Event Independence versus Conditional Probability

26 May 2018

Prologue To Event Independence versus Conditional Probability

In the disciplines of probability, event independence is quiet an important concept. The inference from conditional probability come out with the result that one event occurrence is said to not related to another event, if the two events are independent.

Definition: Event Independence

An event $A$ is said independent of event $B$, if
$\;\;\;\;P(A\vert B)$=$P(A)$.

Event Independence Equivalence

By the definition of event independence, we can have an equivalence of expression from conditional probability:
$P(A\vert B)$=$\frac {P(A\cap B)}{P(B)}$=$P(A)$
$\Leftrightarrow P(A\cap B)=P(A)\cdot P(B)$

Below lists the basic properties:
➀$P(A\cap B)$=$P(A)\cdot P(B)$ indicates event $A$ is independent of event $B$.
➁by its symmetry, $P(A)\cdot P(B)$=$P(B)\cdot P(A)$=$P(B\cap A)$, event $B$ is independent of event $A$.
➂$P(A\vert B)$=$P(A)$ and $P(B\vert A)$=$P(B)$.

Event Independence Extension

[1] multiple events independence

$P(N_{1}\cap N_{2}\cap…\cap N_{m})$
=$P(N_{1})\cdot P(N_{2})\cdot …\cdot P(N_{m-1})\cdot P(N_{m})$, given that all events $N_{i}$ are all independent.

proof::mjtsai1974

➀$P(N_{m-1}\vert N_{m})$=$\frac {P(N_{m-1}\cap N_{m})}{P(N_{m})}$=$P(N_{m-1})$
Then $P(N_{m-1}\cap N_{m})$
=$P(N_{m-1})\cdot P(N_{m})$.
➁$P(N_{m-2}\vert (N_{m-1}\cap N_{m}))$=$\frac {P(N_{m-2}\cap (N_{m-1}\cap N_{m}))}{P(N_{m-1}\cap N_{m})}$=$P(N_{m-2})$
Then $P(N_{m-2}\cap (N_{m-1}\cap N_{m}))$
=$P(N_{m-2})\cdot P(N_{m-1})\cdot P(N_{m})$.
➂by mathematics induction, could we finally have the equivalence of expression.

[2] independence equivalence relation

Event $A$ is independent of event $B$
$\Leftrightarrow$ event $A^{c}$ is independent of event $B$

proof::mjtsai1974

$1$-$P(A\vert B)$
=$1$-$P(A)$
=$P(A^{c})$
=$P(A^{c}\vert B)$
From the end to the beginning could we prove the inverse direction.

[3] independence of all events

Given event $A$ is independent of event $B$, we can infer out all possible independence in between $A$, $A^{c}$, $B$, $B^{c}$.

proof::mjtsai1974

$P(A\vert B)$=$P(A)$
$\Leftrightarrow P(A^{c}\vert B)$=$P(A^{c})$
$\Leftrightarrow P(B\vert A^{c})$=$P(B)$
$\Leftrightarrow P(B^{c}\vert A^{c})$=$P(B^{c})$
$\Leftrightarrow P(B\vert A)$=$P(B)$
$\Leftrightarrow P(B^{c}\vert A)$=$P(B^{c})$
$\Leftrightarrow P(A\vert B^{c})$=$P(A)$
$\Leftrightarrow P(A^{c}\vert B^{c})$=$P(A^{c})$

We conclude if $A$ is independent of $B$, then $A^{c}$ is independent of $B$, $A$ is independent of $B^{c}$, $A^{c}$ is independent of $B^{c}$.

[4] event and its complemenmt

The prpbability of intersection of any given event and its complement is $0$, that is $P(A\cap A^{c})$=$0$.

Example: 2nd Head following 1st Head

Suppose you are tossing a fair coin, the probability of head and tail are all $\frac {1}{2}$, and each tossing is an independent case.

We’d like to ask for the probability that the 2nd tossing out a head, right after the 1st tossing out a head, then,
➀take the event of 1st tossing as $A_{1}$=$\{H,T\}$, take the event of 2nd tossing as $A_{2}$=$\{H,T\}$.
➁the sample space of these 2 tossing would be $\Omega$=$\{HH,HT,TH,TT\}$. The probability of 2 contiguous heads is $\frac {1}{4}$.
➂the probability that the 2nd tossing out a head, right after the 1st tossing out a head is asking for $P(A_{2}\cap A_{1})$, then
$\frac {1}{4}$
=$P(A_{2}\cap A_{1})$
=$P(A_{2}\vert A_{1})\cdot P(A_{1})$
=$\frac {1}{2}\cdot \frac {1}{2}$
=$P(A_{2})\cdot P(A_{1})$
Thus, we have $P(A_{2}\vert A_{1})$=$P(A_{2})$, it is fully compliant with the given that each tossing is an independent case.

Example: Illustration By Tossing A Fair Die

Suppose you are tossing a fair die, the sample space $\Omega$=$\{1,2,3,4,5,6\}$. We denote the event of numbers smaller than $4$ as $A$=$\{1,2,3\}$, and denote the event of even numbers as $B$=$\{2,4,6\}$.

To evaluate if event $A$ is independent of event $B$:
➀$P(A\cap B)$=$P(\{2\})$=$\frac {1}{6}$
➁$P(A)\cdot P(B)$=$\frac {1}{2}\cdot \frac {1}{2}$=$\frac {1}{4}$
Hence, the event of numbers smaller than $4$ is not independent of the event of even numbers.