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?


1 Answer 1


$$\left(\begin{matrix} -2E[X] & 1\end{matrix}\right) \left(\begin{matrix} V[X] & Cov(X,X^2) \\ Cov(X^2,X) & V[X^2] \end{matrix}\right) \left(\begin{matrix} -2E[X] \\ 1 \\ \end{matrix}\right) $$$$ =\left(\begin{matrix}-2E[X]V[X] + Cov(X^2,X) & -2E[X]Cov(X^2,X)+V[X^2]\end{matrix}\right)\left( \begin{matrix} -2E[X] \\ 1 \end{matrix}\right) =4E^2[X]V[X] -4E[X]Cov(X^2,X)+V[X^2] $$

$\begin{align}V[(X-E[X])^2] &= V[X^2 - 2XE[X] + E^2[X]\\ &= V(X^2) +(-2E(X))^2V(X) +V(E(X^2))\\ &\qquad -2Cov(X^2,2XE[X]) + 2Cov(X^2,E^2(X))\\&\qquad -2Cov(-2XE(X), E^2(X))\\ =&4E^2[X]V[X] -4E[X]Cov(X^2,X)+V[X^2]\end{align}$

  • $\begingroup$ Thank you, that was really straightforward, I don't know why I got stuck in evaluating the term $ cov(X^2,2XE[X])$ $\endgroup$
    – omega
    Commented Nov 16, 2018 at 16:04

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