# Question

Let's say we define the Negative Binomial as follows: $$f(x) = {x+r-1 \choose x} p^x (1-p)^r$$

With mean and variance: $$E(x) = \frac{rp}{1-p} \quad \quad V(x) = \frac{rp}{(1-p)^2}$$

We are given some set of data and need to get the maximum likelihood estimate and the method of moments estimate. How do we do this?

# Attempt at Solution

For method of moments, I did:

M1 = mean(x)
M2 = mean(x^2)
p = 1 - (M1/(M2-M1^2))
r = (1-p)*M1/p
q1b = list(r = r, p = p)
print(q1b)


For MLE I did:

NBd_LLL <- function(x,par) {
return(-sum(log(dnbinom(x,size = par, prob = par))))
}
q1c = optim(par = c(0.85,0.35), NBd_LLL, x = x, method = "L-BFGS-B", lower = c(0.1,0.1), upper = c(1,1))
q1c\$par


But these give very different answers, where did it go wrong?

$$\frac{\mathbb{E}(X)}{\mathbb{V}(X)} = 1-p \quad \quad \quad \frac{\mathbb{E}(X)^2}{\mathbb{V}(X)} = rp.$$
Solving these equations for $$p$$ and $$r$$ gives:
$$p = 1-\frac{\mathbb{E}(X)}{\mathbb{V}(X)} \quad \quad \quad r = \frac{\mathbb{E}(X)^2}{\mathbb{V}(X)-\mathbb{E}(X)}.$$
Substituting the sample mean $$\bar{X}$$ and the sample variance $$s_X^2$$ gives the MOM estimators:
$$\hat{p} = 1-\frac{\bar{X}}{s_X^2} \quad \quad \quad \hat{r} = \frac{\bar{X}^2}{s_X^2-\bar{X}}.$$