조건부 확률질량함수,conditional_pmf
조건부 기대치,conditional_expected_value
조건부 분산,conditional_variance

from http://www.kocw.net/home/search/kemView.do?kemId=1279832 8. Conditional Probability, Independence of Events, Sequential Experiments 1:02:39

Let d.r.v.(discrete random variable) X with pmf P_X and event C with P(C)>0.
→ the conditional probability mass function of X given event C:
def.
(a) the conditional expected value of X given event C:
사건 C가 일어났을 때 X의 조건부 기대치.
pmf 자리에 조건부pmf가 왔음.

(b) the conditional variance of X given event C:
C라는 사건이 일어났을 때 X의 조건부 분산.

Let r.v. X : the maximum number of heads obtained Tom and Jane each flip a fair coin twice.

(a) Find the pmf of X.
Sol.

J＼T	0	1	2
0	0	1	2
1	1	1	2
2	2	2	2

이것은

곱	¼ⓐ	½ⓑ	¼ⓒ
¼	1/16	1/8	1/16
½	1/8	1/4	1/8
¼	1/16	1/8	1/16

ⓐ tom이 앞면의 개수가 0이 나오는 것은, 1/2 * 1/2 = 1/4
ⓑ tom이 앞면의 개수가 한번 나오는 것은, 앞뒤 뒤앞이니까 1/2
ⓒ 앞앞이니까 1/4
so, pmf:

X	0	1	2
PX	1/16	1/2	7/16

9. Cumulative Distribution Function , Probability Density Function

(b) Find the conditional pmf of X=2 given that Jane got one head in two tosses.
사건 "Jane got one head in two tosses"를 $\displaystyle J_H_1$ 로.
Sol.
$\displaystyle P(X=2|J_H_1)=\frac{P(\{X=2\}\cap J_H_1)}{P(J_H_1)}=\frac{\frac18}{\frac12}=\frac14.$
$\displaystyle P(X=1|J_H_1)=\frac{\frac38}{\frac12}=\frac34$
$\displaystyle P(X=0|J_H_1)=\frac{0}{\frac12}=0$

(c) Find $\displaystyle E(X|J_H_1)$ and $\displaystyle VAR(X|J_H_1).$

Sol. Since

X	0	1	2
$\displaystyle P(X|J_H_1)$	0	3/4	1/4

$\displaystyle E(X|J_H_1)=0*0+1*(3/4)+2*(1/4)=5/4.$
$\displaystyle VAR(X|J_H_1)=E(X^2|J_H_1)-E(X|J_H_1)^2$
$\displaystyle =(1^2\times\frac34+2^2\times\frac14)-(\frac54)^2=\frac3{16}$

이상 이산, 이후 연속

누적분포함수,cumulative_distribution_function,CDF
def. For r.v. X, the CDF of X:

Related:
Twin: 확률및랜덤프로세스