Modifying the Poisson Distribution The Poisson Distribution $Pr(M|\lambda)$ is given by $$Pr(M|\lambda) = \frac{e^{-\lambda}\lambda^M}{M!}~~~~M = 0,1,2...$$ with mean $\lambda$. In addition, the distribution has the property that $$\int_{0}^{\infty} Pr(M|\lambda) d \lambda = 1~~~~~~~~~~~~~~(*)$$ as is proved in the post. I am interested in adapating this Poisson Distribution so that it can include an additional parameter (say $\mu$) which determines the width of the distribution (much like the variance does for the normal distribution). Does anyone have any ideas of how this can be achieved while still maintaining that property $(*)$ holds? 
Note I want to specifically work with the poisson distrbution rather than the normal distribution due to how the shape of the poisson distribution at zero.  
 A: Consider the following construction:
Let $U\sim \mathcal{U}[a,b]$ and $N\sim \text{Poisson}(U)$.
Through easy calculations we can see that:
1) $\mathbb{E}(N\mid U)\sim \mathcal{U}[a,b]$
2) $\text{Var}(N\mid U)\sim \mathcal{U}[a,b]$
Now, we can calculate
$\mathbb{E}(N)=\mathbb{E}(\mathbb{E}(N\mid U))=\mathbb{E}(\text{`}\mathcal{U}[a,b]\text{´})=\dfrac{a+b}{2}$, and
$\text{Var}(N)=\text{Var}(\mathbb{E}(N\mid U))+\mathbb{E}(\text{Var}(N\mid U))=\text{Var}(\text{`}\mathcal{U}[a,b]\text{´})+\mathbb{E}(\text{`}\mathcal{U}[a,b]\text{´})=\frac{(b-a)^2}{12}+\frac{a+b}{2}$.
Now, suppose that you want a random variable with mean $\mu$ and variance $\sigma^2$. With the previous construct, you must find $a,b$ such that:


*

*$\mu=\frac{a+b}{2}$

*$\sigma^2=\frac{(b-a)^2}{12}+\frac{a+b}{2}=\frac{(b-a)^2}{12}+\mu$
For solve this system, calculate:


*

*$b+a=2\mu$

*$b-a=\sqrt{12(\sigma^2-\mu)}$


And then:


*

*$b=\frac{2\mu+\sqrt{12(\sigma^2-\mu)}}{2}$

*$a=\frac{2\mu-\sqrt{12(\sigma^2-\mu)}}{2}$
Obs: this construction works if $\sigma^2\geq\mu$, but if you must that the random variable comports as Poisson, this is essential.
