Unbiased estimator of the ratio of variances

Question

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.

Answer 1

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.

Answer 2

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!