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Consider the Poisson distribution with an unknown parameter $\mu$ given by $$f(x) = \frac{ \mu^x e^{-\mu} }{x!}.$$

If $X_1, X_2, \ldots, X_n$ are $\require{cancel}\xcancel{n \text{ samples}}$ a sample of size $n$ drawn from the Poisson distribution, determine the maximum likelihood estimator for $\mu$.

I'm having trouble taking the derivative since there's a factorial on the bottom of the equation. Thanks!

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    $\begingroup$ The factorial is a constant, i.e. it does not depend on $\mu. \qquad$ $\endgroup$ – Michael Hardy Feb 28 '18 at 18:03
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Remember that the likelihood function is a function of the parameters not the data $x_i$. Your likelihood function is given by: $$\mathcal L =\prod_{i=1}^n\frac{\mu^{x_i}e^{-\mu}}{x_i!}$$ We take the logarithm to make differentiation easier: $$\log \mathcal L=\sum_i \left(x_i\log(\mu)-\mu-\log(x_i!)\right)$$ Take the derivative with respect to $\mu$ to obtain: $$\frac{\partial}{\partial \mu}\log \mathcal L = \sum_i\left(\frac{x_i}{\mu}-1\right)=\frac{1}{\mu}\sum_i x_i - n$$ Set the derivative to zero to identify where the maximum of the likelihood function occurs (check for yourself that this is indeed a maximum): \begin{align} \frac{\partial}{\partial \mu}\log \mathcal L =0&\Longrightarrow \frac{1}{\mu}\sum_i x_i = n\\ &\Longrightarrow \mu=\frac{1}{n}\sum_i x_i \end{align} So your maximum likelihood estimator is: $$\hat \mu = \frac{1}{n}\sum_{i=1}^nx_i$$

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