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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 $A = \Omega$.

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Alex
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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.

Did
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  • 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 08:52
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    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 08:58
  • 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 $EX | \mathcal{G} = 3$ and so $E[X | \mathcal{G}]$ is the same for each $\omega \in B$. – Alex Sep 10 '12 at 09:03
  • See other comments of mine. – Did Sep 10 '12 at 09:52
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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:

  • $E[E[X\mid\mathcal{G}];\emptyset]=0=E[X;\emptyset]$,
  • $E[E[X\mid\mathcal{G}];\{1,3,5\}]=3\cdot\frac{3}{6}=\frac{3}{2}=E[X;\{1,3,5\}]$,
  • $E[E[X\mid\mathcal{G}];\{2,4,6\}]=4\cdot\frac{3}{6}=2=E[X;\{2,4,6\}]$,
  • $E[E[X\mid\mathcal{G}];\Omega]=\frac72=E[X;\Omega]$.
Did
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Stefan Hansen
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  • 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 07:30
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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 $A \in \mathcal{G}$!

Alex
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  • 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 09:02
  • 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 09:04
  • @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 09:12
  • 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 09:17
  • @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 09:27
  • Indeed. $1\ne{1}$. In your example $\Omega\cap2^\Omega=\varnothing$ hence $G\cap\Omega=\varnothing$. – Did Sep 10 '12 at 09:51