# asymptotic variance for an estimator

We have the negativ binomial pdf

$f(y|r,p)=\binom{r+y-1}{y}p^r(1-p)^y$ with $,y \ge0,p\in(0,1)$

Y is the number of failures before success number r.

Assume $r \ge 0$ is a known integer.

I found the MLE to be $\hat p=\frac{r}{r+\bar Y_n}$

How can I find the asymptotic variance for $\hat p$ ?

I think it has something to do with the expression $\sqrt n(\hat p-p)$ but I am not entirely sure how any of that works.

$$\tag{1}\label{1}\sqrt{n}(\hat p-p)=\sqrt{n}\left(\frac{r}{r+\bar Y_n}-p\right)=\sqrt{n}\left(\frac{r(1-p)-p\bar Y_n}{r+\bar Y_n}\right).$$ Next, the numerator converges in distribution by CLT to some normal distribution: $$\sqrt{n}\left(r(1-p)-p\bar Y_n\right)=-\sqrt{n}\left(\frac{\sum_{i=1}^n pY_i}{n} - \mathbb E\left[pY_1\right]\right)\xrightarrow{d} N(0,\text{Var}[pY_1])$$ And the denominator converges in probability: $$r+\bar Y_n \xrightarrow{p} r+ \mathbb E[Y_1] = r+r\frac{1-p}{p}=\frac{r}{p}.$$

Then use Slutsky's theorem and find the variance of limiting normal distribution in (\ref{1}). The asymptotic variance that you need is the variance of limiting distribution. Or it can also be the variance of limiting distribution divided by $n$, in dependence on local definition given in your course.

Or you can use the following method:

By Central Limit Theorem, $$\sqrt{n}(\bar Y_n-\theta)\xrightarrow{d} N(0,\text{Var}Y_1)$$ with $\theta=\mathbb EY_1=r\frac{1-p}{p}$. Then we can use $$\hat p = g(\bar Y_n)=\frac{r}{r+\bar Y_n},$$ where $g(x)=\frac{r}{r+x}$.

The derivative $g'(x)$ at the point $x=\theta=r\frac{1-p}{p}$ is not zero, and then $$\sqrt{n}(\hat p-p)=\sqrt{n}\left(g(\bar Y_n)-g(\theta)\right)\xrightarrow{d} N\left(0,\, \text{Var}Y_1\cdot[g'(\theta)]^2\right).$$

• @stat333 Do you know such things as Central Limit Theorem and Law of Large Numbers? The first two convergences are by this fundamental theorems. You can also use Delta-method directly: $$\sqrt{n}(\bar Y_n-\theta)\xrightarrow{d} N(0,\text{Var}Y_1)$$ with $\theta=\mathbb EY_1=r\frac{1-p}{p}$. Then use $$\hat p = g(\bar Y_n)=\frac{r}{r+\bar Y_n},$$ where $g(x)=\frac{r}{r+x}$.
– NCh
Apr 6, 2017 at 11:25
• @stat333 I edit the answer and add this method also.
– NCh
Apr 6, 2017 at 11:37