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Consider two samples $X_{1}, \cdots ,X_{m}$ and $Y_{1}, \cdots ,Y_{n}$ where $X_{i} \thicksim N(\mu_{1}, \sigma_{1}^2), i.i.d.$ and $Y_{j} \thicksim N(\mu_{2}, \sigma_{2}^2), i.i.d.$.

Say that both $\mu_{1}$ and $\mu_{2}$ are unknown. Then, what will be an unbiased estimator of the ratio of variances? I mean, unbiased estimator of this; $$\frac{\sigma_{2}^2}{\sigma_{1}^2}$$ And how to proof that?

All of statistics textbooks that I have don't explain the things above. I would appreciate if you help me.

update(2019/11/22) I found a book that shows the unbiased estimator of the ratio of variances in the same condition as I wrote above. The book says, $$\frac{\sum_{i=1}^{n}(Y_i-\bar{Y})^2/(n-1)}{\sum_{i=1}^{m}(X_i-\bar{X})^2/(m+1)}$$ is the unbiased estimator that I want to know. But I can't show its unbiassedness. I tried to use Jensen's inequality, but my friend pointed out it's not effective. Also, @StubbornAtom gave me advice that it is not an unbiased estimator.

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    $\begingroup$ I suppose you’re specifically interested in the finite-sample case. So this might not answer your question but you can get an asymptotically unbiased estimator by dividing $\hat\sigma_2^2$ by $\hat\sigma_1^2$. This works because of Slutsky’s theorem. en.wikipedia.org/wiki/Slutsky%27s_theorem $\endgroup$ – Student Nov 9 '19 at 3:52
  • $\begingroup$ Add the self-study tag. $\endgroup$ – Michael R. Chernick Nov 9 '19 at 4:28
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    $\begingroup$ Assuming you know the basic properties of variances from normal random variables, ;ook up the variance of the F distribution and then you will be able to write down the expectation of the usual estimator of the ratio of variances, and from that seeing how to unbias it is simple. You can derive the expected value of an F-distributed random variable from the formula for the kth moment given there. As for proving that formula, you could attempt it yourself and then if you don't get anywhere, perhaps ask a question about that. $\endgroup$ – Glen_b -Reinstate Monica Nov 9 '19 at 6:50
  • $\begingroup$ @Glen_b I'll try it. Thank you. $\endgroup$ – dff Nov 10 '19 at 1:31
  • $\begingroup$ I found a book that shows the unbiased estimator of the ratio of variances in the same condition as I wrote above. The book says, $$\frac{\sum_{i=1}^{n}(Y_i-\bar{Y})^2/(n-1)}{\sum_{i=1}^{m}(X_i-\bar{X})^2/(m+1)}$$ is the unbiased estimator that I want to know. But I can't show its unbiassedness. I tried to use Jensen's inequality, but my friend pointed out it's not effective. $\endgroup$ – dff Nov 20 '19 at 1:51
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Assuming the two samples are independent of each other.

Use the fact that $\frac{(m-1)S_1^2}{\sigma_1^2}\sim \chi^2_{m-1}$ and $\frac{(n-1)S_2^2}{\sigma_2^2}\sim \chi^2_{n-1}$, where $S_1^2=\frac1{m-1}\sum\limits_{i=1}^m (X_i-\overline X)^2$ and $S_2^2=\frac1{n-1}\sum\limits_{i=1}^n (Y_i-\overline Y)^2$ are the sample variances.

Since the samples are independent, so are $S_1^2$ and $S_2^2$.

A reasonable guess for an estimator of the ratio of variances is the ratio of the sample variances. Getting unbiasedness from there is simple.

We already have $E(S_2^2)=\sigma_2^2$. So find an unbiased estimator of $1/\sigma_1^2$ based on $S_1^2$. For that, start with $E[1/S_1^2]$ to get $E\left[\frac{c}{S_1^2}\right]=\frac{1}{\sigma_1^2}$ for some constant $c$. Use the independence of $S_1^2$ and $S_2^2$ to eventually get

$$E\left[\frac{cS_2^2}{S_1^2}\right]=E\left[S_2^2\right]E\left[\frac{c}{S_1^2}\right]=\frac{\sigma_2^2}{\sigma_1^2}$$

The easier way to do this is to construct an F distribution from the independent chi-square distributions. We have

$$\frac{(n-1)S_2^2/(\sigma_2^2(n-1))}{(m-1)S_1^2/(\sigma_1^2(m-1))}=\frac{S_2^2/\sigma_2^2}{S_1^2/\sigma_1^2}\sim F_{n-1,m-1}$$

Taking expectation immediately leads to the answer.

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  • $\begingroup$ Strange downvote. $\endgroup$ – StubbornAtom Nov 19 '19 at 2:04
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Using Slutsky's theorem of large sample theory you will get an assymptotic unbiased estimator! Else if you want to get an estimator of it then find out the MLEs and use the invariance property!

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    $\begingroup$ The invariance property tells you how to find the MLE of $\sigma_1^2/\sigma_2^2$, but says nothing (on its own) about the bias of this estimator. $\endgroup$ – knrumsey Nov 18 '19 at 22:04
  • $\begingroup$ Yeah! that's why i told that if you want to get an estimator $\endgroup$ – Kushal Bhattacharya Nov 19 '19 at 4:21
  • $\begingroup$ The question is how to find an unbiased estimator. Which I think is addressed nicely in the other answer. $\endgroup$ – knrumsey Nov 19 '19 at 4:23
  • $\begingroup$ Yeah! absolutely $\endgroup$ – Kushal Bhattacharya Nov 19 '19 at 4:24

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