The square root of weighted sum of chi-squared distribution

Question

Let $X\sim\chi_m^2$ and $Y\sim\chi_n^2$ be two independent variables. How to calculate or estimate the expectation of $\sqrt{aX+bY}$, where $a,b>0$?

Answer 1

$\chi_m^2$ variables are also known as $\Gamma(m/2, 2)$, using the shape-scale parameterization of the gamma distribution.

Scaled gamma distributions are themselves gamma, with the scale parameter scaled equally. So $a X \sim \Gamma(m / 2, 2 a)$, $b Y \sim \Gamma(n / 2, 2 b)$.

The sum of independent gammas with the same scale is itself gamma, so that if $b = a$, then $a X + a Y \sim \Gamma(\frac{n+m}{2}, 2 a)$. In that case, its square root is a Nakagami distribution, with mean $$\mathbb E \sqrt{a X + a Y} = \frac{\Gamma(\frac{n+m+1}{2})}{\Gamma(\frac{n+m}{2})} \sqrt{2 a}.$$

If $a \ne b$, I don't think there's such a neat answer. You can find or approximate the distribution of $a X + b Y$ in various ways, outlined in the answers to this question:

• whuber's answer gives a way to get the exact distribution which you could numerically integrate to get the expected square root.
• kjetil's answer gives code for a numerical approximation to the pdf, which again you could numerically integrate for the expected square root.

• Paul's answer uses the Welch-Satterthwaite equation to approximate the sum as a gamma. Using that approximation, we get

  $$a X + b Y \stackrel{approx}{\sim} \Gamma\left( \frac{(m a + n b)^2}{2 (m a^2 + n b^2)}, 2 \frac{m a^2 + n b^2}{m a + n b} \right)$$

  which leads to

  $$ \mathbb E \sqrt{a X + b Y} \approx \frac{\Gamma\left( \frac{(m a + n b)^2}{2 (m a^2 + n b^2)} + \frac12 \right)}{\Gamma\left( \frac{(m a + n b)^2}{2 (m a^2 + n b^2)} \right)} \sqrt{ 2 \frac{m a^2 + n b^2}{m a + n b}} .$$

In any case, you should probably check your answer against a simple Monte Carlo simulation for the parameters you're interested in.