}{x!\sigma\sqrt{2\pi}}\int_{0}^{\infty}\lambda_\ast^{x-1} e^{-\lambda_\ast} e^{-\frac{(log(\lambda_\ast-\mu)^2}{2\sigma^2} }\text{d}\lambda_\ast, \quad x=0,1,2,\dotsc $$ I know that the mean of a Poisson is $\lambda$, and the mean of a lognormal distribution is $e^{(\mu+\frac{\sigma^2}{2})}$, but I do not know how to calculate the mean of the above PLN pdf.

If a general expression would not be feasible, I am finally trying to get the mean for $\mu=2.32$ and $\sigma= 1.31$.

Answer 1

You can find this using the Law of iterated expectation:

$$\mathbb{E}[x] = \mathbb{E}_{\lambda}[\mathbb{E}_x[x|\lambda]]$$

where the subscripts denote what the expectation is taken with respect to.

In your case, $\mathbb{E}_x[x|\lambda] = \lambda$, so, substituting, we obtain:

$$\mathbb{E}[x] = \mathbb{E}_{\lambda}[\lambda]=\text{e}^{\mu+{\sigma^2 \over 2}}$$

i.e., the mean of $\lambda$.