Degree of Freedom

CFA Level I Quant
#1

Hello All,

I came across this question:

If a random variable X = (Variance of “k” Sample Means)/ (Average of “k” Sample Variances), what are the degrees of freedom for this random variable?

Assume that each of the “k” sample mean is calculated by averaging “n” random variables. Moreover, in the denominator, each of the “k” sample variance is calculated from “n” random variables.

The OA is (k-1)/(k(n-1)). I understand that the degrees of freedom of sample mean of n random variables is n-1, and that the degrees of freedom of sample variance of n random variables is n-1. Hence, the denominator of the OA makes sense to me because we will be calculating (n-1) times k. However, I am not sure why the numerator of OA has k-1. Shouldn’t it be (k)*(n-1)? Each sample mean in the numerator will have n-1 degree of freedom. Averaging such k sample means should give us (k)*(n-1) degree of freedom. Isn’t it?

Can someone please explain this to me?

I am really confused. I would appreciate any help.

Thanks in advance.

#2

The official answer seems stupid: if, say, k = 5 and n = 100, then the formula gives 4/(5×99) = 0.008081 degrees of freedom? Absurd.

I think it should be k(n – 1) – 1, so if, say, k = 5 and n = 100, then the formula gives 5(99) – 1 = 494.

You have 500 random variables, and you’ve calculated 6 statistics (5 means of the samples, and one mean of the means), so you should have 500 – 6 = 494 degrees of freedom.

#3

Hello S2000magician,

Thank you so much for your response. I truly appreciate it.

Best regards

#4

My pleasure.