How to find an expression of the variance of a Poisson-Lognormal distribution?

Question

I am using a model for the number of goods in a supermarket cart with a Poisson-lognormal distribution (a lognormal mixture of Poissons).

I would like to find an expression of the variance of this pmf (probability mass function), $$f(x;\mu,\sigma)=\frac{1}{x!\sigma\sqrt{2\pi}}\int_{0}^{\infty}\lambda^{x-1} e^{-\lambda} e^{\frac{(\log(\lambda)-\mu)^2}{2\sigma^2} }\text{d}\lambda,\quad x=0,1,2, \dotsc $$ I know from this post Mean of a Poisson-Lognormal Distribution (PLN) that the mean is $\text{e}^{\mu+{\sigma^2 \over 2}}$. I know that

$\operatorname{Var}(X)=\mathbb{E}[X^2]-(\mathbb{E}[X])^2=\mathbb{E}[X^2]-\text{e}^{2\mu+\sigma^2}$

Is it possible to use the Law of iterated expectation or similar law to find $\mathbb{E}[X^2]$ and an expression for the variance for the Poisson-Lognormal distribution $f(x;\mu,\sigma)$?.

Answer 1

You can use the law of total variance which is analogue to the double expectation theorem. If we have $$ \DeclareMathOperator{\E}{\mathbb{E}} N \mid \Lambda=\lambda \sim \mathcal{Po}(\lambda) \\ \Lambda \sim \mathcal{logNormal}(\mu,\sigma^2) $$ we find using lognormal properties $$ \E N=\E \left[ \E N\mid \Lambda\right] =\E \Lambda =e^{\mu+\sigma^2/2} $$ and $$ \DeclareMathOperator{\V}{\mathbb{V}} \V N = \E \V N \mid \Lambda + \V \E N \mid \Lambda=e^{\mu+\sigma^2/2}+[e^{\sigma^2}-1] e^{2\mu+\sigma^2} $$