1.4 Chain Rule

Chain Rule

For a finite number of events $A_1,...,A_N\subseteq \Omega$ in the same probability space, the chain rule states that the probability of all $N$ events occurring is equal to the product of the conditional probabilities of each event given the events that came before it, such that
$\mathbb{P}[{\cap }_{i=1}^N{A}_i]=\mathbb{P}[A_1]\times \mathbb{P}[A_2\mid A_1]\times \mathbb{P}[A_3\mid A_1\cap A_2]\times ...\times \mathbb{P}[A_N\mid {\cap }_{i=1}^{N-1}{A}_i].$

Intuition.

For two events, the chain rule is simply
$\mathbb{P}[A\cap B]=\mathbb{P}[A]\times \mathbb{P}[B\mid A]=\mathbb{P}[B]\times \mathbb{P}[A\mid B],$
which looks very similar to a reordering of the conditional probability theorem. We can say that the probability of both $A$ and $B$ happening is equal to the scaled probability of $B$ happening given that $A$ has already happened. Let’s expand our use case for 3 events using :

\mathbb{P}[A \cap B \cap C] &= \mathbb{P}[(A \cap B) \cap C] \\ &= \mathbb{P}[A \cap B] \times \mathbb{P}[C \mid A \cap B] \\ &= \mathbb{P}[A] \times \mathbb{P}[B \mid A] \times \mathbb{P}[C \mid A \cap B]. \end{align}$$ This comes from the fact that sets are associative (as seen in the first line of the equation). Therefore, we can implicitly substitute $D = A \cap B$, and first solve for the two-event scenario. # Examples >[!example] Monty Hall Revisited > >Let's figure out the Monty Hall problem using another set of events. For $i \in \{ 1, 2, 3 \}$: >- $C_i$ is the event that the contestant chooses door $i$. >- $P_i$ is the event that the prize is behind door $i$. >- $H_i$ is the event that the host opens door $i$. > >Considering the rules from the previous instance, what is the probability of the contestant choosing door 1, the prize is behind door 2, and the host opens door 3 ($C_1 \cap P_{2} \cap H_3$)? We have: >$$\mathbb{P}[C_1 \cap P_2 \cap H_3]=\mathbb{P}[C_1] \times \mathbb{P}[P_2 | C_1] \times \mathbb{P}[H_3 | C_1 \cap P_2]$$ >Which is equal to $\frac{1}{3} \times \frac{1}{3} \times 1 = \frac{1}{9}$. >Note that if the contestant chooses the same door that the prize is behind, then $\mathbb{P}[C_1 \cap P_2 \cap H_3]=\frac{1}{3} \times \frac{1}{3} \times \frac{1}{2} = \frac{1}{18}$ Since the host has two doors with goats to pick from. > >Going back to the original question, noting that $\mathbb{P}[C_1 \cap H_3]=\mathbb{P}[C_1] \times \mathbb{P}[H_3 | C_1]=\frac{1}{3} \times \frac{1}{2}=\frac{1}{6}$, >$$\mathbb{P}[P_2 | C_1 \cap H_3] = \frac{\mathbb{P}[C_1 \cap P_2 \cap H_3]}{\mathbb{P}[C_1 \cap H_3]}=\frac{\frac{1}{9}}{\frac{1}{6}}=\frac{2}{3}.$$ >If instead the contestant chooses door 2, and the prize is behind door 2, then >$$$$