# Two approaches for finding a MLE in a binomial setting

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).

• This question appears to be missing several crucial pieces. First, is $y_i$ a random variable or not? If so, what do you know about its distribution? Second, you haven't described any data, so how can you possibly estimate anything?
– whuber
Jan 14, 2019 at 21:28
• Sorry it wasn't clear enough. I fixed it. Jan 15, 2019 at 4:54
• Are you estimating $p_0$ for just one box, or for each box? If the latter, do you assume that the boxes have the same probability of defective items or not?
– Ben
Jan 15, 2019 at 7:50
• $p_0$ for a box. The question assumes items and boxes are all independent. Again, $p_0$ is the probability that a box has no defective items. Jan 15, 2019 at 15:15

Well, the two approaches give different answers. For instance, suppose you have two boxes containing two items each, and each has one defective item; then with the first approach you'll estimate that Pr(one item defective)=1/2 and therefore Pr(nothing in box defective)=1/4, but with the second both your $$x$$ variables will be 0, so $$\bar x=0$$ too, so your estimate will be 0.
The second approach is throwing away a lot of information, and the likelihood you're computing is not that of the observations you have (which include the actual values $$y_i$$) but that of something at one remove from those observations: the defect-free-ness of the boxes. So it won't be the Right Thing except by good luck, and the difference between the first and second answers indicates that you didn't get lucky :-).
The hypotheses of the Lehmann–Scheffé theorem call for the estimator to (1) be unbiased and to depend on the data via the value of a (2a) complete, (2b) sufficient statistic. It doesn't look to me as if you've established any of these conditions; in particular, sufficiency (2b) says that no other information that can be extracted from the data tells you anything further about $$p_0$$ once you've found $$\bar x$$. Well, consider two scenarios where you have (say) 100 boxes containing 100 items each. In scenario A, each box contains 99 defective items and 1 good item. In scenario B, each box contains 99 good items and 1 defective item. These scenarios have the same value of $$\bar x$$, and of all the $$x_i$$, but wouldn't you make very different guesses about $$p_0$$ in the two cases?