Conditional expectation of product of Normal variate given their sum Given $$X_1,\ldots,X_n\stackrel{\text{iid}}{\sim}\mathcal N(0,1)$$ I would like to compute the conditional expectation $$\mathbb E\Big[\prod_{i=1}^n X_i \Big| X_1+\cdots+X_n=x\Big]$$ for statistical reasons.
 A: Partial answer. I hope that my computations are right.
Let $A_n = (a^{(n)}_{i,j})_{1 \le i,j \le n}$ be any orthogonal matrix with size $n \times n$ and with last row equal to $n^{-1/2} (1 ~ \ldots ~ 1)$. Set
$$\left(\begin{array}{c}
Y_1 \\
\ldots \\
Y_n \end{array}\right)
= A_n
\left(\begin{array}{c}
X_1 \\
\ldots \\
X_n \end{array}\right).$$
Then $Y_1,\ldots,Y_n$ are still independent random variables with law $\mathcal{N}(0,1)$, and $Y_n = n^{-1/2} (X_1+\cdots+X_n)$.
Since $A_n^{-1} = A_n^\top$, we are led to compute
$$\mathbb{E}\Big[\prod_{i=1}^n \sum_{j=1}^n a^{(n)}_{j,i}Y_j ~\Big|~ Y_n = y\Big],$$
with $y=n^{-1/2}x$. By the independence of $Y_1,\ldots,Y_n$ , this is equal to
\begin{eqnarray*}
\mathbb{E}\Big[\prod_{i=1}^n \Big( \sum_{j=1}^{n-1} a^{(n)}_{j,i}Y_j + a^{(n)}_{n,i}y\Big)\Big]
&=&\mathbb{E}\Big[\prod_{i=1}^n \Big( \sum_{j=1}^{n-1} a^{(n)}_{j,i}Y_j + n^{-1}x\Big)\Big] 
\end{eqnarray*}
If we expand the product, we get a polynomial of $x$ with leading term $n^{-n}x^n$ and having the same parity as $n$. The coefficient of $x^{n-2}$ is
$$n^{-n+2}\sum_{1 \le i_1<i_2 \le n}\sum_{j=1}^{n-1} a^{(n)}_{j,i_1} a^{(n)}_{j,i_2} = n^{-n+2}\sum_{1 \le i_1<i_2 \le n} \Big([(A^{(n)})^\top A^{(n)})]_{i_1,i_2} - n^{-1} \Big) = n^{-n+2}{n \choose 2}(-n^{-1}) = -n^{-n+2} \frac{n-1}{2}.$$
The coefficients of $x^{n-4},x^{n-6},\ldots$ are harder to compute.
