Recurrence for expected length of Gaussian vector

Question

Let $g_k \sim N(0, I_{k \times k})$ be a a standard $k$-dimensional Gaussian vector.

Denote by $\|g\|$ the $2$-norm of $g$. By explicit integration, it is not hard to see that $$ \mathbb E \|g_k\| = \frac{\sqrt 2 \Gamma\left(\frac{k+1}{2}\right)}{\Gamma\left(\frac k 2\right)}\,, $$ where $\Gamma$ is the Gamma function.

In particular, the above expression implies \begin{equation*} \mathbb E \|g_k\|\mathbb E\|g_{k+1}\| = k\,. \end{equation*}

This formula gives a nice recurrence for the expected length of a standard Gaussian. The recurrence is so nice that I'd like to see a slick proof of this fact, if one exists.

Question: Is there an elegant proof of the recurrence $\mathbb E \|g_k\|\mathbb E\|g_{k+1}\| = k$, one that involves no explicit integration?

Answer 1

In dimension $k$ there is a constant $C_k$ such that $$C_k \int_0^{\infty}r^{k-1}e^{-r^2/2}dr = 1.$$ It is not necessary to compute these constants as we will see. Now the expression for the product of expectations in radial form is $$\mathbb{E} \left(\lVert g_k \rVert\right)\mathbb{E} \left(\lVert g_{k+1} \rVert\right)=C_k\int_0^{\infty}r^k e^{-r^2/2}dr \cdot C_{k+1}\int_0^{\infty}r^{k+1} e^{-r^2/2}dr=$$ $$C_k\int_0^{\infty} r^2 r^{k-1} e^{-r^2/2} dr = \mathbb{E} \left(\lVert g_k \rVert^2 \right) = \mathbb{E} \left(X_1^2+\ldots+X_k^2\right)=k \mathbb{E} \left(X_1^2\right)=k.$$ Here $X_1,\ldots,X_k$ are iid standard normal variables.

Answer 2

A "slick" proof can be constructed using polar coordinates for the $k$-dimensional integral. With $r=\|g_k\|$ we define, for any function $f(r)$, $$ \langle f(r) \rangle = \int_0^\infty {\rm e}^{-\frac{r^2}2}f(r)\ {\rm d}r \ ,$$ so that $$ \mathbb E f(r) = \langle r^{k-1}f(r) \rangle \ .$$ We then have $$\mathbb E \|g_k\| = \frac{\langle r^k \rangle}{\langle r^{k-1} \rangle} \quad {\rm and} \quad \mathbb E \|g_k\|^2 = \frac{\langle r^{k+1} \rangle}{\langle r^k \rangle} \ ,$$ and as a consequence $$ \mathbb E \|g_k\| \ \mathbb E \|g_{k+1}\| = \frac{\langle r^k \rangle}{\langle r^{k-1} \rangle} \frac{\langle r^{k+1} \rangle}{\langle r^k \rangle} = \mathbb E \|g_k\|^2 = k \ .$$