Using method of moments on binomial distribution

Question

Hi, I am stuck on this particular question. Any help? Thanks.

Answer 1

The mean and variance for such a binomial can be found in terms of $n$ and $\theta$.

Find the analytical expressions and equate them to those of your sample. You have two variables and two equations, solve for $n$ and $\theta$.

Answer 2

Because $\mu = E(X) = np$ and $E(\bar X)$ estimates $\mu,$ we can estimate $np$ by $\bar X.$

Because $\sigma^2 = np(1-p)$ and $E(s^2)$ estimates $\sigma^2,$ we can estimate $1-p$ by $s^2/\bar X$ and then $p$ by $\hat p = 1-s^2/\bar X.$

Finally, we can estimate $n$ by $\hat n = \bar X/\hat p.$ But $\hat n$ will not necessarily be an integer, so round it to the nearest integer $\tilde n,$ and then adjust $\tilde p = \bar X/\tilde n.$

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Here is a simulated example based on $m = 100$ observations from $\mathsf{Binom}(n, p),$ where we are pretending that $n$ and $p$ are unknown. [Because this is a simulated sample, I know the real values of $n$ and $p$ so we can see how well the scheme above works for actual data.]

From my sample, I got $\bar X = 3.99$ and $s^2 = 2.5757,$ so $\hat p = 1 - s^2/\bar X = 0.3545$ and then $\hat n = \bar X/\hat p = 11.2562,$ which I round to $\tilde n = 11.$ Then $\tilde p = \bar X/\tilde n = 0.3627.$

In truth, the data were simulated using $n = 12$ and $p = 1/3.$ I could easily have contrived an example that 'turns out better', but to be realistic this is about the accuracy one can expect from 100 observations. Even if we know $n = 12$ and take a sample of $m = 100$ then we can imagine a random variable $Y \sim \mathsf{Binom}(1200, p)$ and a 95% margin of error for $p$ might be as large as $\pm 2\sqrt{1/4800}$ or $\pm 0.029.$ And having to estimate $n$ in addition would only make the margin of error larger.