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$X$ is distributed as a Poisson random variable with parameter $\lambda$ if it has the probability mass function:

$f(x; \lambda) = \dfrac{e^{-\lambda}\lambda^x}{x!},\ x = 0,1,2,\ldots$

Find the maximum-likelihood estimator for the parameter $\lambda$ based on a random sample of size $n$.

I understand the theory behind MLEs, but am not sure how to go about solving this problem. I did originally take the natural log, but wasn't sure what to do afterwards. I'm very new to this.

Rosie E
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1 Answers1

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The probability of observing $(x_1,\ldots,x_n)$ is $$L=f(x_1;\lambda) \ldots f(x_n; \lambda) = \frac{e^{-n \lambda} \lambda^{\sum\limits_i x_i} }{x_1! \ldots x_n!}.$$ Taking the log, you get $$\ell = -n \lambda + \left(\sum\limits_i x_i\right) \log \lambda + C,$$ where $C$ is terms which do not depend on $\lambda$. Now, maximize $\ell$ with respect to $\lambda$ by differentiating it with respect to $\lambda$, setting it equal to zero and checking with the first or second derivative test that you do indeed have a maximizer (alternatively note that $\ell$ is concave in $\lambda$ so it is indeed the maximizer).

Did
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Batman
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  • Ok, so I took the derivative and just got the ln (λ). Then when I set that equal to zero, I got one. Is this correct? – Rosie E Jun 30 '14 at 21:05
  • No, $\frac{\partial \ell}{\partial \lambda} = -n + \frac{1}{\lambda} (\sum_i x_i)$. Now, set that equal to 0 and solve for $\lambda$. – Batman Jul 01 '14 at 04:09
  • http://math.stackexchange.com/questions/2154336/checking-that-the-mle-for-the-poisson-parameter-is-really-a-maximum It's related to your answer, which I think it's incomplete. Could you please help me? – An old man in the sea. Feb 21 '17 at 09:08