Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to internalise the measure theoretic definition of conditional expectation.

Consider a fair six-sided die. Formally the probability space is $(\{1, 2, 3, 4, 5,6\}, \mathcal{P}(1, 2, 3, 4, 5, 6), U)$ where $U$ is the discrete uniform distribution. Let the real-valued random variable map the identity map on the sample space so that $X(\omega) = \omega$.

Byron Schmuland answered this question in a way that gives a lot of intuition. Suppose that after the die is rolled you will be told if the value is odd or even. Then you should use a rule for the expectation that depends on the parity. However I still don't see how to formalise his point.

Let the conditioning $\sigma$-field be $\mathcal{G} = \{\emptyset, \Omega, \{1, 3, 5\}, \{2, 4, 6\}\}$ as this includes the events that the value is even or odd. My question is, what is a full and formal description of $E(X | \mathcal{G})$.

Is it this? \begin{equation} E(X | \mathcal{G}) = \begin{cases} 0 & \mbox{if $A = \emptyset$} \\ 3.5 & \mbox{if $A = \Omega$} \\ 3 & \mbox{if $A = \{1, 3, 5\}$} \\ 4 & \mbox{if $A = \{2, 4, 6\}$} \end{cases} \end{equation}

In particular I feel unsure about the cases where $A = \emptyset$ and $A = \Omega$.

share|improve this question
add comment

3 Answers 3

up vote 3 down vote accepted

A crucial point is that $Y=\mathrm E(X\mid \mathcal G)$ is a random variable (hence I really do not know what the identities at the end of your post could mean), which is entirely determined by two conditions. One first asks that

$Y:\Omega\to\mathbb R$ is measurable with respect to $\mathcal G$.

Since $\mathcal G\subset \mathcal F$, this is really a supplementary condition when compared to the condition of being measurable with respect to $\mathcal F$, as any random variable is.

In your case, $\mathcal G=\sigma(B)$ with $B=\{2,4,6\}$ hence one knows a priori that $Y=b\mathbf 1_B+c$ for some $b$ and $c$. To compute $b$ and $c$, one uses the other condition on $Y$, besides being measurable with respect to $\mathcal G$, which is that

$\mathrm E(X;C)=\mathrm E(Y;C)$ for every $C$ in $\mathcal G$.

Here, $\mathrm E(Y;B)=u$ with $u=\mathrm E(X;B)$ and $\mathrm E(Y)=v$ with $v=\mathrm E(X)$. Since $\mathrm P(B)=\frac12$, $\frac12(b+c)=u$ and $\frac12b+c=v$, which yields $b$ and $c$. Thus, $\mathrm E(X\mid \mathcal G)(\omega)=c$ if $\omega=1$, $3$ or $5$ and $\mathrm E(X\mid \mathcal G)(\omega)=b+c$ if $\omega=2$, $4$ or $6$. Numerically, $u=2$ and $v=\frac72$ hence $b=1$ and $c=3$, and $$ \mathrm E(X\mid\mathcal G)=\mathbf 1_B+3. $$ Note: Let me strongly advise anyone interested in these matters to read the wonderful little book Probability with martingales by David Williams.

share|improve this answer
    
Thanks @did. I see that the expectation must be the same for all $\omega \in B$ but how did you know to write $Y(\omega) = b \textbf{1}_B(\omega) + c$? –  Alex Sep 10 '12 at 8:52
1  
You are welcome. The expectation must be the same for every $C$ in $G$, that is, $E(X;C)=E(Y;C)$. The phrase *the same for all $\omega\in B$* is meaningless. One knows that $Y=b1_B+c$ because these are the only random variables measurable for $G$. More generally, if $G$ is generated by a partition $(B_k)$ then every $Z$ measurable for $G$ is $Z=\sum\limits_kb_k1_{B_k}$ for some scalars $b_k$. (In the present case, $b1_{B}+b'1_{\Omega\setminus B}$ is equivalent to $b1_B+c$.) –  Did Sep 10 '12 at 8:58
    
did, can I ask what you mean by saying "the same for all $\omega \in B$ is meaningless"? For example isn't the asnwer by @StefanHansen defining the expectation for each $\omega \in \Omega$. If $\omega \in B = \{1, 3, 5\}$ we use $E[X | \mathcal{G}](\omega) = 3$ and so $E[X | \mathcal{G}]$ is the same for each $\omega \in B$. –  Alex Sep 10 '12 at 9:03
    
See other comments of mine. –  Did Sep 10 '12 at 9:52
add comment

The conditional expectation is a random variable, and therefore your notation does not quite fit. I would write it as: $$ E[X\mid\mathcal{G}](\omega)= \begin{cases} 3, \quad \omega\in\{1,3,5\},\\ 4, \quad \omega\in\{2,4,6\}. \end{cases} $$ Now all you have to check is that this random variable satisfies the conditions for being the conditional expectation. That is, $E[X\mid\mathcal{G}]$ is $\mathcal{G}$-measureable and $$ E[E[X\mid\mathcal{G}]\,;A]=E[X\, ;A], $$ for every $A\in\mathcal{G}$. Here $E[\;; A]$ denotes integration over the set $A$. Now you just have to check each of the cases:

  • $E[X\mid\mathcal{G}]^{-1}(B)=\{1,3,5\}$ if $3\in B$ and $4\notin B$,
  • $E[X\mid\mathcal{G}]^{-1}(B)=\{2,4,6\}$ if $3\notin B$ and $4\in B$,
  • $E[X\mid\mathcal{G}]^{-1}(B)=\Omega$ if both $3,4\in B$,
  • $E[X\mid\mathcal{G}]^{-1}(B)=\emptyset$ if $3,4\notin B$,

and hence $E[X\mid\mathcal{G}]$ is $\mathcal{G}$-measureable. Now we have to check the second condition:

  • $E[E[X\mid\mathcal{G}];\emptyset]=0=E[X;\emptyset]$,
  • $E[E[X\mid\mathcal{G}];\{1,3,5\}]=3\cdot\frac{3}{6}=\frac{3}{2}=E[X;\{1,3,5\}]$,
  • $E[E[X\mid\mathcal{G}];\{2,4,6\}]=4\cdot\frac{3}{6}=2=E[X;\{2,4,6\}]$,
  • $E[E[X\mid\mathcal{G}];\Omega]=\frac72=E[X;\Omega]$.
share|improve this answer
    
Note that this is in no way the most optimal way of showing it (this is just checking the definition by "brute-force") - for a neater approach just look at did's answer above. –  Stefan Hansen Sep 10 '12 at 7:30
add comment

The answers by @did and @StefanHansen are both great. I thought I would explain my key misunderstanding in case it helps someone else.

A random variable is a function $Y: \Omega \to \mathbb{R}$. Even though we condition on $\mathcal{G}$ we only define $Y$ on $\Omega$! $\mathcal{G}$ doesn't contain elements of $\Omega$ but only subsets; even if for example $\{1\} \in \mathcal{G}$. I was confused and thought that since we were conditioning on $\mathcal{G}$ that we needed to define $Y$ for each $A \in \mathcal{G}$!

share|improve this answer
    
You are confusing elements of $\Omega$ with subsets of $\Omega$. The elements of $G$ are subsets of $\Omega$, not (a priori) elements of $\Omega$. –  Did Sep 10 '12 at 9:02
    
I am not sure where I am confusing elements of $\Omega$ with subsets of $\Omega$. For example when I write $\mathcal{G} = \{\emptyset, \Omega\}$ aren't I using $\mathcal{G} \subset \Omega$ because in particular $\emptyset \not \in \Omega$. And when I write $A \not \in \Omega$ isn't it also clear that the elements of $G$ are subsets of $\Omega$ and not (necessarily) elements of $\Omega$ because $\{1, 3, 5\}$ is a subset (but not an element) of $\Omega$? –  Alex Sep 10 '12 at 9:04
    
@did if there is something in particular that I have gotten confused, then I would like to alter it so as not to confuse anyone else; however at the moment I still can't see it! –  Alex Sep 10 '12 at 9:12
    
Your In fact $G$ may not even include $\omega\in\Omega$* leaves the impression that you believe that it may happen (or that the usual situation is) that some $\omega\in\Omega$ are elements of $G$ or parts of $G$. This never happens. See also *not (necessarily) in your comment. If you are aware that $G$ is a subset of $2^\Omega$ (the set of subsets of $\Omega$), never a subset of $\Omega$, then everything is fine. –  Did Sep 10 '12 at 9:17
    
@did, maybe I was a little confused. You are saying for example that $\{1\}$ can be in $\mathcal{G}$ but even then it is as a subset of $\Omega$ and not as an element of $\Omega$. –  Alex Sep 10 '12 at 9:27
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.