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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!

nicefella
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1 Answers1

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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.

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

lightfish
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  • 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! – nicefella Feb 23 '14 at 23:37
  • 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. – lightfish Feb 24 '14 at 00:37
  • 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

    $$\exp{-n\lambda+\sum x_i\log\lambda-n\sum\log x_i! -n\log(1-e^{-\lambda})}$$

     – lightfish Feb 24 '14 at 00:38