Expected norm of first $d$ coordinates of the $n$-dimensional sphere In the report "An Elementary Proof of a Theorem of Johnson and Lindenstrauss" by "Sanjoy Dasgupta and Anupam Gupta" they make the claim that
\begin{equation}
\mathbb{E}(\lVert\frac{(y_1,\dots,y_d)}{\lVert Y \rVert}\rVert^2) = \frac{d}{n}
\end{equation}
where $Y = (y_1,\dots,y_n)$ consists of n i.i.d. standard Gaussian random variables. They claim that $\frac{(y_1,\dots,y_d)}{\lVert Y \rVert}$ is simply a point on the n-dimensional sphere (this is clear to me) and from that the statement immediately follows (not clear to me).
Can someone explain me why this holds?
 A: This is a classic example of linearity of expectation. Your random variable
is
$$X_d=\frac{Y_1^2+\cdots+Y_d^2}{Y_1^2+\cdots+Y_n^2}$$
where the $Y_i$ are independent standard normal variables. That is
$$X_d=Z_1+\cdots+Z_d$$
where
$$Z_i=\frac{Y_i^2}{Y_1^2+\cdots+Y_n^2}.$$
By symmetry, the expectations of the $Z_i$ are the same:
$E(Z_1)=\cdots=E(Z_n)=\lambda$ say. But
$$Z_1+\cdots+Z_n=\frac{Y_1^2+\cdots+Y_n^2}{Y_1^2+\cdots+Y_n^2}=1$$
so that
$$E(Z_1)+\cdots+E(Z_n)=E(Z_1+\cdots+Z_n)=1,$$ and $\lambda=1/n$.
Then
$$E(X_k)=E(Z_1+\cdots+Z_k)=E(Z_1)+\cdots+E(Z_k)=k\lambda=\frac kn.$$
A: Call $A(d,n)$ this quantity. By symmetry, for any subset $I$ of size $d$ in $\{1,\dots,n\}$, we have $E(\|\frac{(y_i)_{i\in I}}{\|Y\|}\|^2)= A(d,n)$.
$$ \sum_I E(\|\frac{(y_i)_{i\in I}}{\|Y\|}\|^2) = \binom{n}{d} A(d,n),$$
where the sum is taken on all subsets $I$ of size $d$. But also,
$$ \sum_I E(\|\frac{(y_i)_{i\in I}}{\|Y\|}\|^2) =\sum_I E(\frac{1}{\|Y\|^2} \sum_{i\in I}|y_i|^2) = E(\frac{1}{\|Y\|^2} \sum_I\sum_{i\in I}|y_i|^2).$$
In the latter sum, the term $|y_1|^2$ appears exactly $\binom{n-1}{d-1}$ times (to create a set $I$ containing $1$, it suffices to choose $d-1$ index in $\{2,\dots,n\}$), and so does $|y_2|^2$, etc. Hence, we have
$$ \sum_I E(\frac{1}{\|Y\|^2} \sum_{i\in I}|y_i|^2) = \binom{n-1}{d-1}E(\frac{1}{\|Y\|^2} \sum_{i=1}^n|y_i|^2)= \binom{n-1}{d-1}.$$
Hence, $\binom{n}{d} A(d,n)= \binom{n-1}{d-1}$ and $A(d,n)=\frac d n$.
