# Is there a formula for the expectation and variance of a ratio of sampling variances?

I know the formulas for the expectation and variance of the sample variance, difference between two sample means and proportions.

Sample Variance

$$E(S^2) = σ^2$$

$$V(S^2) = 2\sigma^4/(n-1)$$

Difference Between Two Sample Means

$$E(\bar X_1-\bar X_2) = μ_1 - μ_2$$

$$V(\bar X_1-\bar X_2) = (σ_1^2 / n_1 + σ_2^2 / n_2)$$

Difference Between Two Sample Proportions

$$E(\bar X_1-\bar X_2) = p_1 - p_2$$

$$V(\bar X_1-\bar X_2) = (p_1(1 - p_1) / n_1 + p_2(1 - p_2) / n_2)$$

I am interested to know the formulas of $$E(S_1^2/S_2^2)$$ and $$V(S_1^2/S_2^2).$$

• Some of your formulas were damaged in pasting them here. Please make sure I did not introduce anything that changes your meaning. The formula for $V(S^2)$ was especially unclear may require your attention. // Also please state the population distributions for each part. – BruceET Aug 7 at 5:36

In your last sentence about $$S_1^2/S_2^2,$$ do you intend these to be sample variances of normal data? if so, please see Wikipedia on F-distributions.

There you will find the following: If $$S_1^2$$ is the variance of a normal sample of size $$n_1$$ and $$S_2^2,$$ is, independently, the variance of a normal sample of size $$n_2,$$ where both normal populations have the same variance $$\sigma^2 = \sigma_1^2 = \sigma_2^2,$$ then $$\frac{S_1^2/\sigma_1^2}{S_2^2/\sigma_2^2}=\frac{S_1^2}{S_2^2} \sim \mathsf{F}(n_1-1,n_2-1),$$

which has mean $$\frac{n_2-1}{n_2-3},$$ for $$n_2>3.$$

You should be able to adjust the result for populations with different variances $$\sigma_1^2$$ and $$\sigma_2^2.$$ Wikipedia also gives the variance of this distribution, which depends on both sample sizes.

For illustration, the simulation below computes $$F=S_1^2/S_2^2$$ for a million pairs of independent samples size $$n_1=n_2=20$$ from $$\mathsf{Norm}(\mu=100,\sigma=15).$$ The histogram shows the million 'variance ratios' along with the density function of $$\mathsf{F}(19,19).$$

set.seed(806)
f = replicate(10^6, var(rnorm(20, 100, 15))/var(rnorm(20, 100, 15)))
mean(f); var(f)
[1] 1.118323    # approx E(F) = 19/17 = 1.118
[1] 0.3176105   # approx V(F)
19/17
[1] 1.117647


hist(f, prob=T, br=50, col="skyblue2")

• (1) The formula for the variance of an F dist'n is messy, but it depends only on the numerator and denominator degrees of freedom, denoted $d_1$ and $d_2,$ respectively. For our F-ratio, $d_1 = n_1 - 1 = 19, d_2 = n_2 - 1 = 19.$ (2) The mean depends only on $d_2.$ (3) If you have $\sigma_1 \ne \sigma_2,$ then you need to use $\frac{S_1^2/\sigma_1^2}{S_2^2/\sigma_2^2} = \frac{\sigma_2^2}{\sigma_1^2}\cdot\frac{S_1^2}{S_2^2}.$ (4) Depending on how much you know about F-dist'ns, you may need to supplement Wikipedia with an intermed. level stat text. My goal to give helpful clues, not complete Ans. – BruceET Aug 8 at 16:16