Method of moments for estimating $n$ in a binomial r.v. - is the estimator the UMVUE?

Question

I have derived the binomial Method of Moments (MoM) estimator for $Binomial(\theta, p)$, where $p$ is known.

Furthermore, I have calculated the variance of my unbias estimator.

Now, I am trying to determine if the MoM estimator is the UMVUE. The approach I'm familiar with is computing the Information $I(\theta)$ and using CR-LB for comparison. However, I can't compute derivatives of a combinatorial term. This leads me to believe that such a question cannot be answered.

Is the UMVUE for $n$ in a Binomial r.v. known? If so, how can I compute it?

Answer 1

Especially in the case where neither $n$ (your $\theta)$ nor $p$ is known, this is a notoriously difficult problem, you might want to look at journal articles retrieved by a Internet search for estimate binomial n; articles with Olkin or Rubin as co-author may be of particular interest.

I will leave the formal discussions of sufficiency, unbiasedness, and UMVUE to you, but the discussion below will show some complications that need to be addressed.

Suppose you have $N$ observations $X_1, X_2, \dots, X_N$ from $\mathsf{Binom}(n, p),$ with $p$ known. Then the method of moments estimator of $n$ is $\bar X/p,$ rounded to the nearest integer.

The rounded MME performs reasonably well for large or moderate $p,$ as illustrated in the simulation below in R, which looks at a million samples of size $N = 100$ from $\mathsf{Binom}(n=20, p = .5).$

set.seed(2020);  B=10^6
n.est = replicate(B, round(mean(rbinom(100, 20, .5))/.5))
summary(n.est)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
    18      20      20      20      20      22 
var(n.est)
[1] 0.2566687
table(n.est)/B
n.est
      18       19       20       21       22 
0.000453 0.126148 0.746043 0.126905 0.000451 
mn= min(n.est); mx=max(n.est)
hist(n.est, prob=2, br = ((mn-1):mx)+.5, col="skyblue2")

Results are better with $p = .9.$ In particular, notice the smaller variance of the MME. [if $p=0.1,$ then the variance of the MME is much larger. (No simulation shown.)]

set.seed(2020);  B=10^6
n.est = replicate(B, round(mean(rbinom(100, 20, .9))/.9))
summary(n.est)
    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
     19      20      20      20      20      21 
var(n.est)
[1] 0.0007419143
table(n.est)/B
n.est
      19       20       21 
0.000518 0.999258 0.000224 
mn= min(n.est); mx=max(n.est)
hist(n.est, prob=2, br = ((mn-1):mx)+.5, col="skyblue2")

However, the MME is clearly not always the best estimator of $n.$ When $p$ is large, then the largest number of Successes $X_{N}$ observed out of $N$ observations from $\mathsf{Binom}(n, p)$ may be a better estimator. In particular, here are results from a million samples of size 100 from \mathsf{Binom}(n=20, p=.85), using the maximum observed number of successes. For large $p$ and sufficiently large $N,$ this estimator seldom underestimates $n$ and never overestimates. It's variance is nearly $0,$ but it is not exactly unbiased.

set.seed(2020);  B=10^6
n.100 = replicate(B, max(rbinom(100, 20, .9)))
summary(n.100)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
     19      20      20      20      20      20 
var(n.100)
[1] 4.99998e-06  # variance very nearly 0
table(n.100)/B
n.100
      19       20 
0.000005 0.999995   # no overestimates 
mn= min(n.100); mx=max(n.100)
hist(n.100, prob=T, br = ((mn-1):mx)+.5, col="skyblue2")