# Expectation and distribution of ratio of correlated Gamma/Chi-square random variables

This question is very similar to: Distribution of the ratio of dependent chi-square random variables

But the big difference is what happens when we don't have standard normal variables.

I want to know the expectation (and potentially other moments/distribution) of:

$$W = \frac{X_1^2}{X_1^2 + \sum_{i=2}^{n} X_i^2}$$

where $$X_i \sim N(0, \sigma_i^2)$$ for $$i = 1, ..., n$$.

If all the variances of the $$X$$-s are equal, then this reduces to a ratio of chi-square random variables and the distribution of $$W$$ is a beta distribution. However, when the variances are not the same, I can't see a way to use that result.

Is the answer a simpe textbook result, or does it need to be worked out from scratch? If it's the latter, what's the best way to go about it? I haven't done this before.

Another way to phrasing it is what is the distribution of $$W = \frac{U}{U+ V}$$ where $$U \sim \Gamma(\frac{N -1}{2}, 2\sigma^2)^{\text{is this correct?}}$$, but then what is $$V$$? I guess that's another question: if $$V = \sum_i \sigma_i ^2 Z_i^2$$ is the sum of a bunch of scaled standard normal variables, what is its distribution? Is it some generalised form of the Gamma distribution?

Given some commenters suggested that this may not have a closed form solution, here is an alternative question. With large $$n$$, $$\sum_{i=2}^{n} X_i^2$$ tends towards normally distributed. So the alternative question is this:

If we have

$$W_2 = \frac{Y^2}{Y^2 + a + bZ}$$

where $$Y \sim N(0, 1)$$ and $$Z \sim N(0, 1)$$ are independent and standard normal, and $$a$$ and $$b$$ are positive constants, does this have a closed form solution?

• I do not think there is a closed form solution in the general case. Jul 19, 2021 at 9:20
• @Xi'an Yeah the expectation looks like a complicated integral, hard to put it into a symbolic integrator since it has an arbitrary number of dimensions. I updated my question with a simplified form (if we assume n is large and the second chi-square tends to a normal distribution). Do you have any thoughts on that? Jul 19, 2021 at 12:31
• Re "with large $n$ ... normally distributed:" Not in general. You need to make assumptions about the $\sigma_i^2.$
– whuber
Jul 19, 2021 at 12:46
• Yeah I know. But firstly, if the variances are finite and if let's say they all go as $\sigma_i^2 / N$ for some "roughly similar" $\sigma_i$, then as $N \rightarrow \infty$ the CLT works right? And anyway, this is the only useful approximation I could think of. Tried running the integral in Wolfgram, but it says the compute time was too long and I needed a pro account, so I assume this probably means that even the approximation doesn't have a closed form solution >.<. Jul 19, 2021 at 13:04
• Where does the correlation come into play? Your initial description of the issue doesn't mention correlation and one might think that all of the $X_i^2$ random variables are independent.
– JimB
Jul 19, 2021 at 15:58

I don't see where you've described any correlation structure (despite the title using the term "correlated"). So maybe starting small with $$n=2$$ and assuming independence will be a start.

Using Mathematica (or probably just basic algebra) one can find the distribution of

$$W_2=X_1^2/(X_1^2+X_2^2)$$

where $$X_i \sim N(0,\sigma_i^2)$$.

dist = TransformedDistribution[x^2, x \[Distributed] NormalDistribution[0, s]];
distw2 = TransformedDistribution[x1/(x1 + x2),
{x1 \[Distributed] dist /. s -> \[Sigma],
x2 \[Distributed] dist /. s -> \[Sigma]}];
Mean[distw2]


with the result being

$$\frac{\sigma_1}{\sigma_1+\sigma_2}$$

That suggests that maybe for general $$n$$, there might be a simple formula for the mean. Either further manipulations or simulations could give support or squash that hope.

The variance of $$W_2$$ is $$\frac{\sigma_1 \sigma_2}{2 (\sigma_1+\sigma_2)^2}$$.

The density of $$W_3=X_1^2/(X_1^2+X_2^2+X_3^2)$$ is

$$-\frac{\sigma_1^2 \sigma_3 E\left(-\frac{(w-1) \sigma_1^2 \left(\sigma_2^2-\sigma_3^2\right)}{\left((w-1) \sigma_1^2-w \sigma_2^2\right) \sigma_3^2}\right)}{\pi \sqrt{w \left(\sigma_2^2 w-\sigma_1^2 (w-1)\right)} \left(\sigma_1^2 (w-1)-\sigma_3^2 w\right)}$$

where $$\sigma_2 \leq \sigma_3$$, $$E(.)$$ is the complete elliptic integral, and found with the following Mathematica code:

dist2 = TransformedDistribution[x2 + x3,
Assumptions -> 0 < s <= s];

dist123 = TransformedDistribution[x1/(x1 + x23),
x23 \[Distributed] dist2}];
Simplify[PDF[dist123, w], Assumptions -> s > 0]


I have not found a closed-form representation for the mean when $$n=3$$ but when the variances are known, then numerical integration should easily find the mean and other desired moments.

Special case

Suppose $$\sigma_1^2=\sigma^2_U$$ and $$\sigma_i^2=\sigma^2_V$$ for $$i=2,\ldots,n$$. Then $$W=U/(U+V)$$. The mean of that distribution can be found with

dist = TransformedDistribution[
u/(u + v), {u \[Distributed] GammaDistribution[1/2, 2 \[Sigma]u^2],
v \[Distributed] GammaDistribution[(n - 1)/2, 2 \[Sigma]v^2]}]

mean = FullSimplify[Mean[dist], Assumptions -> {\[Sigma]u > 0, \[Sigma]v > 0}]


$$\frac{(n+1) \sigma_u^2 \, _2F_1\left(-\frac{1}{2},\frac{n}{2};\frac{n+2}{2};1-\frac{\sigma_v^2}{\sigma_u^2}\right)-\left((n-1) \sigma_v^2+\sigma_u^2\right) \, _2F_1\left(\frac{1}{2},\frac{n}{2};\frac{n+2}{2};1-\frac{\sigma_v^2}{\sigma_u^2}\right)}{n \sigma_u \sigma_v}$$

where $$_2 F_1$$ is the hypergeometric function.

Specifying values of $$n$$ results in great simplifications:

FullSimplify[mean /. n -> 2, Assumptions -> {\[Sigma]u > 0, \[Sigma]v > 0}]


$$\frac{\sigma_U}{\sigma_U+\sigma_V}$$

FullSimplify[mean /. n -> 6, Assumptions -> {\[Sigma]u > 0, \[Sigma]v > 0}]


$$\frac{\sigma_U^2 (3 \sigma_U+\sigma_V)}{3 (\sigma_U+\sigma_V)^3}$$

• Ah yeah sorry the title is misleading; I mean correlated in the sense that the denominator as a whole is correlated with the numerator; though really neither the numerator or denominator are chi-square distributed. Jul 19, 2021 at 16:21
• That's interesting and a bit surprising actually, since my first (incorrect) intuition for this problem was that the answer would be the ratio of the variances (i.e. $\frac{\sigma_1^2}{\sigma_1^2 + \sum \sigma_i^2}$). Since asking my question, I ran numerical tests with $\sum_{i=1} \sigma_i^2 = 1$, a fixed $\sigma_1$ and all the other $\sigma_i^2 = \frac{1 - \sigma_1^2}{N-1}$ for $i = 2, ..., N$, and the results "look" like they tend to an asymptote, and numerically it looks like it doesn't depend on N as long as N is large, which is why I thought the second ratio would be easier. Jul 19, 2021 at 16:30
• For a large $n$, you might consider using a gamma distribution with the same mean and variance as the sum of the $X_i^2$ ($i=2,\ldots,n$) random variables. That will result in an explicit approximation for the mean (as I'm not so sure that you'll get that assuming a normal distribution).
– JimB
Jul 19, 2021 at 18:15
• Yes, it is to use a gamma rather than a normal to estimate the sum of gammas with different variances. It would be a gamma that has the same mean and variance as the sum of the gammas with different variances.
– JimB
Jul 20, 2021 at 12:02
• Sorry. Just added a definition of the $_2 F_1$ hypergeometric function.
– JimB
Jul 20, 2021 at 12:36