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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.

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    $\begingroup$ 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. $\endgroup$ – Michael May 11 at 22:45
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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.

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  • $\begingroup$ 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 $\{X=2\} \cup \{X=3\}\cup\{X=4\}\cup...\cup\{X=12\}$ $\endgroup$ – Yandle May 12 at 0:05
  • $\begingroup$ 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 $\omega=(8,9,9,7)$? $\endgroup$ – Michael May 12 at 1:29
  • $\begingroup$ Next, can you specify two different outcomes $\omega$ and $v$ such that $X(\omega)\neq X(v)$ but $T(\omega)=T(v)$? $\endgroup$ – Michael May 12 at 1:51
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    $\begingroup$ 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. $\endgroup$ – Yandle May 12 at 3:20
  • $\begingroup$ 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? $\endgroup$ – Yandle May 12 at 3:32

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