# What's the expectation of square root of Chi-square variable?

Setting $$X_1,X_2,\ldots \overset{d}{\sim} \mathcal{N}(0,1)$$ $$V = X_1^2 + \ldots + X_n^2$$

Now I would like to find the expectation of $\sqrt{V}$. My biggest problem is I don't know how to write down the density of $\sqrt{V}$. Could someone help me derive it?

The distribution of $V$ is $\chi^2(n)$. The square root of a $\chi^2(n)$ random variable is a $\chi(n)$ random variable.
Let the random variable $V$ have the chi-square distribution with $n$ degrees of freedom with probability density function $$f_V(v) = \begin{cases} \frac{v^{n/2-1} e^{-v/2}}{2^{n/2} \Gamma\left(\frac{n}{2}\right)}, & v \geq 0; \\ 0, & \text{otherwise}. \end{cases}$$ By the transformation technique, using the transformation $Y=g(V)=\sqrt{V}$, we have $$f_Y(y)=f_V(g^{-1}(y))\left|\frac{dv}{dy}\right|=\frac{(y^2)^{n/2-1} e^{-y^2/2}}{2^{n/2} \Gamma\left(\frac{n}{2}\right)}|2y|=\frac{y^{n-1} e^{-y^2/2}}{2^{n/2-1} \Gamma\left(\frac{n}{2}\right)}\qquad y>0$$ which is the probability density function of the chi distribution with $n$ degrees of freedom.
The expectation is $$\Bbb{E}(Y)=\int_{-\infty}^{\infty}yf_Y(y)\textrm{d}y=\int_{0}^{\infty}\frac{y^{n} e^{-y^2/2}}{2^{n/2-1} \Gamma\left(\frac{n}{2}\right)}\textrm{d}y=\sqrt{2}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)}$$
First note that each $X^{2}_i\sim Gamma(\frac{1}{2},\frac{1}{2})$ for each $i$ (i.e. chi square with df=1) now to find distribution of $\sum_{i=1}^{n}X^2_{i}$ we see that using moment generating functions and fact that each are independent of each other that $$M_{\sum_{i=1}^{n}X^2_{i}}=M(t)^{n}=\left(\frac{\frac{1}{2}}{\frac{1}{2}-t}\right)^{\frac{n}{2}}$$ thus $\sum_{i=1}^{n}X^2_{i}\sim Gamma(\frac{n}{2},\frac{1}{2})$ thus $V=\sum_{i=1}^{n}X^2_{i}$ thus $$E(\sqrt{V})=\int_0^\infty \frac{\frac{1}{2}^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}v^{\frac{1}{2}}v^{\frac{n}{2}-1}e^{-\frac{1}{2}x}dx=\frac{\frac{1}{2}^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}\frac{\Gamma(\frac{n+1}{2})}{\frac{1}{2}^{\frac{n+1}{2}}}\int^\infty_0\frac{\frac{1}{2}^{\frac{n+1}{2}}}{\Gamma(\frac{n+1}{2})}v^{\frac{n+1}{2}-1}e^{-\frac{1}{2}x}dx=\frac{\sqrt{2}\Gamma(\frac{n+1}{2})}{\Gamma(\frac{n}{2})}$$
• Nice catch, just simplified wrong second to last step it should be multiplied by $sqrt{2}$ at end Feb 15 '16 at 16:28