5

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.

Xi'an ні війні
  • 726
  • I would have thought for large $n$ this is going to be close to $0$, but for small $n$ it will be more interesting (for example with $n=2$, it is clearly negative when $x \approx 0$ but positive when $|x|$ is large). It is clearly an even function of $x$ when $n$ is even and an odd function of $x$ when $n$ is odd; simulations suggest that it might be an $n$-degree polynomial of $x$. – Henry Jun 16 '22 at 10:47
  • 1
    For $n=2$, one has $E[X_1X_2\mid X_1+X_2]=\left(\frac{X_1+X_2}{2}\right)^2-\frac12$, recalling the MVUE of $\theta^2$ when $X_i\sim N(\theta,1)$. Similar calculation is possible for larger $n$, but I don't know if there is a pattern. – StubbornAtom Jun 16 '22 at 11:29
  • 1
    For $n=1$ it is obviously $x$.

    Empirically for $n=2$ it seems as if it might be $-\frac12 +\left(\frac {x}2\right)^2$ and justified by @StubbornAtom.

    For $n=3$ it seems to be about $-\frac{ x}{3} + \left(\frac {x}3\right)^3$.

    For $n=4$ it might not be far from $\frac{3}{16}- \frac32 \left(\frac {x}4\right)^2 +\left(\frac {x}4\right)^4$, and for $n=5$ possibly $\frac35 \frac{ x}{5} - 2\left(\frac {x}5\right)^3+\left(\frac {x}5\right)^5$. The ending $\cdots -\frac{n-1}{2}\left(\frac xn\right)^{n-2}+\left(\frac xn\right)^{n}$ is as in Christophe Leuridan's answer.

     – Henry Jun 16 '22 at 11:46
  • 1
    For $n=6$ I seems to be getting something like $-\frac{1}{15}+ \frac54 \left(\frac {x}6\right)^2 -\frac52\left(\frac {x}6\right)^4+\left(\frac {x}6\right)^6$ but some of this might be wishful thinking in the interpretation of noisy results. – Henry Jun 16 '22 at 11:52
  • @Henry: Yes I had obtained the same for n=1,2,3. The idea would be to get a n order polynomial with only even (or odd) powers. – Xi'an ні війні Jun 16 '22 at 13:55
  • @Henry: actually, as discussed in the original question, the product is diverging with $n$. – Xi'an ні війні Jun 17 '22 at 06:50
  • @Xi'anнівійні It seems to me looking at those curves that, conditioned on the sum $x$, the expectation of the product is converging towards $0$ as $n$ increases. The CV question seems different: it has a positive mean rather than $0$ and does not condition on the sum – Henry Jun 17 '22 at 07:46

1 Answers1

3

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.

Christophe Leuridan
  • 9,143