I am stuck on this problem and help would be greatly appreciated! I have the following PMF (a modified Poisson Distribution).

\begin{align*} \frac{\lambda^x e^{-\lambda}}{x!(1 - e^{-\lambda})} \end{align*} for some $\lambda >0$ and $x=1,2,3...$

I am supposed to first fine the $mean$ of the distribution and then find the MLE (Maximum Likelihood Estimator).

So for the $mean$, I am not sure how to proceed as the only thing I can think of is to take the $Expectation$ of the PMF, but that would be quite complicated since we have a fraction with factorials in the denominator.

As for the MLE, for $n$ observations, we have the following I believe:

$L(x_1...x_n,\lambda)=\prod_{i=1}^n pmf = \begin{align*} \frac{\lambda^{nx} e^{{-\lambda}n}}{x!^n(1 - e^{-\lambda})^n} \end{align*} $

Is that the correct approach? Then I would have to take the $ln$ of $L$ and solve for $\lambda$ by setting the equation to $0$. Is that the correct approach? Thanks so much for your help!

I greatly appreciate it!


For the expectation,

$$\displaystyle E[X]=\sum_{x=1}^{\infty} x \frac{\lambda^x e^{-\lambda}}{x!(1-e^{-\lambda})}=\frac{e^{-\lambda}}{1-e^{-\lambda}}\sum_{x=1}^{\infty} \frac{\lambda^{x-1}}{(x-1)!}\lambda$$ and then if you recognize what this is, use the fact that $\sum_{x=0}^{\infty} k^x/x!=e^k$.

For the likelihood, you have $X_i$'s as the random variables.

$$\frac{\lambda^{\sum x_i}e^{-n\lambda}}{\prod_{i=1}^nx_i!^n (1-e^{-\lambda})^n}$$

  • $\begingroup$ Thanks but I am still quite confused. For the expectation, why is the $\lambda$ to the power of $x-1$? Also for the likelihood, why is the product sign there? You already have $x_i^{n}$ so is that not redundant? It would be great if you could please be a bit more detailed! Thanks so much! $\endgroup$ – nicefella Feb 23 '14 at 23:37
  • $\begingroup$ For the expectation, I want the summation to be in the power series form of $e^\lambda$, so that's why I made the exponent of $\lambda$ to be the same as the denominator $(x-1)!$. $$\sum_{x=1}^{\infty}\frac{\lambda^x}{(x-1)!}=\sum\frac{\lambda ^{x-1}}{(x-1)!}\lambda=\lambda\sum_{y=0}^{\infty}\frac{\lambda^y}{y!}$$. I just realized that $x$ starts from 1 here so I will change the above answer. $\endgroup$ – lightfish Feb 24 '14 at 0:37
  • $\begingroup$ For the likelihood, you want the joint density of $X_1,X_2,\ldots,X_n$, and since the $X_i$'s are independent, then you multiply each of their densities together, which is why you get the product as you wrote above. But since the $X_i$'s are each different, we need to write $$\exp\{-n\lambda+\sum x_i\log\lambda-n\sum\log x_i! -n\log(1-e^{-\lambda})\}$$ $\endgroup$ – lightfish Feb 24 '14 at 0:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.