Asymptotic distribution of a measure of homogeneity

Question

For an exam preparation I'm trying to solve the following question, but I get stuck. The question is

One measure of the homogeneity of a multinomial population with $k$ cells and probabilities $p = (p_1,\cdots,p_k)$ is the sum of the squares of the probabilities $S(p) = \sum_{i=1}^k p_i^2$. Note that $\frac{1}{k}\leq S(p)\leq 1$, with higher values indicating greater heterogeneity. Given a sample of size $n$ from this population, we may estimate $S(p)$ by $S(\hat{p})$, where $\hat{p} =$ $(\hat{p}_1,\cdots,\hat{p}_k)$ and $\hat{p}_i$ is the proportion of the observations that fall in cell $i$. What is the asymptotic distribution of $S(\hat{p})$?

Following a proof for Neyman's $\chi^2$, I define random variables $X_{ij}$ such that $P(X_{ij} = 1) = p_j$ and $P(X_{ij} = 0) = 1 - p_j$. Then $EX_{ij} = p_j$ and $EX_i = p$ where $p = (p_1,\cdots,p_k)$. Furthermore the covariance matrix of $X_1,...,X_n$ is given by $\Sigma = \begin{bmatrix} p_1(1-p_1) & -p_1p_2 & \cdots & -p_1p_c\\ -p_1p_2 & p_2(1-p_2) & \cdots & -p_2p_c \\ \vdots & \vdots & & \vdots \\ -p_1p_c & -p_2p_c & \cdots & p_c(1-p_c)\end{bmatrix}$.

Then $S(\hat{p}) = \sum_{i=1}^k \begin{pmatrix} \frac{n_i}{n} \end{pmatrix}^2$ where $n_i$ denotes the number of trials that result in outcome $i$ for $i = 1,...,c$. So $S(\hat{p}) = \bar{X}_n^T\bar{X}_n$.

Starting here I get stuck trying to find an asymptotic distribution for $S(\hat{p})$. I know that $\sqrt{n}(\bar{X}_n - p) \xrightarrow{\mathcal{L}} \mathcal{N}(0, \Sigma)$, but I fail to see how I can derive the asymptotic distribution of $\bar{X}_n^T\bar{X}_n$ from here. I hope someone can help me out!

Answer 1

Use the multivariate form of the delta method for $g: \mathbb R^k \to \mathbb R$ defined by $x\mapsto x^\intercal x$. So, for $\sqrt n (\bar X_n - p)\stackrel{d}{\longrightarrow}N(0,\Sigma)$, we have $$\nabla g_{(p)}=2p\,\,\,\,\, \text{and} \,\,\,\,\, \sqrt n (\bar {X}_n^\intercal\bar X_n - p^\intercal p)\stackrel{d}{\longrightarrow}N(0,\nabla g_{(p)}^\intercal\Sigma\nabla g_{(p)})= N(0,4p^\intercal\Sigma p).$$