Asymptotic Relative Efficiency of Sample Mean and Median

Question

I'm following some online lecture notes on AREs and don't understand where a certain value came from. Consider a distribution function $F$ with a density function $f$ symmetric about $\theta$. We're considering two estimators, $\bar{X}_n$ (sample mean) and $\operatorname{Med} \lbrace X_1, X_2, \ldots, X_n\rbrace$

For large sample size $n$, I know that:

$$\bar{X}_n \sim N\left(\theta, \frac{\sigma_F^2}{n}\right) \text{ and } \operatorname{Med} \left\lbrace X_1, X_2, \ldots, X_n\right\rbrace \sim N\left(\theta, \frac{1}{4f(\theta)^2n}\right)$$

Hence it follows that $\operatorname{ARE}(\operatorname{Med}, \bar{X}) = 4f(\theta)^2\sigma_F^2$.

Now it says suppose $F \sim N(\theta, \sigma_F^2)$. Then $\operatorname{ARE}( \operatorname{Med}, \bar{X}) = \frac{2}{\pi} = 0.64$

Where is the $\frac{2}{\pi}$ coming from? I certainly believe it to be true, but I don't understand how that comes from the $4f(\theta)^2\sigma_F^2$.

Answer 1

When $F$ has normal$(\theta,\sigma_F^2)$ distribution, the density is $$ f(x)=\frac1{\sqrt{2\pi\sigma_F^2}} \exp- {(x-\theta)^2\over 2\sigma_F^2} $$ so its value at $x=\theta$ is $$ f(\theta)=\frac1{\sqrt{2\pi\sigma_F^2}}. $$