<s>[INST] 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.
 [/INST] This answer expands @Glen_b's answer.


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