Conditional Expectation problem

[horseshoe]

So this seemed pretty straightforward at first read but has stumped me.

Suppose X and Y are random variables, both with finite means, and suppose Y = E[X|Y] a.s. and X = E[Y|X] a.s. Prove that X = Y a.s.

I'm given the hint to consider E[X - Y; X > c, Y <= c] + E[X - Y; X <= c, Y <= c] but am pretty stumped on this. Any help/hints are appreciated!

 

[diegosanaz]

This is not very formal, but what I did is to express \(E[x|y]=\sum x P_xy P_y\) which is equal to \(\sum y P_y\). Where \(P_xy\) is the probability of the intersection. If you do this for the other equation you will get that for both of them to be true \(P_xy=1\).

 

[inferno]

have a proof ignoring "almost sure" treatment:

Consider any Y=y s.t. P(X!=y) > 0. (we will arrive at a contradiction)

we know E(X-Y|Y=y)=0.
so, E(X-y|X>y).P(X>y)+E(X-y|X<y).P(X<y)=0 (1)

From (1), either both P(X<y) and P(X>y) >0.

Now we will divide (X<y) in two parts (X<y1) and (y1<=X<y) such that (y1<=X<y) has positive probability.

If we can do the division then E(X-Y|Y=y1)=0 gives:
E(X-y1|X<y1).P(X<y1)+E(X-y1|y1<=X<y).P(y1<X<y)+E(X-y1|X>y).P(X>=y) (2)
Check (2) > LHS of (1) = 0
contradiction.

Consider all sets An={X belongs to [y-n, y-(n-1)}
countable union of An = (X<y)
So some An should have positive prob.
for that n call y-n=y1.

Sorry I can not think of an elegant proof right now!

 

[dsturm]

Another way to see why their inequality implies an inherent contradiction comes from measure theory.

Let \((\Omega, \mathcal{F}, \mathbb{P})\) be a probability space.

The relevant (intuitive) lemma is, for any \(\mathcal{G}\subset\mathcal{F}\), and a \(\mathcal{G}\)-measurable function \(X\):
\[ \int_{A\in \mathcal{G}} \mathbb{E}\left[ X |\mathcal{G} \right] \ d\mathbb{P}(\omega)= \int_{A\in\mathcal{G}} X(\omega) \ d\mathbb{P}(\omega) \]

Another way of saying this is,

\(\mathbb{E}[X|A]\cdot\mathbb{P}[A] = \mathbb{E}\left[X1_{A} \right]\)

Consider the random variable \(Z(\omega) = X(\omega)-Y(\omega) \)

Suppose, for the sake of contradiction, that there is a non-empty set

\(\begin{align}A &:=\{\omega:Z(\omega) > 0 \} \\ &\Longleftrightarrow\int_{A} X(\omega)-Y(\omega) \ d\mathbb{P}(\omega) >0 \\ &\Longleftrightarrow \int_A \mathbb{E}[Y|\sigma(X)]\ d \mathbb{P}(\omega) > \int_A Y(\omega) \ d \mathbb{P}(\omega)\end{align}\)
Where the first term in the final inequality is another way of expressing the problem statement. Now, since \(Z\) was measurable on \(\mathcal{A}\), so must \(X\) (and \(Y\)) be, therefore \(A\subset \sigma(X)\) presenting a contradiction. The same argument holds for \(B:=\{\omega:Z(\omega)<0\}\).

Hence \(X\) and \(Y\) are indistinguishable wherever they have non-trivial measure (i.e., almost surely).

Hope this helps.