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Let $X_1,...X_n$ be iid $N(0,\sigma^2)$ where $\sigma$ is unknown. My task is to find the BLUE for $\sigma$ within the set of linear functions of $|X_i|$ for $i=1,...,n$. Here is my work thus far:

First, let $T=\sum_{i=1}^nc_i|X_i|$ be an estimator for $\sigma$. Since $T$ is assumed to be unbiased, it must be true that $E[T]=\sigma$, and by the linearity of $E[.]$, it must also be true that: $$E[|X|]\sum_{i=1}^nc_i=\sigma.$$

By performing the appropriate transformation of $X$, it can be shown via the integration method that $E[| X|]=\sigma\sqrt\frac{2}{\pi} $.

At this point, I don't really know what to do. I can solve for $\sum_{i=1}^nc_i$, but I'm not even sure what this quantity represents, or if it's what I'm supposed to be looking for. Any help appreciated.

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We have

\begin{align} E_{\sigma}(T)&=\sum_{i=1}^n c_iE_{\sigma}(|X_i|) \\&=\sigma\sqrt{\frac{2}{\pi}}\sum_{i=1}^n c_i \end{align}

For $T$ to be an unbiased estimator of $\sigma$, we must have

$$\sqrt{\frac{2}{\pi}}\sum_{i=1}^n c_i=1$$

Or, $$\sum_{i=1}^n c_i=\sqrt{\frac{\pi}{2}}$$

Now,

\begin{align} \operatorname{Var}_{\sigma}(T)&=\sum_{i=1}^n c_i^2\operatorname{Var}_{\sigma}(|X_i|) \\&=\sum_{i=1}^n c_i^2 \left[E_{\sigma}(X_i^2)-(E_{\sigma}(|X_i|))^2\right] \\&=\sum_{i=1}^n c_i^2 \left[\sigma^2+\frac{2}{\pi}\sigma^2\right] \\&=\sigma^2\left(1+\frac{2}{\pi}\right)\sum_{i=1}^n c_i^2 \end{align}

For $T$ to be the BLUE of $\sigma$, the value of $\operatorname{Var}_{\sigma}(T)$ has to the minimum among the variances of all linear unbiased estimators of $\sigma$.

So your problem boils down to

$$\text{ Minimize }\qquad\sum_{i=1}^n c_i^2 \\\quad\qquad\text{ subject to }\qquad\sum_{i=1}^n c_i=\sqrt{\frac{\pi}{2}}$$

Can you proceed now, using some known inequalities, or by using Lagrange multipliers?


By Cauchy-Schwarz,

$$\sum_{i=1}^n c_i^2\sum_{i=1}^n \left(\frac{1}{\sqrt{n}}\right)^2\ge \left(\sum_{i=1}^n \frac{c_i}{\sqrt{n}}\right)^2$$

Or,

$$\sum_{i=1}^n c_i^2 \ge \frac{1}{n}\left(\sum_{i=1}^n c_i\right)^2$$

Equality holds if and only if

\begin{align} c_i & \propto \frac{1}{\sqrt n}\quad ,\,i=1,2,\ldots,n \\ \iff c_i&=\frac{k}{\sqrt n}\quad\text{ for some }k \end{align}

From the constraint $\sum_{i=1}^n c_i=\sqrt{\frac{\pi}{2}}$, you will get $k$.

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  • $\begingroup$ Is variance a linear functional? I'm a little confused about why you're able to pull out $\sum_{i=1}c_i$ from the $Var(T)$ $\endgroup$ – user443369 Nov 1 '18 at 17:28
  • $\begingroup$ @DavidS In general, $$\operatorname{Var}\left(\sum a_i X_i\right)=\sum a_i^2\operatorname{Var}(X_i)+\sum_{i\ne j}a_ia_j\operatorname{Cov}(X_i,X_j)$$ The covariance terms vanish whenever $X_1,X_2,\ldots$ are independent. $\endgroup$ – StubbornAtom Nov 1 '18 at 17:33
  • $\begingroup$ Got it. Also, you say that I'm minimizing $\sum_{i=1}^nc_i^2$, but aren't I supposed to be minimizing the variance of $T$, with respect to $\sum_{i=1}^nc_i^2$? $\endgroup$ – user443369 Nov 1 '18 at 20:29
  • $\begingroup$ @DavidS Minimizing $\operatorname{Var}(T)$ is equivalent to minimizing $\sum c_i^2$ (the multiplier $\sigma^2\left(1+\frac{2}{\pi}\right)$ is just a constant). And obviously, we are minimizing with respect to $c_1,c_2,\ldots,c_n$. Using Lagrange multipliers, for example, you would get the optimal $c_i$'s for which $\sum c_i^2$ is minimized subject to the given constraint. $\endgroup$ – StubbornAtom Nov 1 '18 at 20:37
  • $\begingroup$ Not familiar with the Lagrange method but am I correct in saying that I need to differentiate $T$ with respect to each $c_i$ separately? $\endgroup$ – user443369 Nov 1 '18 at 23:51

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