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I started learning statistics and in my homework i should find the Maximum Likelihood Estimate. The function is $f_x(x)=e^{-\lambda n}\prod_{i=1}^n \frac{\lambda^{x_i}}{x_i!}$

Now i take the log-likelihood:
$$\ln{e^{-\lambda n}}+ln\prod_{i=1}^n \frac{\lambda^{x_i}}{x_i!}= -\lambda n +\sum_{i=1}^n (\ln{(\lambda^{x_i})}-\ln{(x_i!)})= -\lambda n+\sum_{i=1}^n \ln{(\lambda^{x_i})}-\sum_{i=1}^n \ln{(x_i!)}$$

The problem is now that i don't know what $\sum\ln{(x_i!)}$ is. Can someone help me here ?

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    $\begingroup$ $\sum \ln(x_i!)$ doesn't matter, you have to maximize wrt $\lambda$. $\endgroup$
    – roger
    Jun 26, 2013 at 18:11
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    $\begingroup$ Thanks for the answer, so that means that i can just leave it and go on and do the derivation, then i get: $-n+\frac{1}{\lambda}\sum x_i$ $\endgroup$
    – Devid
    Jun 26, 2013 at 18:14
  • $\begingroup$ yep, that's it. $\endgroup$
    – roger
    Jun 26, 2013 at 18:39

1 Answer 1

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By @roger $∑ln(xi!)$ doesn't matter, you have to maximize wrt λ

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