Example: Toss a coin twice. Letting $\mathbb P$ be a probability measure, suppose $\mathbb P(HH)=p^2,\mathbb P(HT)=\mathbb P(TH)=p(1-p), \mathbb P(TT)=(1-p)^2.$ I would like to answer the following questions:

  1. Define $Y$ to be the number of heads in the example. Derive the $\sigma$-field generated by $Y$.

  2. Find the expectation of the random variable $Y$ from the previous exercise, and also the conditional expectation of $Y$ given $\mathcal F_{1}$ s.t $\mathcal F_{1}=\{\varnothing,\{HH,HT\},\{TH,TT\},\Omega\}$. Check that in this case $E[E[Y|\mathcal F_{1}]]=E[Y]$.

My answers: the sample space is $\Omega=\{HH,HT,TH,TT\}$. I think the answer to 1 is $Y=\{\{TT\},\{HT,TH\},\{HH\}\}$, the $\sigma$-field generated by $Y$ is $\mathcal F(Y)=\{\varnothing,\{TT\},\{HH,HT,TH\},\{HT,TH\},\{HH,TT\},\{HH\},\{HT,TH,TT\},\Omega\}.$

But for question 2 I could only get $E(Y)=2p$. How is the conditional expectation computed?


2 Answers 2


The formal definition of conditional expectation is that $E[Y|\mathcal{F}_1]$ is any random variable measurable with respect to $\mathcal{F}_1$ having the property that

$$\int_F E[Y|\mathcal{F}_1](\omega)d\mathbb{P}(\omega) = \int_F Y(\omega) d\mathbb{P}(\omega)$$

for all $\mathcal{F}_1$-measurable sets $F$.

In the present case, this definition invites us to inspect all the measurable subsets $F$ with respect to $\mathcal{F}_1$, which you already computed in the first problem. The trick is to begin with the smallest, most basic $\mathcal{F}_1$-measurable sets (apart from the empty set), which are $\{HH, HT\}$ and $\{TH, TT\}$. Although we don't yet know $E[Y|\mathcal{F}_1]$, we can use the right hand side to compute its integrals. Because neither of these events can be decomposed (nontrivially) into smaller ones, the conditional expectation must have a constant value on each one. For example, writing $$E[Y|\mathcal{F}_1](HH) =E[Y|\mathcal{F}_1](HT) = z,$$

the definition gives

$$\eqalign{ zp &= zp^2 + zp(1-p) \\ &=E[Y|\mathcal{F}_1](HH) \mathbb{P}(HH) +E[Y|\mathcal{F}_1](HT) \mathbb{P}(HT)\\ &=\int_{\{HH, HT\}} E[Y|\mathcal{F}_1](\omega)d\mathbb{P}(\omega)\\ &= \int_{\{HH, HT\}} Y(\omega) d\mathbb{P}(\omega)\\ &= Y(HH)\mathbb{P}(HH) + Y(HT)\mathbb{P}(HT) \\ &= 2p^2 + 1p(1-p)= p+p^2, }$$

whence we deduce

$$z = \frac{p+p^2}{p} = 1 + p.$$

A similar calculation for $F = \{TH, TT\}$ (do it!) establishes that

$$E[Y|\mathcal{F}_1](TH) =E[Y|\mathcal{F}_1](TT) = p.$$

There is a simple intuition to support these abstract calculations: $\mathcal{F}_1$ records the information available after flipping the coin the first time. If it comes up heads, the possible events (which have only partially occurred!) are $HH$ and $HT$. We already have one head and there is a chance $p$ that the second flip will be a head. Thus, at this stage, our expectation of $Y$ equals $1$ (for what has happened) plus $1\times p$ (for what could happen yet), summing to $1+p$. If instead the first flip is tails, we have seen no heads yet but there is still a chance of $p$ of seeing a head on the second flip: the expectation of $Y$ is just $0 + 1\times p = p$ in that case.

As a check, we can compute

$$\eqalign{ E[E[Y|\mathcal{F}_1]] &= \int_\Omega E[Y|\mathcal{F}_1](\omega)d\mathbb{P}(\omega) \\ &= (1+p)\mathbb{P}(E[Y|\mathcal{F}_1]=1+p) + (p)\mathbb{P}(E[Y|\mathcal{F}_1]=p)\\ & = (1+p)\mathbb{P}(\{HH, HT\}) + (p) \mathbb{P}(\{TH, TT\})\\ &= (1+p)(p^2 + p(1-p)) + p((1-p)p + (1-p)^2)\\ &= 2p, }$$

exactly as before.

It should be clear that this is just a laborious way of expressing the idea that there is a $p$ chance at the outset of heads--which has a conditional expectation of $1+p$-- and a $1-p$ chance of tails--which has a conditional expectation of $p$: everything reduces to the elementary calculation of conditional expectations, which needs no sigma fields or integrals. The point of this exercise is to build on that intuition to develop an understanding that will hold up when these sigma algebras get much, much more complicated.

  • 2
    $\begingroup$ I also wrote my answer because that deleted answer bothered me a little bit. A sub-sigma-field represents partial information, conditional expectations are measurable functions. We all should write this on the mirror. I like your intuitive explanation at the end $\endgroup$
    – Zen
    Nov 2, 2013 at 0:49
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    $\begingroup$ @whuber, there is one small part of this that I don't understand: the conditional expectation must have a constant value on each one - what does that mean? $\endgroup$
    – makansij
    Oct 27, 2018 at 18:38
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    $\begingroup$ @Hunle "One" refers back to "events." An event is a set; the conditional expectation is a function; ergo, it must have the same value on each element of the event. $\endgroup$
    – whuber
    Oct 27, 2018 at 18:55
  • 1
    $\begingroup$ @whuber right of course! Then I don't quite understand why we have the following: $E[Y|\mathcal{F}_1](HH) =E[Y|\mathcal{F}_1](HT) = z,$. I can see that the events cannot be further decomposed, but why must the 2 be equal? $\endgroup$ Dec 27, 2018 at 2:27
  • 1
    $\begingroup$ @itztricky Because in every element of $\mathcal{F}_1,$ either both of HH, HT are in it or both are not in it. Ergo, all measurable functions must have the same values on both outcomes. You might be overthinking this. Perhaps it would help to know that this is not a statistical argument--it is just a basic set theory deduction. $\endgroup$
    – whuber
    Dec 27, 2018 at 13:45

Take the sample space $\Omega$ as the Cartesian product $\{H,T\}\times\{H,T\}$, with sigma-field $\mathscr{F}$ equal to the class of all subsets of $\Omega$. The sigma-field generated by $Y$ (denoted by $\sigma(Y)$) is the smallest sub-sigma-field of $\mathscr{F}$ in which $Y$ is measurable. Since $Y\in\{0,1,2\}$, the inverse images $$ Y^{-1}(\{0\}) = \{(T,T)\}\, , \quad Y^{-1}(\{1\}) = \{(H,T),(T,H)\}\, , \quad \quad Y^{-1}(\{2\}) = \{(H,H)\} \, , $$ show that $$ \sigma(Y)=\sigma\left\{\{(T,T)\}, \{(H,T),(T,H)\}, \{(H,H)\}\right\} = \left\{\emptyset,\{(T,T)\}, \{(H,T),(T,H)\}, \{(H,H)\},\{(H,T),(T,H),(H,H)\},\{(T,T),(H,H)\},\{(T,T),(H,T),(T,H)\},\Omega\right\} \, . $$ Define $$ \mathscr{G}=\{\emptyset,\{(H,H),(H,T)\},\{(T,T),(T,H)\},\Omega\}\subset\mathscr{F} \, . $$

The conditional expectation $Z=\mathrm{E}[Y\mid \mathscr{G}]$ is a $\mathscr{G}$-measurable random variable satisfying $$ \mathrm{E}[Y I_A]=\mathrm{E}[Z I_A] \, , \qquad\qquad (*) $$ for every $A\in\mathscr{G}$. The fact that $Z$ is $\mathscr{G}$-measurable entails that it is constant in the atoms of $\mathscr{G}$ (this is the crucial idea). Let $$ Z(H,H) = Z(H,T) = a, \quad Z(T,T) = Z(T,H) = b \, . $$ Taking $A=\{(H,H),(H,T)\}$, relation $(*)$ yields $$ 2 \cdot p^2 + 1 \cdot p(1-p) = a \cdot p^2 + a \cdot p(1-p) \, , $$ implying that $a=1+p$. Similarly, we find $b=p$.

Finally, $$ \mathrm{E}[Y]= 1 \cdot 2 p(1-p) + 2\cdot p^2 = 2p \, , $$ and $$ \mathrm{E}[Z] = (1+p) \cdot (p^2 + p(1-p)) + p \cdot ((1-p)^2 + p(1-p)) = 2p \, , $$ as "expected".

P.S. May I suggest a beautiful related exercise? Let $\Omega=[0,1]$, $\mathscr{F}$ be the Borel subsets of $\Omega$, and $P$ be Lebesgue measure. Let $\mathscr{G}$ be the sub-sigma-field of $\mathscr{F}$ generated by the partition $$\{[0,1/2],(1/2,1]\}\, . $$ Let $X$ be the identity map ($X(\omega)=\omega$). Plot the graph of $\mathrm{E}[X\mid\mathscr{G}]$.

  • $\begingroup$ Isn't it the case that $E[X|\mathscr{G}]$ is unique in your final exercise, because $\mathscr{G}$ is finite? There can be no question of changing $E[X|\mathscr{G}]$ on a nonempty set of measure zero w.r.t. $\mathscr{G}$. $\endgroup$
    – whuber
    Nov 1, 2013 at 22:45

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