# Equivalence sum of conditional expectations given random observations and sum of conditional expectation given order statistics

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

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