$X$ 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.
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).