Sub:

Poisson관련 내용은

푸아송_분포,Poisson_distribution에 적음.

1. 확률
2. 랜덤프로세스
3. Conditional probability mass function
4. ex.
5. 누적분포함수,cdf
6. 확률밀도함수,pdf
7. 또 다른 정리 시도...
8. Textbooks

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2. 랜덤프로세스 ¶

random process = stochastic process

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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

기대값,expected_value

조건부 분산,conditional_variance

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.

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

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

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5. 누적분포함수,cdf ¶

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_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?
$F_X(x)=pF_1(x)+(1-p)F_2(x)$

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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

$f_X(x)=\begin{cases}1/(b-a),&a\le x\le b\\0,&{\rm otherwise}\end{cases}$

$\Rightarrow$

$F_X(x)=\begin{cases}0&x<a\\(x-a)/(b-a)&a\le x \le b\\1&x>b\end{cases}$

example: 지수확률함수,exponential_RV
The transmission time X of messages in a communication system has an

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)

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7. 또 다른 정리 시도... ¶

이산확률변수,discrete_RV	연속확률변수,continuous_RV
이산확률분포,discrete_probability_distribution	연속확률분포,continuous_probability_distribution
확률질량함수,probability_mass_function,PMF	확률밀도함수,probability_density_function,PDF