Question
Let $Z_1, \ldots, Z_n$ be independent and identically distributed standard normal random variables. What are the constants $c_1, \ldots, c_n$ (not all zero) such that the random variable $$M = c_1Z_1 + \ldots + c_nZ_n$$ is independent of $Z_i - M$ for each $i = 1, \ldots, n$?
My working
$\because Z_i \stackrel{\mathrm{iid}}{\sim} \mathcal{N} (0, 1)$
$\therefore M \sim \mathcal{N} (0, c_1^2 + \ldots + c_n^2)$
$\begin{aligned}[t] Cov(Z_i - M,\ M) & = Cov(Z_i,\ M) - Var(M) \\[1 mm] & = \mathbb{E}(Z_iM) - \mathbb{E}(Z_i)\mathbb{E}(M) - Var(M) \\[1 mm] & = \mathbb{E}(Z_iM) - (c_1^2 + \ldots + c_n^2) \\[1 mm] & = 0 \end{aligned}$
$\implies \mathbb{E}(Z_iM) = c_1^2 + \ldots + c_n^2$
However, this is where I am stuck. How do I find the expectation of $Z_iM$? Any intuitive suggestions will be greatly appreciated :)
$Z_1-M,Z_2-M,...,Z_n-M,M$ are jointly normal with mean $0$. So indepedence of $M$ and $Z_i-M$ is equiavlent to $E(Z_i-M)M=0$. But $EZ_iM=\sum c_jEZ_iZ_j=\sum c_jEZ_iEZ_j=c_i$ since $EZ_iZ_j=1$ for $i=j$ and $0$ for $i \neq j$. So the conidtion is $c_i=EM^{2}=\sum c_k^{2}$ for each $i$. Thus $c_i$ is a number $c$ independent of $i$ and $c=nc^{2}$ so, $c=0$ or $c=\frac 1 n$. $c=0$ has been ruled out so $c_i=\frac 1 n$ for all $n$.