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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 $V(S_1^2/S_2^2).$

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  • $\begingroup$ 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. $\endgroup$ – BruceET Aug 7 at 5:36
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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

enter image description here

hist(f, prob=T, br=50, col="skyblue2")
 curve(df(x,19,19), add=T, col="red")
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  • $\begingroup$ Is the variance the one in Wikipedia with many d variables? Also why does the mean in Wikipedia state that d2/(d2 - 2)? I think you really make great explanations but I just have a hard time following especially when you say I should just reference Wikipedia. But I understand how the sampling distribution of the sample variance represent an F-distribution. I just really wanted outright formulas, with all of them being enumerated if possible. $\endgroup$ – AndroidV11 Aug 8 at 10:55
  • $\begingroup$ (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. $\endgroup$ – BruceET Aug 8 at 16:16

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