Conditional expectation and sum of random variables

Question

Let $Y, X_1, . . . , X_n$ be continuous random variables (not necessarily independent) with non negative range, i.e. $P(Y < 0) = 0$ and $P(X_i < 0) = 0$ for $i = 1 \ldots n$, verifying the following property concerning the conditional expectation:

(1) $E[Y | X_1 + ... + X_n = u] \geq E[Y | X_1 + ... + X_n = v]$

for arbitrary positive reals $u \geq v$. Is there anyone who is able to prove that (1) implies

(2) $E[Y | X_1 = w] \geq E[Y | X_1 = z]$

for arbitrary positive reals $w \geq z$ ? Also a counterexample which prove that this is not always the case would be welcome.

Answer 1

Counterexample: Let $X_3$ be a random variable that satisfies $E[X_1|X_3 = u] \geq E[X_1|X_3 = v]$ whenever $u \geq v$. Let $X_2$ be a random variable that does not satisfy $E[X_1|X_2 = u] \geq E[X_1|X_2 = v]$ whenever $u \geq v$. Let $X_4 = -X_2$.

Answer 2

Let $Y=X_2+X_3+X_4$. Here is a counterexample:

Let $X_1=-X_2=X_3=X_4$. Then $E[X_1\mid Y]=Y$ is an increasing function of $Y$ while $E[X_1\mid X_2]=-X_2$ is a decreasing function of $X_2$.

If the random variables must be positive, let $X_1=X_3=X_4$ with values in some bounded positive interval $(0,c)$ and $X_2=c-X_1$. Then $E[X_1\mid Y]=Y-c$ is an increasing function of $Y$ while $E[X_1\mid X_2]=c-X_2$ is a decreasing function of $X_2$.

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(Due to vandalism by the OP, the question that this answer addresses is now impossible to guess. We reproduce it here, correcting some typos, grammar errors and awkward formulations.)

Let X1, X2, X3, X4 be continuous random variables (not necessarily independent) verifying the following property concerning the conditional expectation:

(1) E(X1 | X2 + X3 + X4 = u) ≥ E(X1 | X2 + X3 + X4 = v)

for arbitrary positive reals u ≥ v. Can one prove that (1) implies

(2) E(X1 | X2 = w) ≥ E(X1 | X2 = z)

for arbitrary positive reals w ≥ z ? Counterexamples proving that this is not always the case would also be welcome.