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 $a,b>0$?

  • $\begingroup$ I assume that a and b are known values. $\endgroup$ Apr 23, 2017 at 0:19
  • $\begingroup$ Yes, $a$ and $b$ are known. $\endgroup$ Apr 23, 2017 at 0:30
  • $\begingroup$ Then the given answer by Dougal is excellent. $\endgroup$ Apr 23, 2017 at 0:46

2 Answers 2


$\chi_m^2$ 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.

  • $\begingroup$ Is there any analysis on the accuracy of Welch-Satterthwaite approximation? $\endgroup$ Apr 23, 2017 at 0:05
  • $\begingroup$ @user07001129 I don't know of any offhand, and didn't find any in quick googling, though I didn't look very hard or read the original papers: dx.doi.org/10.2307%2F3002019 and dx.doi.org/10.2307%2F2332510, which may have some. I'd again recommend computing some Monte Carlo estimates for a few parameter values that you care about and seeing how they line up with the approximate answer. $\endgroup$
    – Danica
    Apr 23, 2017 at 18:34

The answer of @Danica provides several approaches for obtaining the distribution of $aX +bY$. This expression typically arises as the quadratic form in normal random variables. For completeness, I would like to add some additional approaches for computing $aX + bX$ numerically, which could then be used to calculate the expected square root numerically.

Imhof (1961) (https://www.jstor.org/stable/2332763) provides an exact numerical method, for which there is an R package available here.

Bodemham and Adams (2016) (https://link.springer.com/article/10.1007/s11222-015-9583-4) evaluates the accuracy of 6 different approximations, including the Satterthwaite-Welch approximation. These are less accurate than Imhof's method, but generally considerably less computationally intensive, e.g. they found that the Satterthwaite-Welch approximation is abouth 50 times faster than Imhof's method. There is an associated Python package called momentchi2 available here.

In my experience Imhof's method is sufficiently fast for 10 to 20 terms, even when I use it in Monte Carlo simulations.

Also see @Danica's answer here, which refers to an approximate approach by Bausch (2013) (http://arxiv.org/pdf/1208.2691.pdf) suitable when the number of terms are large.


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.