확률_및_랜덤_프로세스

Difference between r1.6 and the current

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Poisson관련 내용은 [[VG:푸아송_분포,Poisson_distribution]]에 적음.
 
= 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 $J_H_1$ .
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:
$F_X(x)=P(X\le x),\quad\quad -\infty<x<\infty.$
 
----
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:확률밀도함수,probability_density_function,PDF]]
[[VG:확률분포,probability_distribution]]
[[VG:확률함수,probability_function]]
Twin: [[VG:확률및랜덤프로세스]]
Moved to [[확률및랜덤프세스,probability_and_random_process]]