The expected value of $\ln(X)$ if $X$ is chi square distributed

Question

$X$ is a chi-squared random variable which is the square of two standard normal variables. How can I find the expected value of $\ln(X)$ in this case.

First time using this site, please. Be patient.

Answer 1

Since $X\sim \chi^2(2)$ then $X\sim \exp(1/2)$. So, $\frac12X\sim \exp(1)$ and hence $$μ-σ\ln{\left(\frac12X\right)}\sim GEV(μ,σ,0)$$ with $μ\in\mathbb R, σ>0$, where $GEV$ denotes the Generalized extreme value distribution. Hence $$\mathbb E\left[μ-σ\ln\left(\frac12X\right)\right]=μ+σ\mathbf{γ}\implies \mathbb E[\ln{X}]=\ln2-\mathbf γ$$ where $γ$ is the Euler-Mascheroni constant.

Answer 2

Consider $Z = X^2 + Y^2$ where X and Y are independent standard normals. Then the joint density $f_{(X,Y)}$ of X and Y is the product of the densities and so $$\begin{align*}E[\ln(Z)] &= E[\ln(X^2 + Y^2)] \\ &= \int_{-\infty}^\infty \int_{-\infty}^\infty \ln(x^2 + y^2) f_{(X,Y)}(x,y)\; dx\; dy \\ &= \frac{1}{2\pi}\int_{-\infty}^\infty \int_{-\infty}^\infty \ln(x^2 + y^2)\exp\left(-\frac{x^2 + y^2}{2}\right) \; dx\; dy \\ &= -\gamma + \ln(2) \end{align*}$$

where $\gamma$ is the Euler-Mascheroni constant and the last equation was calculated by Mathematica…