# How to calculate this expectation where the random variable is restricted on a sphere? [closed]

Hello! I have a question about how to calculate the expectation of a quadratic form as follows, where $X$ is a random variable that uniformly distributed on the unit sphere: $$E_X[(\mathbf{x}^\top A\mathbf{x})^2] =\int_{\mathbf{x}\in S}{p(\mathbf{x})(\mathbf{x}^\top A\mathbf{x})^2dS(\mathbf{x})} =\int_{\mathbf{x}\in S}{\frac{1}{4\pi}(\mathbf{x}^\top A\mathbf{x})^2dS(\mathbf{x})}$$ where $S=\{\mathbf{x}\in\mathcal{R}^N|\mathbf{x}^\top\mathbf{x}=1\}$.

If $\mathbf{x}$ is Gaussian, there are some conclusions about the expectation of the quadratic forms. However, I find it difficult to deal with when the variable is distributed on a sphere. When the dimension is 2 or 3, this problem can be solved by representing it with polar coordinate. However, when the dimension is high, such a representation will be rather redundant, how can I calculate this integral then? Please give some help for this problem if you have any idea. Thank you very much!

-

## closed as too localized by fedja, Igor Rivin, Suvrit, Andrés Caicedo, Andy PutmanNov 22 '11 at 4:54

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

What is $p(\boldsymbol{x})$? – Brendan McKay Nov 20 '11 at 14:09
Since we assume that x is uniformly distributed on the unit sphere, here we have $p(x)$ is a constant as $p(x)=\frac{1}{4\pi}$ Thank you! – ppyang Nov 20 '11 at 14:24
a) $Ex_i^2=1/N$. b) $Ex_ix_j=0$. c) I'm voting to close as not a research level question. – fedja Nov 20 '11 at 14:24
Did you add squares just now? It isn't an expectation of a quadratic form any more. I believe Fedja's point is that your function is a polynomial and you can just calculate the expectations of each term and add them up. – Brendan McKay Nov 20 '11 at 15:19
Crossposted: math.stackexchange.com/questions/83931/… – Byron Schmuland Nov 20 '11 at 19:50

Assuming $A$ is diagonalizable in an orthonormal basis with a diagonal of $a_k$'s, one asks for the expectation $C(A)$ of the random variable $$(X^TAX)^2=\left(\sum\limits_ka_kX_k^2\right)^2=\sum\limits_ka_k^2X_k^4+\sum\limits_{k\ne \ell}a_ka_\ell X_k^2X_\ell^2,$$ where the vector $X=(X_k)_{1\leqslant k\leqslant d}$ is uniformly distributed on the Euclidean unit sphere. Hence, $$C(A)=c_d\cdot\sum_ka_k^2+b_d\cdot\sum_{k\neq\ell}a_ka_\ell=(c_d-b_d)\cdot\sum_ka_k^2+b_d\cdot\left(\sum_ka_k\right)^2,$$ with $$c_d=\mathrm E(X_1^4),\qquad b_d=\mathrm E(X_1^2X_2^2).$$ Since the $X_k^2$'s are identically distributed and sum to $\|X\|^2=1$, $\mathrm E(X_1^2)=1/d$ and $$dc_d+d(d-1)b_d=1.$$ Furthermore, a direct computation (1) yields $b_d=1/(d(d+2))$ hence $c_d-b_d=2/(d(d+2))$ and $$C(A)=\frac1{d(d+2)}\cdot\left(2\sum\limits_ka_k^2+\left(\sum\limits_ka_k\right)^2\right).$$ Finally, $$\color{red}{\mathrm E((X^TAX)^2)=\frac{2\text{tr}(A^2)+\left(\text{tr}(A)\right)^2}{d(d+2)}}.$$ Note:
(1) One can write $X$ as $X=Z/\|Z\|$ where the vector $Z=(Z_k)_{1\leqslant k\leqslant d}$ is i.i.d. standard gaussian. Thus, $$X_1^2X_2^2=Z_1^2Z_2^2\|Z\|^{-4}=Z_1^2Z_2^2\,\int_0^{+\infty}\mathrm e^{-t\|Z\|^2}\cdot t\mathrm dt,$$ hence $$X_1^2X_2^2=\int_0^{+\infty}Z_1^2\mathrm e^{-tZ_1^2}\cdot Z_2^2\mathrm e^{-tZ_2^2}\cdot \mathrm e^{-t(Z_3^2+\cdots+Z_d^2)}\cdot t\mathrm dt.$$ Integrating this, $$b_d=\mathrm E(X_1^2X_2^2)=\int_0^{+\infty}u'(t)^2\cdot u(t)^{d-2}\cdot t\mathrm dt\quad\text{with}\quad u(t)=\mathrm E(\mathrm e^{-tZ_1^2}).$$ A standard computation yields $u(t)=(1+2t)^{-1/2}$, hence $u'(t)=-(1+2t)^{-3/2}$ and $$b_d=\int_0^{+\infty}(1+2t)^{-d/2-2}\cdot t\mathrm dt=\left.\frac1{2d(1+2t)^{d/2}}-\frac1{2(d+2)(1+2t)^{d/2+1}}\right|_0^{+\infty}=\frac1{d(d+2)}.$$