# Expectation of $\frac{X_1^4}{(X_1^2 + \cdots + X_d^2)^2}$

Let $$X_1$$, $$X_2$$, $$\cdots$$, $$X_d \sim \mathcal{N}(0, 1)$$ and be independent. What is the expectation of $$\frac{X_1^4}{(X_1^2 + \cdots + X_d^2)^2}$$?

It is easy to find $$\mathbb{E}\left(\frac{X_1^2}{X_1^2 + \cdots + X_d^2}\right) = \frac{1}{d}$$ by symmetry. But I do not know how to find the expectation of $$\frac{X_1^4}{(X_1^2 + \cdots + X_d^2)^2}$$. Could you please provide some hints?

What I have obtained so far

I wanted to find $$\mathbb{E}\left(\frac{X_1^4}{(X_1^2 + \cdots + X_d^2)^2}\right)$$ by symmetry. But this case is different from that for $$\mathbb{E}\left(\frac{X_1^2}{X_1^2 + \cdots + X_d^2}\right)$$ because $$\mathbb{E}\left(\frac{X_i^4}{(X_1^2 + \cdots + X_d^2)^2}\right)$$ may be not equal to $$\mathbb{E}\left(\frac{X_i^2X_j^2}{(X_1^2 + \cdots + X_d^2)^2}\right)$$. So I need some other ideas to find the expectation.

Where this question comes from

A question in mathematics stack exchange asks for the variance of $$\|Ax\|_2^2$$ for a unit uniform random vector $$x$$ on $$S^{d-1}$$. My derivation shows that the answer depends sorely on the values of $$\mathbb{E}\left(\frac{X_i^4}{(X_1^2 + \cdots + X_d^2)^2}\right)$$ and $$\mathbb{E}\left(\frac{X_i^2X_j^2}{(X_1^2 + \cdots + X_d^2)^2}\right)$$ for $$i \neq j$$. Since $$\sum_{i \neq j}\mathbb{E} \left( \frac{X_i^2X_j^2}{(X_1^2 + \cdots + X_d^2)^2}\right) + \sum_i \mathbb{E}\left(\frac{X_i^4}{(X_1^2 + \cdots + X_d^2)^2}\right) = 1$$ and by symmetry, we only need to know the value of $$\mathbb{E}\left(\frac{X_1^4}{(X_1^2 + \cdots + X_d^2)^2}\right)$$ to obtain other expectations.

The distribution of $X_i^2$ is chi-square (and also a special case of gamma).

The distribution of $\frac{X_1^2}{X_1^2 + \cdots + X_d^2}$ is thereby beta.

The expectation of the square of a beta isn't difficult.

Fact 1: If $$X_1$$, $$X_2$$, $$\cdots$$, $$X_n$$ are independent standard normal distribution random variables, then the sum of their squares has the chi-squared distribution with $$n$$ degrees of freedom. In other words, $$X_1^2 + \cdots + X_n^2 \sim \chi^2(n)$$

Therefore, $$X_1^2 \sim \chi^2(1)$$ and $$X_2^2 + \cdots + X_d^2 \sim \chi^2(d-1)$$.

Fact 2: If $$X \sim \chi^2(\lambda_1)$$ and $$Y \sim \chi^2(\lambda_2)$$, then $$\frac{X}{X + Y} \sim \texttt{beta}(\frac{\lambda_1}{2}, \frac{\lambda_2}{2})$$

Therefore, $$Y = \frac{X_1^2}{X_1^2 + \cdots + X_d^2} \sim \texttt{beta}(\frac{1}{2}, \frac{d-1}{2})$$.

Fact 3: If $$X \sim \texttt{beta}(\alpha, \beta)$$, then $$\mathbb{E}(X) = \frac{\alpha}{\alpha + \beta}$$ and $$\mathbb{Var}(X) = \frac{\alpha\beta}{(\alpha + \beta)^2(\alpha + \beta + 1)}$$

Therefore, $$\mathbb{E}(Y) = \frac{1}{d}$$ and $$\mathbb{Var}(Y) = \frac{2(d-1)}{d^2(d+2)}$$

Finally, $$\mathbb{E}(Y^2) = \mathbb{Var}(Y) + \mathbb{E}(Y)^2 = \frac{3d}{d^2(d+2)}.$$

• @NP-hard: It seems that you in fact asked this question in order to be able to answer this question? Why not just mention that? – joriki Jul 10 '16 at 7:22
• @joriki Thanks. I will add the link to the question. – user72637 Jul 10 '16 at 8:47