# Asymptotic distribution of sample variance via multivariate delta method

I was trying to get the asymptotic distribution of sample variance using multivariate delta method and without normality assumption.

So I defined the random vector $$z = \left( \begin{matrix} X \\ X^2 \\ \end{matrix} \right)$$ and the function $$g$$ which acts on components : $$g(z_1,z_2)=z_2 - z_1^2$$ Clearly $$g(E[z])=\sigma_X^2$$.

Taking a random sample, the random vector: $$z_n= \left( \begin{matrix} \frac{1}{n}\sum_i X_i \\ \frac{1}{n}\sum_i X_i^2 \\ \end{matrix} \right)$$ a plugin estimator for $$\sigma_X^2$$ is $$g(z_n)=S_X^2$$ (sample variance).

Knowing that $$z_n$$ converges asymptotically to a normal distribution thanks to central limit theorem: $${{z}_{n}}\xrightarrow{d}N\left( E[z],\frac{1}{n}\left( \begin{matrix} V[X] & \operatorname{cov}(X,{{X}^{2}}) \\ \operatorname{cov}({{X}^{2}},X) & V[{{X}^{2}}] \\ \end{matrix} \right) \right)$$ we can apply multivariate delta method.

I get stuck in the estimation of variance by delta method which should be $$({{\left. \nabla g \right|}_{E[X]}})^t \left( \begin{matrix} V[X] & \operatorname{cov}(X,{{X}^{2}}) \\ \operatorname{cov}({{X}^{2}},X) & V[{{X}^{2}}] \\ \end{matrix} \right){{\left. \nabla g \right|}_{E[X]}}$$ with $${{\left. \nabla g \right|}_{E[X]}} = \left( \begin{matrix} -2E[X] \\ 1 \\ \end{matrix} \right)$$ .

Performing the matrix product as above I end up with an asymptotic variance which does not match the one I found on textbooks , which is: $$V[(X-E[X])^2]$$ What I'm missing?

$$\left(\begin{matrix} -2E[X] & 1\end{matrix}\right) \left(\begin{matrix} V[X] & Cov(X,X^2) \\ Cov(X^2,X) & V[X^2] \end{matrix}\right) \left(\begin{matrix} -2E[X] \\ 1 \\ \end{matrix}\right)$$$$=\left(\begin{matrix}-2E[X]V[X] + Cov(X^2,X) & -2E[X]Cov(X^2,X)+V[X^2]\end{matrix}\right)\left( \begin{matrix} -2E[X] \\ 1 \end{matrix}\right) =4E^2[X]V[X] -4E[X]Cov(X^2,X)+V[X^2]$$
\begin{align}V[(X-E[X])^2] &= V[X^2 - 2XE[X] + E^2[X]\\ &= V(X^2) +(-2E(X))^2V(X) +V(E(X^2))\\ &\qquad -2Cov(X^2,2XE[X]) + 2Cov(X^2,E^2(X))\\&\qquad -2Cov(-2XE(X), E^2(X))\\ =&4E^2[X]V[X] -4E[X]Cov(X^2,X)+V[X^2]\end{align}
• Thank you, that was really straightforward, I don't know why I got stuck in evaluating the term $cov(X^2,2XE[X])$ – omega Nov 16 '18 at 16:04