I'm learning towards an exam in mathematical statistics and I came across the following question. I was wondering if the second approach of solving the question is legitimate. If both are correct, is there a reason to prefer one approach to the other?
In order to test some product, a sample of $n$ packages was collected. Each package contains exactly $k$ identical items. Let $y_i (i=1,...n)$ denote the number of defective items in the i-th box (assume items and boxes are independent). Of course, $y_1,...,y_n$ are binomial random variables. Let $p_0$ denote the probability that a box has no defective items. Find a MLE for $p_0$. Is the MLE UMVUE (uniformly minimal variance unbiased estimator)?
The approach I saw in a solution from some other student was to define $p_k$ as the probability that an item is defective. We then have: $p_o=(1-p_k)^k$ and $y_i \sim Bin(k, p_k)$. We can then write the likelihood: $L(p_k|y)=p_k^{\sum_i{y_i}}(1-p_k)^{nk-\sum_i{y_i}}\prod_i{k\choose{y_i}}$. The resulting MLE for $p_k$ is $\hat{p}_k= \bar{y}/k$. It then follows that the MLE for $p_0$ is $\hat{p}_k= (1-\bar{y}/k)^k$.
My approach was to define a set of Bernoulli variables $x_i = I(y_i = 0)$. The likelihood is then given by: $L(p_0 |x) = p_0^{\sum_i{x_i}}(1-p_0)^{n-\sum_i{x_i}}$ which yields the MLE: $\hat{p_o}=\bar{x}$. Then, due to the Lehmann-Scheffe theorem, the MLE is UMVUE (it is based on a complete statistic).