# 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: $P_X(x|C)=P(X=x|C)=\frac{P(\{X=x\}\cap C)}{P(C)}$

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.

# 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 $J_H_1$ 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

# 5. 누적분포함수,cdf ¶ 누적분포함수,cumulative_distribution_function,CDF

def. For r.v. X, the CDF of X:
[Image_Link]/wiki/cgi-bin/mimetex.cgi?\fs7 F_X(x)=P(X\le x),\quad\quad -\infty

이산확률변수,discrete_RV의 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_RV의 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? $F_X(x)=pF_1(x)+(1-p)F_2(x)$

# 6. 확률밀도함수,pdf ¶

PDF는 CDF의 미분,derivative으로 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 디랙_델타함수,Dirac_delta_function

Properties of PDF (확률밀도함수의 성질) $\bullet\, f_X(x)\ge 0$ (since CDF is nondecreasing) $\bullet\, P[a\le X\le b]=\int_a^b f_X(x)dx$ $\bullet\, F_X(x)=\int_{-\infty}^x f_X(t)dt$ $\bullet\, \int_{-\infty}^{+\infty}f_X(t)dt=1$

A valid pdf can be formed by any nonnegative, piecewise continuous function $g(x)$ that has a finite integral $\int_{-\infty}^{+\infty} g(x)dx=c<\infty \Rightarrow f_X(x)=g(x)/c$

example: 균등확률변수 uniform r.v.
The pdf of the uniform r.v. is given by $f_X(x)=\begin{cases}1/(b-a),&a\le x\le b\\0,&{\rm otherwise}\end{cases}$ $\Rightarrow$ $F_X(x)=\begin{cases}0&xb\end{cases}$

example: 지수확률함수,exponential_RV
The transmission time X of messages in a communication system has an 지수분포,exponential_distribution: $P[X>x]=e^{-\lambda x},\;x>0$
이것의 pdf를 구하기 $F_X(x)=P[X\le x]=1-P[X>x]=1-e^{-\lambda x}$ $f_X(x)=\frac{d}{dx}F_X(x)=\frac{d}{dx}(1-e^{-\lambda x})=\lambda e^{-\lambda x}$
(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

이상은 학부과정, 이하는 대학원과정

An Introduction to Statistical Signal Processing
Gray and Davisson

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