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I'm trying to work out the MSE of an estimator $T$ for the quantile of a normal distribution. So, $$ T(X_1,\dots,X_n) = \overline X + z_q S $$ where $x_q = \mu + z_q \sigma$ and where $z_q$ is a corresponding lower quantile of $Z\sim N(0,1)$.

I know that the MSE of this is $\mathbb{E}((T - x_q)^2)$ but I want the MSE as a function of $\mu,\sigma, n$ and $q$.

What I have so far: $$ \begin{aligned} \text{MSE}(T) &= \mathbb{E}((T - x_q)^2) = \text{Var}(T) + (\mathbb{E}(T)-x_q)^2 \\ &=S+(\hat X -x_q)^2 \end{aligned} $$ But I'm unsure where to go next.

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This is simpler than it looks because for an iid Normal sample, the mean $\bar X$ is independent of the sample standard deviation $S$. Therefore the variance ("MSE") of their linear combination $$T=\bar X + z_\alpha S$$ can be expressed as a linear combination of their variances. Let's find those variances and then do the algebra.

For a standard Normal distribution with $\mu=0$ and $\sigma=1$, the variance of the mean is $\sigma^2/n=1/n$, the expectation of $S^2$ is the standard variance of $\sigma^2=1$, and (this is the hard part) the expectation of $S$ is

$$c(n)=\mathbb{E}(S) = \sqrt{\frac{2}{n-1}} \frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)}.$$

Therefore, using the fact that $\operatorname{Var}(S) = \mathbb{E}(S^2) - \mathbb{E}(S)^2$,

$$\operatorname{Var}(T) = \operatorname{Var}(\bar X) + z_\alpha^2 \operatorname{Var}(S) = \frac{1}{n} + z_\alpha^2(1 - c(n)^2).\tag{1}$$

Because $T$ scales in proportion to $\sigma$ and is merely shifted by $\mu$, its variance for an underlying Normal$(\mu,\sigma^2)$ distribution is $\sigma^2$ times this value.


This can be confirmed via simulation, as shown in the appended R code. After you specify the values of all parameters in the question--$n$, $z_\alpha$, $\sigma$, and $\mu$--and you specify the size of the simulation (as n.sim), the code computes n.sim realizations of $T$, plots their histogram (to confirm it is behaving as expected), and finishes by reporting the variance of these realizations and the theoretical value of its variance (the MSE of $T$) as given by $\sigma^2$ times formula $(1)$.

A computational subtlety attends the calculation of $c(n)$: for $n\gt 343$, $\Gamma(n/2)$ overflows double-precision arithmetic. It is therefore essential to compute the value of $c(n)^2$ in $(2)$ using logarithms of the Gamma function, following the formula

$$c(n)^2 = \frac{2}{n-1}\exp\left(2\left(\log\Gamma\left(\frac{n}{2}\right) - \log\Gamma\left(\frac{n-1}{2}\right)\right)\right).$$

Here is an example of the simulation. After issuing the command set.seed(17) (to initialize the random number generator), its output for $\mu=-40, \sigma=5, n=8, z_\alpha=3$ and $10\,000$ iterations was

Variance is 18.52; expected is 18.59

The values are close. They remain consistently close when the input parameters are varied.

n <- 8
z <- 3
sigma <- 5
mu <- -40
n.sim <- 1e4

set.seed(17)
x <- matrix(rnorm(n*n.sim, mu, sigma), nrow=n)
T <- apply(x, 2, function(y) mean(y) + z * sd(y))
hist(T)

v <- 1/n + z^2 * (1 - 2/(n-1) * exp(2*(lgamma(n/2) - lgamma((n-1)/2))))
message("Variance is ", signif(var(T), 4), "; expected is ", signif(sigma^2 * v, 4))
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  • $\begingroup$ "The theoretical value of the variance (the MSE of $T$)." $\endgroup$ – whuber Oct 26 '16 at 19:07
  • $\begingroup$ Your question provides four numerical inputs: $n,\mu,\sigma, z_q$. As both the formulas and the code demonstrate, the answer therefore depends on those numbers and those numbers only (and it actually doesn't matter what value $\mu$ has). Aren't they precisely the "variables you wanted"? $z_\alpha$ is my name for what you named $z_q$, whose relationship to $q$ you clearly described. If you want to rewrite it, you will have to tell us what it means and how it might be related to $n, \mu$, and $\sigma$. Your question only indicates it's a fixed number. $\endgroup$ – whuber Oct 26 '16 at 19:33
  • $\begingroup$ It doesn't depend on $\mu$ at all, nor does it depend on $\sigma$: you have explicitly stated it's a quantile of a standard Normal distribution. That has nothing to do with $\mu$ or $\sigma$. $\endgroup$ – whuber Oct 26 '16 at 19:41
  • $\begingroup$ As I pointed out in my reply, and again in a comment, the MSE does not depend on $\mu$. $\endgroup$ – whuber Oct 26 '16 at 19:43

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