Estimating the parameters of a Poisson times a constant Suppose that we have a distribution $X = aY$ where the distribution of $Y$ is Poisson{$\lambda$}, what is the maximum likelihood estimator of $a$?
By the method of moments we can estimate $a$ as $\sigma^2_X /\mu_X $. This is because $\mu_X = a \lambda$ and $\sigma^2_X = a^2 \lambda$.
The log-likelihood function is I think:
$$
l(\lambda,a|x_j) = \log \prod_{j=1}^{n} e^{-\lambda} \frac{\lambda^{x_j/a}} {(x_j/a)!} 
$$
$$
=-n\lambda + {\frac{1}{a}}\log{\lambda}\sum_{j=1}^n{x_j}-\sum_{j=1}^n \log{\left(\frac{x_j}{a}!\right)}
$$
However I am having trouble solving this for $a$ after differentiating and setting to zero.
The motivation for this question is that I am trying to learn how to do maximum likelihood for this distribution. I know that $a$ can be estimated by finidng the greatest common divisor of the differences of the observed values of $X$ but I am not interested in finding $a$ that way. Especially because I have some real-world motivation for this question and actual observations would be distributed as $X + \epsilon$ where $\epsilon \sim(0,\sigma_\epsilon^2$).

EDIT: this was an answer to the original form of the question.
A simple answer was staring me in the face although I am not sure if this is likely not the maximum likelihood estimator of $a$ (will happily select an answer that can). If $Y\sim\textrm{Poisson}(\lambda) $ then $\mu_Y=\lambda$ and $\sigma^2_Y = \lambda$. If $X=aY$ then $\mu_X=a\lambda$ and $\sigma^2_X = a^2\lambda$. Hence $\sigma^2_X/\mu_X$ estimates $a$.
Example in R:
> Y = rpois(1000,0.7)
> X = 0.2*Y
> var(X)/mean(X)
[1] 0.2109116

