# Limiting distribution of $n(T_n-4p^3(1-p))$

I want to find the limiting distribution of a $n(T_n-4p^3(1-p))$, where $T_n=\displaystyle\frac{4(n-t)t(t-1)(t-2)}{n(n-1)(n-2)(n-3)}$ with $t=\sum X_i$ is the UMVUE of $4p^3(1-p)$ that I found, where $X_1,\ldots,X_n$ is iid Bernoulli distributed with mean $p$.

I know the delta method can't be used here since the term $T_n$ involves $n$, and the likely way to find the limiting distribution would be to find it directly, but I'm not sure how.

When $n\to\infty$, the random variable $\sqrt{n}\cdot(T_n-4p^3(1-p))$ converges in distribution to a centered normal distribution with variance $\sigma^2=16p^5(1-p)(3-4p)^2$ (note the normalization by $\sqrt{n}$ instead of $n$). Is this your question?
Edit: If $p=\frac34$, the random variable $n\cdot\left(T_n-\frac{27}{64}\right)$ converges in distribution to the distribution of $-\frac{27}{32}Z^2$, where $Z$ is standard normal.
• Yes! although the question had the normalization by $n$ instead of $\sqrt{n}$. So we can use CLT directly on $T_n$ since the $X_i$'s are iid? – lightfish Jun 29 '14 at 15:27
• Rather, directly on $t$, yes. – Did Jun 29 '14 at 15:48
• If $p=3/4$, would it be $n$? Do we use the second order delta method? – lightfish Jun 29 '14 at 16:32