Maybe an MLE of a multinomial distribution?

Question

My question:

I usually don't have much trouble with deriving the MLE from the log-likelihood, but I am a little stumped on an example I found in "In All Likelihood" [1] pg.75.

Could someone show the steps from the log-likelihood to the MLE?

What the book says: Discrete data are usually presented in grouped form. For example, suppose $x_1, ..., x_N$ are iid sample from binomial($n,\theta$) with $n$ known. We first summarize the data where

$$ \begin{array}{c|c c c c} k& 0 & 1 & \dots &n \\ \hline n_k& n_0&n_1 &\dots &n_n\\ \end{array} $$

where $n_k$ is the number of $x_i$'s equal to $k$, so $\sum_k n_k = N$. We can now think of the data ($n_0,\ldots,n_n$) as having a multinomial distribution with probabilities ($p_0,\dots,p_n$) given by the binomial probabilities

$$ p_k = {n\choose k} \theta^k(1-\theta)^{n-k} $$

The log-likelihood is given by

$$ \log L(\theta)= \sum_{k=0}^n n_k\log p_k. $$

We can show that the MLE is $\theta$ is

$$ \hat{\theta} = \frac{\sum_k kn_k}{Nn} $$

with standard error

$$ \operatorname{se}(\hat{\theta}) = \sqrt{\frac{\hat{\theta}(1-\hat{\theta})}{Nn}} $$

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[1] Y. Pawitan, (2001), 'In All Likelihood: Statistical Modelling and Inference Using Likelihood', Oxford University Press.

Answer 1

I think I just figured this out.

$$ \log L(\theta) = \sum_{k=0}^n n_k\log p_k \\ = \sum_{k=0}^n n_k(k\log(\theta) + (n-k)\log(1-\theta)) \\ \frac{\partial}{\partial\theta} = \frac{\sum_{k=0}^n n_kk}{\theta} - \frac{\sum_{k=0}^n n_k(n-k)}{1-\theta} \\ 0 = \frac{\sum_{k=0}^n n_k k}{\theta} - \frac{\sum_{k=0}^n n_k(n-k)}{1-\theta} \\ \frac{\sum_{k=0}^n n_k(n-k)}{1-\theta} = \frac{\sum_{k=0}^n n_k k}{\theta} \\ \frac{\sum_{k=0}^n n_k(n-k)}{\sum_{k=0}^n n_k k} = \frac{1-\theta}{\theta} \\ \frac{\sum_{k=0}^n n_kn}{\sum_{k=0}^n n_k k} - 1 = \frac{1}{\theta}-1 \\ \frac{\sum_{k=0}^n n_kn}{\sum_{k=0}^n n_k k} = \frac{1}{\theta} \\ \theta = \frac{\sum_{k=0}^n n_k k}{\sum_{k=0}^n n_kn} \\ \theta = \frac{\sum_{k=0}^n n_k k}{Nn} $$