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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 $\sigma= 1.31$.

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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 $\lambda$.

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  • $\begingroup$ Thank you for your answer. The mean of the PLN is the same as the lognormal, is that right? $\endgroup$ – pablo Jul 17 at 17:43
  • $\begingroup$ That is correct. Note that this nice relationship only holds if the "outer" distribution, in your case the lognormal, is actually on the mean of the "inner" distribution, in your case the Poisson. Well no doubt there are special cases where you get the same result when this doesn't hold, but it always holds in this case. $\endgroup$ – jbowman Jul 17 at 17:46

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