# Having trouble determining the sample space for conditional expectation problems

My book states the following theorem

Let $$X$$ be a random variable with sample space $$\Omega$$. If $$F_1, F_2, . . . , F_r$$ are events such that $$F_i$$ and $$F_j$$ are disjoint (for $$i$$ not equal to $$j$$) and $$\Omega = \cup_jF_j$$ then $$E(X) = \sum_jE(X|F_j)P(F_j)$$

I understand this to mean that given $$X$$ which can take on a value / outcome from $$\Omega$$ and $$F_1, F_2, . . . , F_r$$ are all pairwise disjoint whose union forms $$\Omega$$, $$E(X)$$ can be computed by the given equation. However, I had trouble understanding the example illustrating this theorem:

Let T be the number of rolls in a single play of craps. We can think of a single play as a two-stage process. The first stage consists of a single roll of a pair of dice. The play is over if this roll is a 2, 3, 7, 11, or 12. Otherwise, the player's point is established, and the second stage begins. This second stage consists of a sequence of rolls which ends when either the player's point or a 7 is rolled. We record the outcomes of this two-stage experiment using the random variables X and S, where X denotes the first roll, and S denotes the number of rolls in the second stage of the experiment (of course, S is sometimes equal to 0). Note that T = S + 1. Then $$E(T) = \sum_{j=2}^{12} E(T|X = j)P(X=j)$$

Here I think T is analogous to the X in the theorem above and each outcome of a roll of the two dices ($$X = j$$) is an event analogous to an $$F_j$$. However, the potential values of T can take on are in the set $$\{1, 2, ... \infty\}$$ while I think the set formed by the union of events $$X = j$$ for all $$j$$ is just $$\{2,3..,12\}$$. I'm confused about what the sample space $$\Omega$$ would be in this example since it appears the values of the random variable and event appear to be drawn from different sets.

• This is called the "law of total expectation" and is very similar to the "law of total probablity." Indeed their partition is $$\Omega = \{X=2\}\cup \{X=3\}\cup ...\cup \{X=12\}$$ Every element of the sample space $\Omega$ has the form $$(\mbox{first roll}, \mbox{other rolls}) = (X, \mbox{other rolls})$$ and so the event $\{X=2\}$ is the set of all such elements of the form $(2, \mbox{other rolls})$. Intuitively we know that $X$ can only take values in $\{2, 3, ..., 12\}$ and so the cases $\{X=2\}, \{X=3\}, ..., \{X=12\}$ are mutually exclusive but collectively exhaustive. – Michael May 11 at 22:45

## 1 Answer

This is called the "law of total expectation" and is similar to the "law of total probability." The sample space $$\Omega$$ is the set of all outcomes $$\omega$$ of the form: $$\omega = (\mbox{first roll}, \mbox{sequence of other rolls})=(X, \mbox{sequence of other rolls})$$ This sample space can be partitioned into events where the first roll is 2, the first roll is 3, ..., the first roll is 12. So the partition is: $$\Omega = \{X=2\} \cup \{X=3\}\cup\{X=4\}\cup...\cup\{X=12\}$$ We know that $$X$$ can only take values in the set $$\{2, ..., 12\}$$ and so the events $$\{X=2\}, \{X=3\}, ..., \{X=12\}$$ are indeed mutually exclusive and collectively exhaustive. The event $$\{X=4\}$$ contains all outcomes that start with a first roll of 4.

The individual events $$\{X=i\}$$ look like this: \begin{align} \{X=2\} &= \{(2)\}\\ \{X=3\} &= \{(3)\}\\ \{X=4\} &= \{(4, 4), (4, 2, 2, 5, 4), (4, 12, 5, 5, 7), (4, 8, 4), ...\} \end{align} and so on. The event $$\{X=4\}$$ has a countably infinite number of outcomes, but all of them are sequences that start with $$4$$ and end with either 4 or 7.

The events $$\{X=2\}, \{X=3\}, \{X=7\}, \{X=11\}, \{X=12\}$$ all contain just one outcome each and so we trivially have \begin{align} E[T|X=2]&=1\\ E[T|X=3]&=1\\ E[T|X=7]&=1\\ E[T|X=11]&=1\\ E[T|X=12]&=1 \end{align} On the other hand, $$E[T|X=4]$$ is equal to 1 plus the expected time to roll either a 4 or a 7.

• My confusion is the values $T$ can take, because in the example $T$ is specified as number of rolls, so I thought it would take on values $\{1,2,...\infty\}$ and it wasn't obvious to me how that is an outcome in the set $\{X=2\} \cup \{X=3\}\cup\{X=4\}\cup...\cup\{X=12\}$ – Yandle May 12 at 0:05
• Both T and X are random variables, meaning they can be determined by the outcome $\omega$. $$\omega=(4,5,6,5,11,2,4)\implies X(\omega)=4,T(\omega)=7$$ Can you tell me $X(\omega)$ and $T(\omega)$ if $\omega=(8,9,9,7)$? – Michael May 12 at 1:29
• Next, can you specify two different outcomes $\omega$ and $v$ such that $X(\omega)\neq X(v)$ but $T(\omega)=T(v)$? – Michael May 12 at 1:51
• Following this logic, $X((8,9,9,7)) = 8$ and $T((8,9,9,7)) = 4$. $\omega = (4,9,9,7)$ and $v = (8,9,9,7)$ should satisfy the second problem. – Yandle May 12 at 3:20
• My book defines random variables as "value of the outcome of a certain experiment" and sample space as "the set of all possible values of [the random variable], or equivalently, the set of all possible outcomes of the experiment". I interpreted this to mean that the random variable must take on a value $\omega \in \Omega$ (kind of like domain of a function), but that's not the case here. $X$ and $T$ both take on values that are not technically an outcome found in $\Omega$. Where is my misunderstanding? – Yandle May 12 at 3:32