# Confusion about greater variance in the numerator for F ratio

I am studying about F ratio and how, as a random variable, it follows F Distribution. So let me explain what confuses me.

This is what the theory says -- We draw two random samples $sample_x$ and $sample_y$ from two different Normally distributed populations with equal variance $\sigma^2$. Let the sample variances of these samples be $s^2_x$ and $s^2_y$ respectively. The fixed sample sizes for $sample_x$ is $n_x$ and the sample size for $sample_y$ is $n_y$. Then if we form the random variable $\frac{\sigma^2_x}{\sigma^2_y}$ , such that the greater variance (whichever is the greater variance in that sample pair) must appear appear as the numerator. This is what I am not able to understand.

If it's a random variable for the sampling distribution for that ratio -- it means , if we draw a random sample pair (x,y) with fixed sizes $n_x$ and $n_y$ many many times from their respective parent populations (say we do it 1000 times e.g.), we will get 1000 pairs of variances i.e. ($s^2_x$,$s^2_y$). Now if we have to draw a histogram for F distribution , we have to calculate 1000 numbers (ratios) out of each of the 1000 variance pairs ($s^2_x$,$s^2_y$). And the theory says that the greater variance has to appear as numerator in the ratio. Now how can it be fixed? Across all the 1000 pairs it may change, in some of the pairs the sample x (the first) may have higher variance, and in some of these the sample y (the second) can have the greater variance. If we have to have a common fixed formula for the random variable (supposedly $\frac{\sigma^2_x}{\sigma^2_y}$ ), how can it change from pair to pair? It has to remain fixed for all the 1000 instances. This is my dilemma.

Can you try to explain?

Thanks, Dhiraj