Why does conditional probability used in mean square error equal zero?

[EdMel]

Hi guys,

I am having trouble showing that \mathbb{E}\left[(Y-\mathbb{E}[Y|X])^{2}\right]=0.

I understand the proof of why E[Y|X] minimizes the mean square error, but I cannot understand why it is then equal to zero.

I tried multiplying out the square to get \mathbb{E}\left[Y^{2}\right]-2\mathbb{E}\left[Y\mathbb{E}[Y|X]\right]+\mathbb{E}\left[\mathbb{E}[Y|X]\mathbb{E}[Y|X]\right]
but have not been able to justify \mathbb{E}\left[Y\mathbb{E}[Y|X]\right]=\mathbb{E}\left[Y^{2}\right]<br /> or \mathbb{E}\left[\mathbb{E}[Y|X]\mathbb{E}[Y|X]\right]=\mathbb{E}\left[Y^{2}\right].

Thanks in advance.

 

[micromass]

Can you tell us your definition of the conditional expectation and what properties you are allowed to use?

 

[Stephen Tashi]

It would also be helpful to improve the notation. If X and Y are random variables and f(X,Y) is a function of them then the notation E f(X,y) is ambiguous. It is not clear whether the expectation is being computed with respect to the distribution of X or the distribution of Y - or perhaps with respect to the joint distribution for (X,Y).

You can use a subscript to denote which distribution is used to compute the expectation. For example, if Y is not a function of X then E_X ( E_Y ( 3Y + 1) ) is the expectation of with respect to the distribution of X of the constant value E_Y(3Y + 1) Hence E_X (E_Y (3Y+1)) = E_Y (3Y+ 1).

 

[FactChecker]

It's hard to prove because it is not true unless (Y-E(Y|X))==0. Are you sure that (Y-E(Y|X)) is supposed to be squared?

 

[Stephen Tashi]

It's hard to prove because it is not true unless (Y-E(Y|X))==0.

And before (Y - E(Y|X)) is equal or not equal to zero, it would have to mean something. How do we interpret Y - E(Y|X) ? Is it a random variable? To realize it , do we realize a value Y = y0 from the distribution of Y and then take the expected value of the constant y0 with respect to the distribution of X ?