for $N$ samples of two correlated random variables $X \sim N\left(0,\sigma_X^2\right)$ and $Y \sim N\left(0, \sigma_Y^2\right)$ with correlation $\rho$, I am analyzing the ratio of the sample variances, $r=\frac{s_Y^2}{s_X^2}$. Currently, I am foccusing on $E\left[r\right] = E\left[\frac{s_Y^2}{s_X^2}\right]$.

I have found that $s_X^2 \sim \Gamma\left(\frac{N-1}{2}, \frac{2 \sigma_X^2}{N-1}\right)$, $s_Y^2 \sim \Gamma\left(\frac{N-1}{2}, \frac{2 \sigma_Y^2}{N-1}\right)$ since $\frac{(N-1)S_X^2}{\sigma_X^2}\sim \chi_{N-1}^2$ (and likewise for $s_Y^2$).

The ratio of the sample variances is therefore the ratio of two dependent gamma distributions. I have found some papers on this topic, for instance [1, 2]. These papers do however always assume the gamma distributions having different shape and same scale parameters (scale 1 to be precise) - just the opposite of my problem.

I have tried using $\frac{N-1}{2 \sigma_X^2} s_X^2 \sim \Gamma \left(\frac{N-1}{2}, 1\right)$, $\frac{N-1}{2 \sigma_Y^2} s_Y^2 \sim \Gamma \left(\frac{N-1}{2}, 1\right)$ - the ratio of these should be beta prime distributed.
Using the formula for the expected value of a beta prime distribution this leads to
$E\left[\frac{s_Y^2}{s_X^2}\right] =E\left[\frac{\frac{N-1}{2 \sigma_X^2}}{\frac{N-1}{2 \sigma_Y^2}} \frac{\frac{N-1}{2 \sigma_Y^2}}{\frac{N-1}{2 \sigma_X^2}} \frac{s_Y^2}{s_X^2}\right] =\frac{\sigma_Y^2}{\sigma_X^2} E\left[\frac{\frac{N-1}{2 \sigma_Y^2}}{\frac{N-1}{2 \sigma_X^2}} \frac{s_Y^2}{s_X^2}\right] =\frac{\sigma_Y^2}{\sigma_X^2} \frac{\alpha}{\beta - 1}$ with $\alpha = \beta = \frac{N-1}{2}$,
which doesn't seem to be the correct result (shouldn't the ratio in some way depend on the correlation?) and which differs from the results of numerical experiments.

How can the expected value of the ratio be calculated? At which point am I mistaken?
I would very much appreciate any help - thank you in advance.

[1] Lee et al.: Distribution of a ratio of correlated gamma random variables. SIAM Journal on Applied Mathematics 36.2 (1979): 304-320.
[2] Tubbs: Moments for a ratio of correlated gamma variates. Communications in Statistics-Theory and Methods 15.1 (1986): 251-259.

  • 1
    $\begingroup$ Because $X/\sigma_X, Y/\sigma_Y \sim N(0,1),$ the distribution of $r$ must be $\sigma_Y^2/\sigma_X^2$ times the solution found by assuming both $X$ and $Y$ have unit variances. $\endgroup$
    – whuber
    Commented Jun 23, 2014 at 13:40
  • $\begingroup$ Assuming $X, Y \sim N\left(0,1\right)$ one has $s_X^2, s_Y^2 \sim \Gamma \left( \frac{N-1}{2}, \frac{2}{N-1} \right)$. Using the reasoning with the beta prime distribution described above, this leads to the same result. I am therefore not sure in which way using unit variances might be useful. Could you please further clarify your comment? $\endgroup$
    – sebastianb
    Commented Jun 23, 2014 at 14:11
  • $\begingroup$ Your question remarks that you have found papers that can solve this problem when the Gamma distributions are correlated and the scale parameters are equal. I have reduced your question to that, so I am supposing you can now consult your references for the answer. $\endgroup$
    – whuber
    Commented Jun 23, 2014 at 14:16
  • 1
    $\begingroup$ I don't follow your latest comment: because $\sigma_X$ and $\sigma_Y$ are parameters, the concept of one "depending" on the other does not seem to apply. $\endgroup$
    – whuber
    Commented Jun 23, 2014 at 14:18
  • 1
    $\begingroup$ @sebastianb You specify that $X$ is Normal, and $Y$ is Normal, and that they are correlated $\rho$, but you do not state whether $(X,Y)$ are jointly Normal ... or what their joint distribution is. I don't see how you can calculate the desired ratio expectation, given dependence, unless the joint pdf is known and stated. $\endgroup$
    – wolfies
    Commented Jun 23, 2014 at 14:59

1 Answer 1


If we assume that the underlying normals are jointly normal, then the result is very simple.

This answer uses results from A. H. Joarder (2007/2009), Moments of the product and ratio of two correlated chi-square variables (open access).

We have

$$r\equiv \frac{s_y^2}{s_x^2} = \frac {\sigma^2_y}{\sigma^2_x}\frac{(n-1)s_y^2/\sigma^2_y}{(n-1)s_x^2/\sigma^2_x}= \frac {\sigma^2_y}{\sigma^2_x} \frac {U_y}{U_x},\;\; U_y \sim \mathcal \chi^2_{(n-1)},\;\; U_x \sim \mathcal \chi^2_{(n-1)}$$


$$E(r) = \frac {\sigma^2_y}{\sigma^2_x} E\left(\frac {U_y}{U_x}\right)$$

In other words, we need the expected value (first raw moment) of the ratio of two correlated chi-squares with equal degrees of freedom.

Note that standardizing the normals does not affect their correlation coefficient, denote it simply $\rho \in (-1,1)$. Then according to the above paper (which references another paper), $$\operatorname{Corr}(U_y, U_x) = \rho^2$$

and by Corollary 3.8 p. 590, if $n-1 >2$ (i.e we need a sample size at least equal to $4$) we have

$$E\left(\frac {U_y}{U_x}\right) = \frac {n-1 -2\rho^2}{n-1-2} = \frac {n-1 -2\rho^2}{n-3}$$


$$E(r) = \frac {\sigma^2_y}{\sigma^2_x}\frac {n-1 -2\rho^2}{n-3}$$

The paper contains also expressions for the next three raw moments, increasing slightly the requirement on the size of the sample. Results appear to depend on the two chi-squares having the same degrees of freedom.

  • $\begingroup$ Thank you, great answer; the expected value from numerical simulations matches the expected value calculated using the formula in your answer. Furthermore, I have gone back to Tubb's "Moments for a ratio of correlated gamma variates" and now came to the same result. However, Tubb imposed the restriction of positive correlation (at least for the general case). Therefore your way using the chi squared distribution is actually more reasonable than using the gamma distribution. $\endgroup$
    – sebastianb
    Commented Jun 24, 2014 at 6:53
  • 1
    $\begingroup$ Isn't it great to be able to verify theoretical results in an artificial environment (simulation), and not, say, having first to build a bridge that will collapse in order to realize that somewhere in the calculations a minus sign was taken as a plus sign? Thanks for the verification! $\endgroup$ Commented Jun 24, 2014 at 8:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.