확률및랜덤프로세스,probability_and_random_process

Difference between r1.7 and the current

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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.
(사건 "Jane got one head in two tosses"를 $J_{H_1}$ 로 표기.)
Sol.
$P(X=2|J_H_1)=\frac{P(\{X=2\}\cap J_H_1)}{P(J_H_1)}=\frac{\frac18}{\frac12}=\frac14.$
$P(X=1|J_H_1)=\frac{\frac38}{\frac12}=\frac34$
$P(X=0|J_H_1)=\frac{0}{\frac12}=0$
$P(X=2|J_{H_1})=\frac{P(\{X=2\}\cap J_{H_1})}{P(J_{H_1})}=\frac{\frac18}{\frac12}=\frac14.$
$P(X=1|J_{H_1})=\frac{\frac38}{\frac12}=\frac34$
$P(X=0|J_{H_1})=\frac{0}{\frac12}=0$

(c) Find $E(X|J_H_1)$ and $VAR(X|J_H_1).$
(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 ||
$E(X|J_H_1)=0*0+1*(3/4)+2*(1/4)=5/4.$
$VAR(X|J_H_1)=E(X^2|J_H_1)-E(X|J_H_1)^2$
||$P(X|J_{H_1})$ ||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





2. 랜덤프로세스

random process = stochastic process


3. 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
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
¼ⓐ½ⓑ¼ⓒ
¼1/16 1/8 1/16
½1/8 1/4 1/8
¼1/16 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

VG:누적분포함수,cumulative_distribution_function,CDF

def. For r.v. X, the CDF of X:
$\displaystyle F_X(x)=P(X\le x),\quad\quad -\infty

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

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

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

6. 확률밀도함수,pdf

PDF는 CDF의 미분,derivative으로 정의.

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

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

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

see VG:디랙_델타함수,Dirac_delta_function

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

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

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


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

Leon-Garcia - see VG: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