Poisson관련 내용은 [[VG:푸아송_분포,Poisson_distribution]]에 적음.

[[TableOfContents]]

= Conditional probability mass function =
조건부 확률질량함수,conditional_pmf
 Conditional + [[VG:확률질량함수,probability_mass_function,PMF]]
 조건부확률질량함수,conditional_probability_mass_function,conditional_PMF
조건부 기대치,conditional_expected_value
 [[VG:기대값,expectation]]
조건부 분산,conditional_variance
 [[VG:분산,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:
 $P_X(x|C)=P(X=x|C)=\frac{P(\{X=x\}\cap C)}{P(C)}$

def.
(a) the '''conditional expected value''' of X given event C:
 $E(X|C)=m_{X|C}=\sum_{\textrm{all }k}x_kP_k(x_k|C)$
사건 C가 일어났을 때 X의 조건부 기대치.
pmf 자리에 조건부pmf가 왔음.

(b) the '''conditional variance''' of X given event C:
 $VAR(X|C)=E\left((X-m_{X|C})^2\right)=E(X^2|C)-\left(E(X|C)\right)^2$
C라는 사건이 일어났을 때 X의 조건부 분산.

= ex. =
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 ||
||P,,X,, ||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 $J_H_1$ 로.
Sol.
$P(X=2|J_H_1)=\frac{P(\{X=2\}\cap J_H_1)}{P(J_H_1)}=\frac{\frac18}{\frac12}=\frac14.$
$P(X=1|J_H_1)=\frac{\frac38}{\frac12}=\frac34$
$P(X=0|J_H_1)=\frac{0}{\frac12}=0$

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

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

이상 이산, 이후 연속

= 누적분포함수,cdf =
[[VG:누적분포함수,cumulative_distribution_function,CDF]]
 
def. For r.v. X, the CDF of X:
 $F_X(x)=P(X\le x),\quad\quad -\infty<x<\infty.$

[[이산확률변수,discrete_RV]]의 CDF
right-continuous, staircase function with jumps
$F_X(x)=\sum_{x_k\le x}p_X(x_k)=\sum_k p_X(x_k) u(x-x_k)$

[[연속확률변수,continuous_RV]]의 CDF
continuous, nonnegative function $f(x)$ 의 적분으로 쓸 수 있음
$F_X(x)=\int_{-\infty}^x f(t)dt$

Mixed R.V의 CDF..... 이게 뭐람?
$F_X(x)=pF_1(x)+(1-p)F_2(x)$

= 확률밀도함수,pdf =
'''PDF'''는 CDF의 [[미분,derivative]]으로 정의.

For [[연속확률변수,continuous_RV]]:
 $f_X(x)=\frac{dF_X(x)}{dx}$

For [[이산확률변수,discrete_RV]]:
 $f_X(x)=\frac{d}{dx}\sum_k p_X(x_k)u(x-x_k)=\sum_k p_X(x_k)\delta(x-x_k)$

참고로 delta function $\delta(t):$
 $u(x)=\int_{t=-\infty}^x \delta(t)dt$

see [[VG:디랙_델타함수,Dirac_delta_function]]

Properties of PDF (확률밀도함수의 성질)
 $\bullet\, f_X(x)\ge 0$ (since CDF is nondecreasing)
 $\bullet\, P[a\le X\le b]=\int_a^b f_X(x)dx$
 $\bullet\, F_X(x)=\int_{-\infty}^x f_X(t)dt$
= 또 다른 정리 시도... =
||[[이산확률변수,discrete_RV]] ||[[연속확률변수,continuous_RV]] ||
||[[VG:이산확률분포,discrete_probability_distribution]] ||[[VG:연속확률분포,continuous_probability_distribution]] ||
||[[VG:확률질량함수,probability_mass_function,PMF]] ||[[VG:확률밀도함수,probability_density_function,PDF]] ||



[[누적분포함수,cumulative_distribution_function,CDF]]

----
Related:
 [[확률,probability]]
 [[VG:확률,probability]]
 [[VG:연속확률분포,continuous_probability_distribution]]
 [[VG:이산확률분포,discrete_probability_distribution]]
 [[VG:전확률정리,total_probability_theorem]]
 [[VG:조건부확률,conditional_probability]]
 [[VG:집합과_확률,set_and_probability]]
 [[VG:확률분포,probability_distribution]]
 [[VG:확률함수,probability_function]]
Twin: [[VG:확률및랜덤프로세스]]