For sample $x_1,...,x_n$~Erlang, $f(x_i|k,\lambda)=\frac{\lambda^kx^{k-1}e^{-\lambda x_i}}{(k-1)!}$
Population has expectation $k/\lambda$ and variance $k/\lambda^2$. You may assume second derivative of log likelihood is negative
Suppose parameter k is known and can be treated as constant. Propose MLE for $\lambda$.
If I could get some help, specifically with the log-likelihood part of the problem and the derivative, that would be appreciated
I have got to $L=e^{-\lambda \sum x_i}\lambda^{kn}\Pi x_i^{k-1}(\frac{1}{(k-1)!})^n$
Hints:
The next stage is $\log_e L = -\lambda \sum x_i + kn \log_e \lambda +\,$ something not involving $\lambda$
Then take the derivative with respect to $\lambda$
Then see for which values of $\lambda$ where the derivative is zero, positive and negative, and so where the log-likelihood is maximised
Which should tell you the maximum likelihood estimator for $\lambda$
It turns out that this will also be the naive estimator for the rate when observing that the sum of $kn$ iid exponential random variables is $\sum x_i$.