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Many books while introducing the regression problem, start with the assertion that any random variable $Y$ can be decomposed into two orthogonal terms $$ Y= E[Y|X]+\epsilon. $$ In the classical statistics $E[Y|X]$ is a shorthand for $E[Y|X=x]$ where $X$ is some "controlled" (non-random) variable. However in econometric research $X$ is a random variable, thus I guess that $E[Y|X]$ is a shorthand for $E[Y|\sigma(X)]$, where $\sigma(X)$ is a sigma algebra generated by $X$.

  1. Is it right interpretation?

Another assertion is that $E[Y|X]$ is an orthogonal projection.

  1. What space does $Y$ projected onto (on $\sigma(X)$?)?

I pretty well understand it from the algebraic point of view when $$ y = \hat{y} + e, $$ and $HY=X(X'X)^{-1}X'y$. In this case the orthogonality of $e$ w.r.t $\hat{y}$ has clear geometric interpretation ($H$ is an orthogonal projection of $y$ onto $C(X)$ and $e \in C(X)^{\perp}$). However, this is a post-hoc approach when we already observed the data points $\{y_i, x_{1i},...,x_{pi}\}_{i=1}^n$, while I'm interested in the stochastic process that generates it.

To sum up, my questions are:

  1. If $X$ is random variable and defined on the same probability space as $Y$, why does an orthogonal decomposition of the kind $$ Y = E[Y|\sigma(X)]+\epsilon=h(X)+\epsilon $$ exists? How can I prove its existence (and uniqueness)? (I know it requires squared integrability of $Y$, but I have non-intuitive explanation how it is suffice for the decomposition to exist).

  2. Are the projections $E[Y|\sigma(X)]$ or $E[Y|X=x]$ project on $\sigma(X)$? If so, does it have any intuitive meaning (like in the linear Algebra analog)

  3. If $\epsilon$ defined on the same probability space, what it means to be orthogonal to $E[Y|\sigma(X)]$?

Would appreciate any help.

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Okay, I'll try my best to answer your three questions.

1 - In fact, there is a whole different way to define $E[Y|X]$, but you can prove that this other definition is just equivalent to $E[Y|\sigma(X)]$. So I guess a simple answer to your first question is yes, in fact: $$E[Y|X]=E[Y|\sigma(X)],$$

2 and 3 - As you said, $Y$ must be in $L^2$, because that is a space with inner product. In fact, if $f$ and $g$ are functions in $L^2(\Omega,\mathcal{F},P)$, we can define an inner product in this space by: $$<f,g>:=E[fg]$$

So that is the inner product where you must prove the orthogonality of $\epsilon$ and $E[Y|X]$. It's not hard to see that you must prove the following statements:

(i)-If $Y\in L^2$ and $X$ is any random variable, then $E[Y|X]\in L^2,$

(ii)-$E[\epsilon E[Y|X]=0$.

Note that $\epsilon=Y-E[Y|X]$, so (i) proves that $\epsilon\in L^2$ as well. Note also that $E[Y|X]$ is a $\sigma(X)$-measurable function, so this projection is just a projection of the function $Y$ into the space of $\sigma(X)$-measurable functions. You can think of $E[Y|X]$ as the $\sigma(X)$-measurable function closest to $Y$ in the norm defined by the inner product above.

I don't know if you will need a lot of mathematical tools to prove (i) and (ii). But I hope my answer can clarify your doubts. If you have any problems with the calculations or proofs, let me know

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  • $\begingroup$ Thank you for your answer! So, to be clear you say that given that $Y^2 \in L^2$, then exist $E[Y|\sigma(X)]$, as such we can simply write $ Y = E[Y|\sigma(X)] + \epsilon$ and the orthogonality of $\epsilon$ and $E[Y|\sigma(X)]$ stems from the fact that $\epsilon = Y-E[Y|\sigma(X)]$. ($E(\epsilon E[Y|\sigma(X)] =E( E(\epsilon E[Y|\sigma(X)] |\sigma(X))=0$)? $\endgroup$ – V. Vancak Feb 19 '17 at 16:07
  • $\begingroup$ Yes, that's exactly it. We know that $E[Y|\sigma(X)]$ always exist, and we can always write $Y$ in such way. There are a few hidden steps in the last calculations, but I think you understood them :) $\endgroup$ – Gabriel Sanfins Feb 19 '17 at 17:03

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