2
$\begingroup$

I've had no problems showing that

$$E[E[Y|X,Z]|Z]=E[Y|Z]$$

by the law of iterated expectation. For the latter I summed over $x$ for a certain value of $Z=z$: $$\begin{align} E[E[Y|X,Z]|Z] &= \sum_x E[Y|X=x,Z=z]\cdot P(X=x|Z=z)\\ &=\sum_{x,y} y\cdot P(Y=y|X=x,Z=z)P(X=x|Z=z)\\ &=\sum_{x,y} y\cdot\frac{P(Y=y,X=x,Z=z)}{P(X=x,Z=z)}\cdot\frac{P(X=x,Z=z)}{P(Z=z)}\\ &=\sum_{x,y} y\cdot\frac{P(Y=y,X=x,Z=z)}{P(Z=z)}\\ &=\sum_{x,y} y\cdot\frac{P(Y=y,X=x,Z=z)}{P(Z=z)}\\ &=\sum_{y} y\cdot\frac{P(Y=y,Z=z)}{P(Z=z)}\\ &=\sum_{y} y\cdot P(Y=y|Z=z)\\ &=E(Y|Z=z) \end{align}$$ However for $E[E[Y|X]|X,Z]=E[Y|X]$ i certainly have to go over $z$ for a certain value of $X=x$ which will be like: $$\begin{align} E[E[Y|X]|X,Z] &= \sum_z E[Y|X=x]\cdot P(Z=z|X=x)\\ &=\sum_{z,y} y\cdot P(Y=y|X=x)\cdot P(Z=z|X=x) \\ &=\sum_{z,y} y\cdot\frac{P(Y=y,X=x)}{P(X=x)}\cdot\frac{P(Z=z,X=x)}{P(X=x)} \end{align}$$ Now I'm kinda stuck....

Thanks in advance! Kind regards. Tim

/ed here the link where the statement comes from http://www.vwl.uni-mannheim.de/mammen/notes5.pdf (see page 4, Theorem 2.4, Section iii)

  • 0
    Are your random variables discrete ?2012-08-23
  • 0
    yes. for the discrete case is good enough for me :). if its easier to show with the continous case thats fine too tough.2012-08-23
  • 0
    I don't think that this holds. Can you explain why, in line 2 of your derivation, $x$ vanishes? How can the Information of $X$ be irrelevant? Are X, Y and Z somehow related or have special properties? Further more: what do you want to prove here? The claim is missing!2012-08-23
  • 0
    Hey. the missing x is indeed a mistake on my part which has been corrected (edited). The x was just not there :>. The reason why i want to show the equaltiy in E[E[Y|X] | X,Z] is that this is just a simplification of another problem i got. The problem im refering to is that if i calculate the conditional expectation of a conditional expectation that (in this case) the additional information about Z is no more needed since it is equal to the conditional expectation of y given the value of x.2012-08-23
  • 0
    in case this is still to abstract here the proper problem i got: Assume that E[e|x,z,d]=E[e|v,d]=E[e|c] (this is a assumption which i dont explain here but for the sake of this problem this is given). Know for y = xb + zd + E[e|c] + u we can show that E[u|x,z,d] = E[u|x,z,c] = 0. For this i need that E[E[e|c]|x,z,d]=E[E[e|c]|x,z,c]=E[e|c].2012-08-23

1 Answers 1

1

One wants to show that $E(T\mid X,Z)=T$ with $T=E(Y\mid X)$. This holds true in full generality since (i) the random variable $T$ is $\sigma(X)$-measurable by definition hence $T$ is $\sigma(X,Z)$-measurable, and (ii) $E(U\mid X,Z)=U$ for every $\sigma(X,Z)$-measurable random variable $U$.

Recall that $E(U\mid V)$ is defined as the (almost surely) unique random variable $W$ such that (1.) $W$ is $\sigma(V)$-measurable, and (2.) $E(W\,\mathbf 1_A)=E(U\,\mathbf 1_A)$ for every $A$ in $\sigma(V)$. Additionally, (1.) is equivalent to (1'.) $W=\varphi(V)$ for some measurable function $\varphi$.

  • 0
    I knew this property. Could you pleas also check if the statement in the "I had no problem showing that..."-section is true?2012-08-24
  • 0
    @vanguard2k To tell you the truth, already the first = sign of the series is a mystery to me... But the result holds.2012-08-24
  • 0
    I can understand why $E(Y|X)$ is $\sigma(X,Z)$-measurable but why should $E(Y|X,Z)$ be $\sigma(Z)$-measurable?2012-08-24
  • 0
    @vanguard2k But $E(Y\mid X,Z)$ is not $\sigma(Z)$-measurable, in general. Why do you ask?2012-08-24
  • 0
    @vanguard2k As I said, $E(Y\mid X,Z)$ is not a priori $\sigma(Z)$-measurable, hence $E(E(Y\mid X,Z)\mid Z)$ is not $E(Y\mid X,Z)$ in general. But $E(E(Y\mid X,Z)\mid Z)=E(Y\mid Z)$ is always true (and is often called the tower property). What I proved in my post uses another property, which is that $E(U\mid V)=U$ as soon as $U$ is $\sigma(V)$-measurable.2012-08-24
  • 0
    Note: my previous comment addresses a now deleted comment by @vanguard2k.2012-08-24
  • 0
    I realized my mistake shortly after posting my comment ;-) everything's fine for me now. Thank you!2012-08-24
  • 0
    hey, sry for this late reply but this sigma-measurable is not smth im really good in. why is by defintion T sigma(X,Z) measurable? because i calculate the expectation of T given X and Z and hence T must be sigma(X,T) measurable? kind regards2012-08-25
  • 0
    *why is by defintion T sigma(X,Z) measurable?* Which brings the question: which definition of conditional expectation do you know and use?2012-08-25
  • 0
    for me the conditional expectation is just a definition - as this was presented to me they did not bother about measurable. the definition i know is E(Y|X=x) = \sum_x x * P(Y=y|X=x) which is similar to the introduction to conditional expecation on wikipedia and what i used to show that E[E[Y|X,Z]|Z]=E[Y|Z].2012-08-25
  • 0
    i now what $E(Y|\sigma(X))$ means (in compare to $E(Y|X)$) regarding events but i dont see why Y is $\sigma(X,Z)$-measurable.2012-08-25
  • 0
    Sorry but you did not define the object E(Y|X), only E(Y|X=x) for every x. So, once again, what is E(Y|X) for you? Note that one needs to know that to even give some sense to an expression like E(E(Y|X,Z)|Z), which is in your post.2012-08-25
  • 0
    hm you are right but the reason i dont define things is that im not used to do it :). so im going to try :) my guess is that my my understanding E(Y|X) is the same as $E(Y|\sigma(X))$ since my understanding from E(Y|X) is that i observe the true expected value of Y for any possible given value of X which is every x in the image of X. ok i rephrase since $\sigma(X)$ contains more then just the value of x. but in the sense of events i kinda mean the same. since im used to express things from a econometric point of view X is for instance just a random variable converning age or height2012-08-25
  • 0
    Call me stubborn if you want but I would be at a loss to prove something about E(E(Y|X,Z)|Z) without knowing what E(Y|X,Z) is... :-) For a definition, see the second paragraph of my post.2012-08-25
  • 0
    my guess is that uve studied math and ppl i worked with are usually in this way :) its just the way u guys think :D if u mean the W thing then i still dont quite get it. for me theres written smth like this: E(U|V) is itself a random variable (this i now) which 1) is $\sigma(V)$-measurable (i dont quite understand what this means) and 2) has the same expected value given a set A which is in $\sigma(V)$ for me $\sigma(V)$ contains events. the simplest example i can thing of is the game heads,tails: $\Omega=\{\text{tails,heads}\}$ and them if $X(\omega)= 1$ if $\omega$ = tails etc...2012-08-25
  • 0
    now i can define $\sigma(X)$ for instance as $\{\{\text{heads,tails}},\{\text{tails}\},\{\text{heads}\},\emptyset\}$ so my understanding of $\sigma(X)$ measurable (for this given example) is that i can measure this W (the conditional expectation) for events which are contained in that $\sigma(X)$2012-08-25
  • 0
    is my 2nd example that E[E[Y|X]|X,Z]=E[Y|X] not the tower property?2012-08-26
  • 0
    No. The tower property is the fact that E(E(X|G)|H)=E(X|H) when H is a sigma-algebra included in the sigma-algebra G. When G is a sigma-algebra included in the sigma-algebra H, E(E(X|G)|H)=E(X|G) because E(Y|H)=Y for every random variable H-measurable and because E(X|G) is H-measurable.2012-08-26
  • 0
    but the outer expectation i.e. E(Y|X) is $\sigma(X,Z)$-measurable. The inner expectation is $\sigma(X)$-measurable and is the inner $\sigma$-algebra not included in the outer?2012-08-26
  • 0
    The fact that E(E(X|G)|H)=E(X|G) when G is a sigma-algebra included in the sigma-algebra H is true but this (true) fact is not called the tower property. I thought my previous comment made this clear.2012-08-26