Practice Problems

Categories
Unit 01 Unit 02 Unit 03 Unit 04 Unit 05 Unit 06 Unit 07
Edit Problem
Unit 01 /
Problem 2:


H conditional probability and the multiplication rule
|
Hard


(Recursive Conditional)

Suppose there are $n$ events, $A_1,A_2,...,A_n$ and the intersection of all these events is non-empty. Write an inductive proof to show that $$P(A_1\cap A_2\cap\ldots \cap A_n)=P(A_1)P(A_2|A_1)P(A_3|A_1\cap A_2)\ldots P(A_n|A_1\cap A_2\cap\ldots\cap A_{n-1}).$$

Solution:

For $n=1$, we have the trivial statement that $P(A_1)=P(A_1)$.Now let's assume that the statement whenever $n < N$. In particular, it is true when n = N-1, i.e.:

$$P(A_1,A_2,...,A_{N-1})=P(A_1)P(A_2|A_1)...P(A_{N-1}|A_1\cap A_2\cap ... \cap A_{N-2})$$

To show the statement holds for $n=N$, let's define an event B, where:

$B=A_1\cap A_2 \cap ... \cap A_{N-1}$

$$ P(A_1,A_2,...,A_{N-1},A_n)=P(B \cap A_N)$$

From Bayes rule we have:

$$P(B \cap A_N)=P(B)P(A_N|B)$$

Substituting $A_1\cap A_2 \cap ... \cap A_{N-1}$ for B,

$$ P(A_1\cap A_2 \cap ... \cap A_{N-1} \cap A_N)=P(A_1\cap A_2 \cap ... \cap A_{N-1})P(A_N|A_1\cap A_2 \cap ... \cap A_{N-1}) $$

Substituting the first equation above, we have the final result,$$\begin{align*}P(A_1\cap A_2 \cap ... \cap A_{N-1} \cap A_N)= \\P(A_1)P(A_2|A_1)...P(A_{N-1}|A_1\cap A_2\cap ... \cap A_{N-2})P(A_N|A_1\cap A_2 \cap ... \cap A_{N-1})\end{align*}$$