Conditional Probability and Independence

Conditional Probability (조건부 확률)

The probability that $C_2$ occurs given that event $C_1$ has occurred is called the conditional probability of $C_2$ given $C_1$ and is defined by 

 

그림으로 표현하자면 아래와 같다.

Conditional probability is a probability? 

조건부확률이 확률의 정의를 만족하는지 살펴보자.

 

Properites of Conditional Probability

 

Ex. Four cards are to be dealt successively, at random and without replacement, from an ordinary deck of playing cards. The probability of receving a spade, a heart, a diamond, and a club, in that order is.. 

 

$C_1$ : 1st - 13 / 52

$C_2$ : 2nd - 13 / 51

$C_3$ : 3rd - 13 / 50

$C_4$ : 4th - 13 / 49

 

(카드 놀이에서 4번을 뽑는데 각각 다른 걸 뽑을 확률을 나타낸 것)

 

 

- Prior probability vs. posterior probability

- Let $C_1, ... C_k$ : k causes of an event.

- P($C_i$) indicates the chance of $i$th cause.

  If known, it is obtained from the past investigation >>>> prior probability 

  (쉽게 말해 알려진 확률을 prior probability)

- P($C_i | C$) indicates the chance of $i$th cause when the event $C$ happened.

  It updates the past information >>>> posterior probability

 

 

Independence

1. Statistically / stochastically independent means independent in a probability sense.
2. Two events A and B are independent if and only if $P(A|B) = P(A|B^c) = P(A)$ or $P(B|A) =P(B|A^c) = P(B)$. Otherwise, they are independent.
3. Events $C_1, C_2, C_3$  are pairwise independents if and only if $P(C_1 \cap C_2) = P(C_1)P(C_2)$, $P(C_1 \cap C_3) = P(C_1)P(C_3)$, $P(C2 \cap C_3) = P(C_2)P(C_3)$.

 

Mutual Independence (모든 경우가 독립이면)

 

ex. Pairwise independence does not imply mutual independence

1,2,3,4가 적혀있는 spinner를 2번 돌렸다. 

$C_1$은 두 번 돌린 spinner의 숫자의 합이 5가 되는 사건이고, $C_2$는 첫 번째 돌렸을 때 1이 나오면 되는 사건이고, $C_3$는 두 번째 돌렸을 때 4가 나오면 되는 사건이다. 

 

표를 통해서 모든 경우의 수를 생각해보면

1st \ 2nd	1	2	3	4
1	(1, 1)	(1, 2)	(1, 3)	(1, 4)
2	(2, 1)	(2, 2)	(2, 3)	(2, 4)
3	(3, 1)	(3, 2)	(3, 3)	(3, 4)
4	(4, 1)	(4, 2)	(4, 3)	(4, 4)

 

$C_1$ > 빨간색

$C_2$ > 파란색

$C_3$ > 민트색

 

Then $P(C_i) = 1/4, i = 1, 2, 3$, and for $i \neq j$, $P(C_i \cap C_j) = 1/16$.

Thus, C_1, C_2, C_3 are pairwise independent.

 

But, $C_1 \cap C_2 \cap C_3$ is the event that (1, 4) is spun and its probability is 1/16.

$P(C_1 \cap C_2 \cap C_3) = 1 / 4 \times 1/4 \times 1/4 \neq 1 / 16$

 

이것이 의미하는 것이 두 사건이 독립이라고 해서 mutual independence가 성립한다고 보장할 수 없다를 예제를 통해서 확인할 수 있다.