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


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

  • $\begingroup$ Is this exponential family? $\endgroup$ Commented Sep 15, 2019 at 18:42
  • $\begingroup$ @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). $\endgroup$
    – pablo
    Commented Sep 15, 2019 at 18:50

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


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