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 ${\displaystyle A}$ .

An Urn A has 1 black ball and 2 white balls and another Urn 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 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,

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

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

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

• René L. Schilling (2021), Measure, Integral, Probability & Processes - Probab(ilistical)ly the Theoretical Minimum (1 ed.), Technische Universität Dresden, Germany, ISBN 979-8-5991-0488-9{{citation}}: CS1 maint: location missing publisher (link)
• William Feller (1968), An Introduction to Probability Theory and Its Applications, vol. I (3 ed.), New York / London / Sydney: Wiley, ISBN 978-0-471-25708-0
• Russell, Stuart J.; Norvig, Peter (2003), Artificial Intelligence: A Modern Approach (2nd ed.), Upper Saddle River, New Jersey: Prentice Hall, ISBN 0-13-790395-2, p. 496.