Suppose $X_1,...,X_n$ are independent and identically distributed random variables defined on some probability space $(\Omega, \mathcal{A}, P)$. Define $Y=\sum_{i=1}^{n}X_i$. If we denote the corresponding order statistics as $X_{n,n} \geq X_{n-1,n} \geq ... \geq X_{1,n}$ then one can also write $Y=\sum_{i=1}^{n} X_{i,n}$.

I was wondering whether the following also holds: $$\sum_{i=1}^{n}\mathbb{E}(Y|X_i) \overset{?}{=} \sum_{i=1}^{n}\mathbb{E}(Y|X_{i,n})$$ My intuition would be yes but I have no idea how to prove it.


1 Answer 1


Not necessarily equal.
Look at the case $n=i=2$ with $X_i$ iid standard normal.
$\phi$ and $\Phi$ are the density and distribution functions for standard normal random variable.

$E[Y|X_1]=X_1$ and $E[Y|X_2]=X_2$

But, $$E[Y|X_{2,2}]=E[X_1|X_{2,2}]+X_{2,2}=\frac{\int_{-\infty}^{X_{2,2}}y\phi(y)dy}{\Phi(X_{2,2})}+X_{2,2}=\frac{-\phi(X_{2,2})}{\Phi(X_{2,2})}+X_{2,2}$$ and $$E[Y|X_{1,2}]=\frac{-\phi(X_{1,2})}{\Phi(X_{1,2})}+X_{1,2}$$ So, $$E[Y|X_{1,2}]+E[Y|X_{2,2}]=\frac{-\phi(X_1)}{\Phi(X_1)}+X_1+\frac{-\phi(X_2)}{\Phi(X_2)}+X_2$$ which is not necessarily equal to $E[Y|X_1]+E[Y|X_2]=X_1+X_2$


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