Chain rule (probability)

For two events ${\displaystyle A}$ and ${\displaystyle B}$, the chain rule states that

${\displaystyle \mathbb {P} (A\cap B)=\mathbb {P} (B\mid A)\mathbb {P} (A)}$,

where ${\displaystyle \mathbb {P} (B\mid A)}$ denotes the conditional probability of ${\displaystyle B}$ given ${\displaystyle A}$.

A Jar A has 1 black ball and 2 white balls, and another Jar B has 1 black ball and 3 white balls. Suppose we pick an urn at random and then select a ball from that urn. Let event ${\displaystyle A}$ be choosing the first urn, i.e. ${\displaystyle \mathbb {P} (A)=\mathbb {P} ({\overline {A}})=1/2}$, where ${\displaystyle {\overline {A}}}$ is the complementary event of ${\displaystyle A}$. Let event ${\displaystyle B}$ be the chance we choose a white ball. The chance of choosing a white ball, given that we have chosen the first urn, is ${\displaystyle \mathbb {P} (B|A)=2/3.}$ The intersection ${\displaystyle A\cap B}$ then describes choosing the first urn and a white ball from it. The probability can be calculated by the chain rule as follows:

${\displaystyle \mathbb {P} (A\cap B)=\mathbb {P} (B\mid A)\mathbb {P} (A)={\frac {2}{3}}\cdot {\frac {1}{2}}={\frac {1}{3}}.}$

For events ${\displaystyle A_{1},\ldots ,A_{n}}$ whose intersection has not probability zero, the chain rule states

${\displaystyle {\begin{aligned}\mathbb {P} \left(A_{1}\cap A_{2}\cap \ldots \cap A_{n}\right)&=\mathbb {P} \left(A_{n}\mid A_{1}\cap \ldots \cap A_{n-1}\right)\mathbb {P} \left(A_{1}\cap \ldots \cap A_{n-1}\right)\\&=\mathbb {P} \left(A_{n}\mid A_{1}\cap \ldots \cap A_{n-1}\right)\mathbb {P} \left(A_{n-1}\mid A_{1}\cap \ldots \cap A_{n-2}\right)\mathbb {P} \left(A_{1}\cap \ldots \cap A_{n-2}\right)\\&=\mathbb {P} \left(A_{n}\mid A_{1}\cap \ldots \cap A_{n-1}\right)\mathbb {P} \left(A_{n-1}\mid A_{1}\cap \ldots \cap A_{n-2}\right)\cdot \ldots \cdot \mathbb {P} (A_{3}\mid A_{1}\cap A_{2})\mathbb {P} (A_{2}\mid A_{1})\mathbb {P} (A_{1})\\&=\mathbb {P} (A_{1})\mathbb {P} (A_{2}\mid A_{1})\mathbb {P} (A_{3}\mid A_{1}\cap A_{2})\cdot \ldots \cdot \mathbb {P} (A_{n}\mid A_{1}\cap \dots \cap A_{n-1})\\&=\prod _{k=1}^{n}\mathbb {P} (A_{k}\mid A_{1}\cap \dots \cap A_{k-1})\\&=\prod _{k=1}^{n}\mathbb {P} \left(A_{k}\,{\Bigg |}\,\bigcap _{j=1}^{k-1}A_{j}\right).\end{aligned}}}$

For ${\displaystyle n=4}$, i.e. four events, the chain rule reads

${\displaystyle {\begin{aligned}\mathbb {P} (A_{1}\cap A_{2}\cap A_{3}\cap A_{4})&=\mathbb {P} (A_{4}\mid A_{3}\cap A_{2}\cap A_{1})\mathbb {P} (A_{3}\cap A_{2}\cap A_{1})\\&=\mathbb {P} (A_{4}\mid A_{3}\cap A_{2}\cap A_{1})\mathbb {P} (A_{3}\mid A_{2}\cap A_{1})\mathbb {P} (A_{2}\cap A_{1})\\&=\mathbb {P} (A_{4}\mid A_{3}\cap A_{2}\cap A_{1})\mathbb {P} (A_{3}\mid A_{2}\cap A_{1})\mathbb {P} (A_{2}\mid A_{1})\mathbb {P} (A_{1})\end{aligned}}}$.

We randomly draw 4 cards (one at a time) without replacement from deck with 52 cards. What is the probability that we have picked 4 aces?

First, we set ${\textstyle A_{n}:=\left\{{\text{draw an ace in the }}n^{\text{th}}{\text{ try}}\right\}}$. Obviously, we get the following probabilities

${\displaystyle \mathbb {P} (A_{1})={\frac {4}{52}},\qquad \mathbb {P} (A_{2}\mid A_{1})={\frac {3}{51}},\qquad \mathbb {P} (A_{3}\mid A_{1}\cap A_{2})={\frac {2}{50}},\qquad \mathbb {P} (A_{4}\mid A_{1}\cap A_{2}\cap A_{3})={\frac {1}{49}}}$.

Applying the chain rule,

${\displaystyle \mathbb {P} (A_{1}\cap A_{2}\cap A_{3}\cap A_{4})={\frac {4}{52}}\cdot {\frac {3}{51}}\cdot {\frac {2}{50}}\cdot {\frac {1}{49}}={\frac {24}{6497400}}}$.

Let ${\displaystyle (\Omega ,{\mathcal {A}},\mathbb {P} )}$ be a probability space. Recall that the conditional probability of an ${\displaystyle A\in {\mathcal {A}}}$ given ${\displaystyle B\in {\mathcal {A}}}$ is defined as

${\displaystyle {\begin{aligned}\mathbb {P} (A\mid B):={\begin{cases}{\frac {\mathbb {P} (A\cap B)}{\mathbb {P} (B)}},&\mathbb {P} (B)>0,\\0&\mathbb {P} (B)=0.\end{cases}}\end{aligned}}}$

Then we have the following theorem.

For two discrete random variables ${\displaystyle X,Y}$, we use the events${\displaystyle A:=\{X=x\}}$ and ${\displaystyle B:=\{Y=y\}}$ in the definition above, and find the joint distribution as

${\displaystyle \mathbb {P} (X=x,Y=y)=\mathbb {P} (X=x\mid Y=y)\mathbb {P} (Y=y),}$

or

${\displaystyle \mathbb {P} _{(X,Y)}(x,y)=\mathbb {P} _{X\mid Y}(x\mid y)\mathbb {P} _{Y}(y),}$

where ${\displaystyle \mathbb {P} _{X}(x):=\mathbb {P} (X=x)}$ is the probability distribution of ${\displaystyle X}$ and ${\displaystyle \mathbb {P} _{X\mid Y}(x\mid y)}$ conditional probability distribution of ${\displaystyle X}$ given ${\displaystyle Y}$.

Let ${\displaystyle X_{1},\ldots ,X_{n}}$ be random variables and ${\displaystyle x_{1},\dots ,x_{n}\in \mathbb {R} }$. By the definition of the conditional probability,

${\displaystyle \mathbb {P} \left(X_{n}=x_{n},\ldots ,X_{1}=x_{1}\right)=\mathbb {P} \left(X_{n}=x_{n}|X_{n-1}=x_{n-1},\ldots ,X_{1}=x_{1}\right)\mathbb {P} \left(X_{n-1}=x_{n-1},\ldots ,X_{1}=x_{1}\right)}$

and using the chain rule, where we set ${\displaystyle A_{k}:=\{X_{k}=x_{k}\}}$, we can find the joint distribution as

${\displaystyle {\begin{aligned}\mathbb {P} \left(X_{1}=x_{1},\ldots X_{n}=x_{n}\right)&=\mathbb {P} \left(X_{1}=x_{1}\mid X_{2}=x_{2},\ldots ,X_{n}=x_{n}\right)\mathbb {P} \left(X_{2}=x_{2},\ldots ,X_{n}=x_{n}\right)\\&=\mathbb {P} (X_{1}=x_{1})\mathbb {P} (X_{2}=x_{2}\mid X_{1}=x_{1})\mathbb {P} (X_{3}=x_{3}\mid X_{1}=x_{1},X_{2}=x_{2})\cdot \ldots \\&\qquad \cdot \mathbb {P} (X_{n}=x_{n}\mid X_{1}=x_{1},\dots ,X_{n-1}=x_{n-1})\\\end{aligned}}}$

For ${\displaystyle n=3}$, i.e. considering three random variables. Then, the chain rule reads

${\displaystyle {\begin{aligned}\mathbb {P} _{(X_{1},X_{2},X_{3})}(x_{1},x_{2},x_{3})&=\mathbb {P} (X_{1}=x_{1},X_{2}=x_{2},X_{3}=x_{3})\\&=\mathbb {P} (X_{3}=x_{3}\mid X_{2}=x_{2},X_{1}=x_{1})\mathbb {P} (X_{2}=x_{2},X_{1}=x_{1})\\&=\mathbb {P} (X_{3}=x_{3}\mid X_{2}=x_{2},X_{1}=x_{1})\mathbb {P} (X_{2}=x_{2}\mid X_{1}=x_{1})\mathbb {P} (X_{1}=x_{1})\\&=\mathbb {P} _{X_{3}\mid X_{2},X_{1}}(x_{3}\mid x_{2},x_{1})\mathbb {P} _{X_{2}\mid X_{1}}(x_{2}\mid x_{1})\mathbb {P} _{X_{1}}(x_{1}).\end{aligned}}}$