# 확률및랜덤프로세스,probability_and_random_process

Sub:
Poisson관련 내용은 푸아송_분포,Poisson_distribution에 적음.

# 2. 랜덤프로세스 ¶

random process = stochastic process

# 3. Conditional probability mass function ¶

조건부 확률질량함수,conditional_pmf
Conditional + 확률질량함수,probability_mass_function,PMF
조건부확률질량함수,conditional_probability_mass_function,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 PX and event C with P(C)>0.
→ the conditional probability mass function of X given event C:
$\displaystyle P_X(x|C)=P(X=x|C)=\frac{P(\{X=x\}\cap C)}{P(C)}$

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

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

# 4. 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 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 표기.)
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

# 5. 누적분포함수,cdf ¶

누적분포함수,cumulative_distribution_function,CDF

def. For r.v. X, the CDF of X:
(x<0인 경우는 생략)

# 8. Textbooks ¶

Probability, Random Variables and Random Signal Principles, 4th Edition
Peyton Peebles Jr

Probability and Random Processes with Application to Signal Processing
Stark and Woods

Leon-Garcia - see 확률및랜덤프로세스

이상은 학부과정, 이하는 대학원과정

An Introduction to Statistical Signal Processing
Gray and Davisson

Probability, Random Variables, and Stochasitc Processes, 4th edition
Papoulis and Pillai