Transforming a Poisson distribution into a power law

Question

Consider the probability mass function of the Poisson distribution given a mean $\lambda$:

\begin{equation} \mathbb{P}\left(Y=k|\lambda\right)=\frac{e^{-\lambda} \lambda^{k}}{k !} \end{equation}

By assuming that $\lambda$ itself is a random variable, is it possible to transform the distribution of the random variable $Y$ into a power law? In other words, is there a pdf $f(\lambda)$ that turns $\mathbb{P}\left(Y=k\right)$ into the pmf of a power law:

\begin{align} \mathbb{P}\left(Y=k\right)=\int_0^\infty \mathbb{P}\left(Y=k|\lambda\right)f(\lambda)\mathrm{d}\lambda \end{align}

Also, does this type of manipulation have a name? I am sure this must be a common question, but I could not find typical examples that englobe this case.

Edit: After further readings, it seems that a Poisson process with random intensity is called a Cox process or a double stochastic Poisson process.

Answer 1

This is not possible, at least not if you want a power law distribution with finite mean and finite variance. A Poisson distribution with a random intensity is overdispersed, meaning that the variance is greater than the mean. A power law distribution $p_s(k)=k^{-s}/\zeta(s)$, $s>3$ has a variance smaller than the mean.