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).

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    $\begingroup$ 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? $\endgroup$
    – whuber
    Commented Jan 14, 2019 at 21:28
  • $\begingroup$ Sorry it wasn't clear enough. I fixed it. $\endgroup$
    – Nir B
    Commented Jan 15, 2019 at 4:54
  • $\begingroup$ 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? $\endgroup$
    – Ben
    Commented Jan 15, 2019 at 7:50
  • $\begingroup$ $p_0$ 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. $\endgroup$
    – Nir B
    Commented Jan 15, 2019 at 15:15

1 Answer 1


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?

  • $\begingroup$ Ok, so I get why the second approach throws away information from the y's (your example was a good intuitive explanation). The second approach gives more extreme results due to it's ignorance of the values of y. I wonder how close both approaches asymptotically... $\endgroup$
    – Nir B
    Commented Jan 15, 2019 at 8:36

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