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

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 $$f(x;\mu,\sigma)$$?.

• Is this exponential family? Commented Sep 15, 2019 at 18:42
• @user0 I am not sure. I know that I can sample the pdf $f$ in 3 steps: 1) compute a normally distributed value, 2) take the exponent (this will sample the lognormal) and then 3) sample a Poisson distribution with parameter the exponent value obtained in 2). Commented Sep 15, 2019 at 18:50

## 1 Answer

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 $$\DeclareMathOperator{\V}{\mathbb{V}} \V N = \E \V N \mid \Lambda + \V \E N \mid \Lambda=e^{\mu+\sigma^2/2}+[e^{\sigma^2}-1] e^{2\mu+\sigma^2}$$