Asymptotic correlation between sample mean and sample median

Question

Suppose $X_1,X_2,\cdots$ are i.i.d. $N(\mu,1)$. Show that the asymptotic correlation between sample mean and sample median (after suitably centering and renormalization) is $\sqrt{\frac{2}{\pi}}$.

Answer 1

Obtain this paper, written by T.S. Ferguson, a professor at UCLA (his page is here).

It derives the joint asymptotic distribution for the sample mean and sample median.

To be specific, let $\hat X_n$ be the sample mean and $\mu$ the population mean, $Y_n$ be the sample median and $\mathbb v$ the population median. Let $f()$ be the probability density of the random variables involved ($X$) Let $\sigma^2$ be the variance. Then Ferguson proves that

$$\sqrt n\Big [\left (\begin{matrix} \hat X_n \\ Y_n \end{matrix}\right) - \left (\begin{matrix} \mu \\ \mathbb v \end{matrix}\right)\Big ] \rightarrow_{\mathbf L}\; N\Big [\left (\begin{matrix} 0 \\ 0 \end{matrix}\right) , \Sigma \Big]$$

$$ \Sigma = \left (\begin{matrix} \sigma^2 & E\left(|X-\mathbb v|\right)\left[2f(\mathbb v)\right]^{-1} \\ E\left(|X-\mathbb v|\right)\left[2f(\mathbb v)\right]^{-1} & \left[2f(\mathbb v)\right]^{-2} \end{matrix}\right)$$

Then the asymptotic correlation of this centered and normalized quantity is (abusing notation as usual)

$$\rho_{A}(\hat X_n,\, Y_n) = \frac {\text {Cov} (\hat X_n,\, Y_n)}{\sqrt {\text{Var}(\hat X_n)\text{Var}(Y_n)}} = \frac {E\left(|X-\mathbb v|\right)\left[2f(\mathbb v)\right]^{-1}}{\sigma\left[2f(\mathbb v)\right]^{-1}} = \frac {E\left(|X-\mathbb v|\right)}{\sigma}$$

In your case, $\sigma = 1$ so we end up with

$$\rho_{A}(\hat X_n,\, Y_n) = E\left(|X-\mathbb v|\right)$$

In your case, the population follows the normal with unitary variance, so the random variable $Z= X-\mathbb v$ is $N(0,1)$. Then its absolute value follows the (standard) half normal distribution, whose expected value is $$ E(|Z|) =\sigma\sqrt {\frac{2}{\pi}} = \sqrt {\frac{2}{\pi}}$$ since here $\sigma =1$. So $$\rho_{A}(\hat X_n,\, Y_n) = \sqrt {\frac{2}{\pi}}$$

Added note: It can be seen that the result does not depend on $\sigma =1$ since $\sigma$ cancels out from nominator and denominator.