# isotropic density function

A random vector $$X \in \mathbb{R}^m$$ is said to be $$\textit{isotropic}$$ if its density function can be written as $$\begin{equation*} f(x) = \tilde{f}(\|x\|) \end{equation*}$$, so how can i proof the following

If $$X$$ is isotropic and $$g: \mathbb{R}\rightarrow \mathbb{R}$$ then $$$$\mathbb{E}\left[g(\|X\|)\right] = \int_{\mathbb{R}^m}g(\|x\|)f(x)dx = \frac{2\pi^{m/2}}{\Gamma(m/2)}\int_{0}^{\infty}g(r)\tilde{f}(r)r^{m-1}dr.$$$$.

• Try to spot the surface of the unit sphere in the last integral. – Xi'an Sep 6 '19 at 12:10

You can use hyperspherical coordinates to do this in a methodical (but longer) way if you don't recognize things like the surface area of an $$m$$-dimensional unit sphere per @Xian's comment. Starting with $$I = \int_{\mathbb R^m} g(\|x\|)\tilde f(\|x\|)\,\text dx$$ we can put this in hyperspherical coordinates by letting $$r = \|x\|$$ and then the remaining angles are all completely integrated over, so using the Jacobian of $$r^{m-1}\sin^{m-2}\theta_1\sin^{m-3}\theta_2\dots\sin \theta_{m-2}$$ we get $$I = \int_{\theta_{m-1} = 0}^{2\pi} \int_{\theta_{m-2} = 0}^\pi \dots \int_{\theta_1=0}^\pi\int_{r=0}^\infty g(r)\tilde f(r) r^{m-1}\sin^{m-2}\theta_1\sin^{m-3}\theta_2\dots\sin \theta_{m-2}\,\text d\theta_{m-1} \text d\theta_{m-2} \dots \text d\theta_1 \text dr$$ which can be split into separate integrals so $$I = 2\pi \cdot \left[\prod_{k=1}^{m-2} \int_{0}^\pi \sin^{k}\theta \,\text d\theta\right]\cdot \int_0^\infty g(r)\tilde f(r) r^{m-1}\,\text d r.$$

I'm going to assume $$m \geq 4$$ because the other cases can be handled with the more usual polar and spherical coordinates and I don't want to have to make exceptions for these cases.

For the product of sine integrals, we can look up the values in e.g. the wikipedia article on particular definite integrals. That article integrates over $$[0,\pi/2]$$ but splitting our integrals into $$[0,\pi/2]\cup(\pi/2, \pi]$$ and substituting $$u = x-\pi/2$$ in the second one, we have $$\int_0^\pi \sin^k(x)\,\text dx = 2\int_0^{\pi/2} \sin^k(x)\,\text dx$$ and can then use the formulas as written.

The key thing to note is the way that successive products annihilate. The product starts at $$k=1$$ so consider the successive terms $$k = 2n-1$$ and $$k=2n$$: $$\int_0^\pi \sin^{2n-1}(x)\,\text dx \cdot \int_0^\pi \sin^{2n}(x)\,\text dx \\ = 4 \cdot \frac \pi 2 \cdot \frac{1}{2n} = \frac \pi n$$

From this we see that if $$m$$ is even, so $$m-2$$ is too, then $$2\pi \prod_{k=1}^{m-2} \int_{0}^\pi \sin^{k}\theta \,\text d\theta = 2\pi \frac{\pi^{m/2-1}}{(m/2-1)!} = \frac{2 \pi^{m/2}}{\Gamma(m/2)}$$ as desired.

If $$m$$ is odd then we have $$2\pi \prod_{k=1}^{m-2} \int_{0}^\pi \sin^{k}\theta \,\text d\theta = \\ 2\pi \left[\prod_{k=1}^{m-3} \int_{0}^\pi \sin^{k}\theta \,\text d\theta\right]\int_0^\pi \sin^{m-2}(x)\,\text dx \\ = 2\pi \frac{\pi^{(m-3)/2}}{((m-3)/2)!} \cdot 2 \int_0^{\pi/2} \sin^{m-2}(x)\,\text dx.$$ Letting $$m-2 = 2k+1$$ so $$k = \frac{m-3}{2}$$ I have $$2\pi \frac{\pi^{(m-3)/2}}{((m-3)/2)!} \cdot 2 \int_0^{\pi/2} \sin^{m-2}(x)\,\text dx = \\4\frac{\pi^{k+1}}{k!} \frac{(2k)!!}{(2k+1)!!}.$$

I'll leave this to you to prove, but you can prove the following result to get this in the form we want (I'd recommend using induction):

Lemma: for $$k=1,2,3\dots$$, $$\frac{(2k)!!}{(2k+1)!!k!} = \frac{\sqrt \pi}{2\Gamma\left(k + \frac 32\right)}.$$

Using this lemma and substituting $$k = \frac {m-3}2$$ in, you should be able to finish the proof.

• Oh thanks, I am really appreciate for your fantastic answer – mhmt Sep 12 '19 at 9:00