# Probability that one sample variance is greater than another

I'm practicing for a stats exam and I came upon a tough problem. Here's how it goes (Translated from french so translation may not be perfect):

We have 2 independent samples, the first is size 10 from a normally distributed population with variance 4 and the second is size 5 from a normally distributed population with variance 2.

Find an expression for the probability that the sample variance of the second sample is greater than the sample variance of the first one.

I don't really know where to start. I thought about using the chi-squared formula for sample variance but I haven't found anything so far.

Thanks for your help :)

## 1 Answer

Let $X_1,\ldots,X_{10}$ be the first sample, $X_i\sim N(a,4)$. Then for (unbiased) sample variance $$S_X^2=\frac{1}{9}\sum_{i=1}^{10}\left(X_i-\bar X\right)^2$$ we know that $$\frac{9S_X^2}{4}\sim \chi^2_9.$$ Let $Y_1,\ldots,Y_5$ be the second sample independent from the first one. Here $Y_i\sim N(b,2)$ For its (unbiased) sample variance $$S_Y^2=\frac{1}{4}\sum_{i=1}^{5}\left(Y_i-\bar Y\right)^2$$ we know that $$\frac{4S_Y^2}{2}\sim \chi^2_4.$$

Note also that the following ratio of two appropriately scaled independent chi-squared r.v.'s is $F$-distributed: $$\dfrac{\frac{\chi^2_9}{9}}{\frac{\chi^2_4}{4}}=\dfrac{\frac19\frac{9S_X^2}{4}}{\frac14\frac{4S_Y^2}{2}}=\frac{S_X^2}{2S_Y^2}\sim F_{9,4}$$ We need to find $$\mathbb P(S_X^2 < S_Y^2) = \mathbb P\left(\frac{S_X^2}{2S_Y^2}<\frac12\right)=F_{9,4}\left(\frac12\right)=0,178196058830063$$ The last value can be found using any tool which can found probabililties for $F$-distribution.

Note that if sample variances are biased, you will need small changes in constant factors.