Expectation of uniformly distributed random variables on unit sphere in $\mathbb{R}^n$

Question

Random vector $X = ( X_1, X_2, \dots, X_n) $ is uniformly distributed on the unit sphere in $\mathbb{R}^n$

Find $M_2 =E[X_1^2], M_4= E[X_1^4]$

In this case, author asks to find out second and fourth moments of random variable X.

Author's Solution: Since $X_i$ are identically distributed and $\displaystyle\sum_{i=1}^n X^2_i =1. $ It readily follows that $M_2= \displaystyle\frac{1}{n}$

For $M_4$, first note that by squaring $\displaystyle\sum_{i=1}^n X^2_i =1 $ and taking expectation we obtain $$n M_4 + n(n-1)V_{12}=1 \tag {1}\label{(1)}$$ where $V_{12}=E\left[X_1^2X_2^2\right]$.

Thus, to complete the solution, we need another relation between $M_4$ and $V_{12}$.

To this end, note that $(X_1, X_2)$ and $\left(\displaystyle\frac{X_1 + X_2}{\sqrt{2}}, \displaystyle\frac{X_1-X_2}{\sqrt{2}}\right)$ are identically distributed. Hence $E\left[\displaystyle\frac{(X_1 +X_2)^2 (X_1 -X_2)^2}{4}\right] = E\left[X_1^2X_2^2\right]$ which after expanding gives $M_4 = 3V_{12}$

Combined with \eqref{(1)}, this yields $M_4= \displaystyle\frac{3}{n(n+2)}$

Author's remark(1)

A common way to generate uniform distribution on the sphere is to take $X_i=\displaystyle\frac{Z_i}{\sqrt{Z_1^2 +Z_2^2 +\dots, +Z_n^2}}$, where $Z_i$ are independent standard Normal random variables.

---

My thoughts

We know $V_{12}=E[X_1^2X_2^2]= E\left[\displaystyle\frac{X_1^4 -2X_1^2X_2^2 +X_2^4}{4}\right]$

If we substitute the aforesaid value of $V_{12}$ in \eqref{(1)} we get $$nE[X_1^4] + n^2\frac{E[X_1^4]}{4}- 2\cdot n^2\frac{E[X_1^2X_2^2]}{4} + n^2\frac{E[X_2^4]}{4} -n\frac{E[X_1^4]}{4} + 2\cdot n \frac{[X_1^2X_2^2]}{4} - n\frac{E[X_2^4]}{4}=1$$

Now how to proceed further?

Note: I performed further computations but I didn't get $E[X_1^4]=3V_{12}$. That's why I stuck at this step.

---

New thoughts (after reopening of the post and publication of the accepted answer):

Combined with \eqref{(1)}, $M_4 = 3V_{12}$ we get $M_4= \displaystyle\frac{3}{n(n+2)}$

Working:

We know $V_{12}=E[X_1^2X_2^2]= E\left[\displaystyle\frac{X_1^4 -2X_1^2X_2^2 +X_2^4}{4}\right]= \displaystyle\frac{M_4}{3}$

If we substitute $V_{12}=E\left[\displaystyle\frac{X_1^4 -2X_1^2X_2^2 +X_2^4}{4}\right]$ in \eqref{(1)} we get $$nE[X_1^4] +\frac{n(n-1)}4\left(E(X_1^4)-2E(X_1^2X_2^2)+E(X_2^4)\right)=1.$$

If you solve for $M_4= E[X_1^4]$ we get $$\left[ E[X_1^4] - E[X_1^2X_2^2]\right] =\displaystyle\frac{2}{n(n+2)}$$. Substituting $V_{12}=\displaystyle\frac{M_4}{3}$ we get, $M_4=\displaystyle\frac{3}{n(n+2)}$

Answer 1

Expanding $E\left[\displaystyle\frac{(X_1 +X_2)^2 (X_1 -X_2)^2}{4}\right] = E\left[X_1^2X_2^2\right]$ and using that $E[X_2^4]=E[X_1^4]=M_4$ and $E\left[X_1^2X_2^2\right]=V_{12}$ gives immediately $V_{12}=\frac{M_4}3$. It is this expression of $V_{12}$ which the author substitued in $n M_4 + n(n-1)V_{12}=1$.

Edit 5 days later to expand this answer, hopefully letting you understand why you (old and new) thoughts were not adequate:

• You began your "Workings" with "We know $V_{12}=[\dots=\dots]= \displaystyle\frac{M_4}{3}$". No, we don't know it yet.
• You insisted on substituting $$V_{12}=E\left[\displaystyle\frac{X_1^4 -2X_1^2X_2^2 +X_2^4}{4}\right]\tag {2}\label{(2)}$$ in $$n M_4 + n(n-1)V_{12}=1 \tag1\label{(1bis)}$$ This is overcomplicating things and the author's method was much simpler: he first expanded \eqref{(2)} as follows: $$V_{12}=\frac{M_4-2V_{12}+M_4}4,$$ which simplifies to $V_{12}=\frac{M_4}3$, and then substituted this (not \eqref{(2)}) in \eqref{(1bis)}, resulting in $$n M_4 + n(n-1)\frac{M_4}3=1$$ i.e. $M_4=\frac3{n(n+2)}$.

Answer 2

I was looking at a similar problem very recently. Write the vector in spherical coordinates, where your $X_1$ corresponds to $x_1 = \cos\varphi_1$ in the notation of Wikipedia. Then integrate $$\int_0^{2\pi}\int_{[0,\pi]^{n-2}}\cos^k\varphi_1\sin^{n-2}\varphi_1\ldots\sin\varphi_{n-2}\mathrm d \varphi,$$ to get the $k$th moment. The integral with respect to the components $\varphi_i$ for $i > 1$ equates to the surface area of the sphere of dimension $n-2$. Then you're left with the integral with respect to $\varphi_1$.