This is a problem from Hogg & Tanis, 8th edition, p. 282:

Let $X_1, X_2, ... X_n$ be a random sample of size $n$ from a normal distribution. Show that an unbiased estimator of $\sigma$ is $cS$, where


I used the fact that $U=\frac{n-1}{\sigma^2}S^2$ is $\chi^2(n-1)$ and looked for $E[\sqrt{U}]$. So —

$$E[\sqrt{U}]=\int_0^\infty \sqrt{u}\frac{u^{(n-1)/2}e^{-u/2}}{\Gamma\left(\frac{n}{2}\right)\cdot2^{n/2}}du$$

After combining terms and throwing in a bunch of extra (but apparently necessary) terms, this simplifies to

$$\frac{\Gamma\left(\frac{n}{2}\right)\sqrt{2}}{\Gamma\left(\frac{n-1}{2}\right)}\int_0^\infty f(v)dv$$

where $V\sim \chi(n)$; the integral, therefore, evaluates to $1$. Since $U=\frac{n-1}{\sigma^2}S^2$ — and this step seems way too easy; for some reason, I have the impression I should be suspicious of maneuvers like this —


and so


Moving the constants to the RHS and multiplying both sides by $c$ yields the desired result, that $E[cS]=\sigma$.

My question is: Does that work?

As far as I can tell, this $\frac{n-1}{\sigma^2}S^2$ business appeared fully formed and without any explanation in the text. I don't know what it is, I don't know what to do with it, I probably don't even know what I know. Can anyone suggest some sources where I could get some more insight into this thing? Apparently it's called "sample variance", but looking that up in my text's index isn't entirely helpful.

(There's another part to this question, on the convergence of $c$, I'm totally stuck on; I'll be posting that question separately.)

  • $\begingroup$ You don't show all your steps. Can you show how you computed the density of $\sqrt{U}$ before you calculated the expectation? In general, $P(g(X)\leq x) \neq F(g(x))$ but $F(g^{-1}(x))|\frac{dx}{dg}|$..you need the Jacobian term to keep the probability integratig to 1. You may have already done this and other steps, but it would help if you showed all your work if you want us to evaluate your derivation. $\endgroup$ – user76844 Jan 29 '14 at 5:23
  • $\begingroup$ @Eupraxis1981: Early on I considered a transformation of variables, but I didn't end up doing that, either because I couldn't figure out a way to do so or because my professor later gave my class a hint on the first step. I'm computing $E[g(X)]$, which is equal to $\int_{-\infty}^\infty g(x)f(x)dx$ where $f(x)$ is the pdf. As I wrote, I'm still figuring out what's going on, so am I missing something here? $\endgroup$ – dmk Jan 29 '14 at 19:55
  • $\begingroup$ @Eupraxis1981: Beyond that, the steps I left out were those involving the manipulation of fractions; most of this occurred between setting up the integral for $E[\sqrt{U}]$ and introducing $f(v)$. In the end, I seem to have gotten the desired result, which would be unlikely if I multiplied by the wrong the number somewhere. What I'm wondering is if my reasoning looks sound. $\endgroup$ – dmk Jan 29 '14 at 20:02
  • $\begingroup$ Isee. Then your reasoning appears correct. The step you are suspicious of looks fine to me. U is defined as the distribution of the sample variance, so the square root of that value will give the distribution of the sample standard deviation. You correctly calculated the expected value of $\sqrt{U}$ and so have the bias of the sample standard deviation. $\endgroup$ – user76844 Jan 29 '14 at 21:17
  • $\begingroup$ In finding $E(g(x))=\int_{-\infty}^\infty g(x)f(x)dx$ what is the p.d.f used?Is it the pdf in en.wikipedia.org/wiki/Chi_distribution? Can you please explain a bit more as how you computed $E[\sqrt{U}]=\int_0^\infty \sqrt{u}\frac{u^{(n-1)/2}e^{-u/2}}{\Gamma\left(\frac{n}{2}\right)\cdot2^{n/2}}du$ $\endgroup$ – clarkson Apr 22 '14 at 3:08

$$E(cS) = cE(S) = c\frac {\sigma}{\sqrt{n-1}}E\left(\frac {\sqrt{n-1}}{\sigma}S\right)$$

The variable inside the expected value now follows a chi-distribution with d.f. $k=n-1$, so

$$E(cS) = c\frac {\sigma}{\sqrt{n-1}}\sqrt{2}\,\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)}$$


$$E(cS) = \frac{\sqrt{n-1}\Gamma\left(\frac{n-1}{2}\right)}{\sqrt{2}\Gamma\left(\frac{n}{2}\right)}\frac {\sigma}{\sqrt{n-1}}\sqrt{2}\,\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)} = \sigma$$

So you're good.

  • $\begingroup$ Is chi square and chi are two different distributions.I only knew of chi square distribution $\endgroup$ – clarkson Apr 22 '14 at 3:10
  • $\begingroup$ @clarkson Yes they are. I have provided the wikipedia link for a first reference. $\endgroup$ – Alecos Papadopoulos Apr 22 '14 at 3:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.