1. Conditional probability mass function ¶
조건부 확률질량함수,conditional_pmf
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
Conditional + 확률질량함수,probability_mass_function,PMF
조건부확률질량함수,conditional_probability_mass_function,conditional_PMF
조건부 기대치,conditional_expected_value조건부확률질량함수,conditional_probability_mass_function,conditional_PMF
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:
(a) the conditional expected value of X given event C:
pmf 자리에 조건부pmf가 왔음.
→ 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.2. 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.
이것은
ⓐ 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:
9. Cumulative Distribution Function , Probability Density Function
Sol.
J\T | 0 | 1 | 2 |
0 | 0 | 1 | 2 |
1 | 1 | 1 | 2 |
2 | 2 | 2 | 2 |
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 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.
Sol.
$\displaystyle P(X=2|J_H_1)=\frac{P(\{X=2\}\cap J_H_1)}{P(J_H_1)}=\frac{\frac18}{\frac12}=\frac14.$
$\displaystyle P(X=1|J_H_1)=\frac{\frac38}{\frac12}=\frac34$
$\displaystyle P(X=0|J_H_1)=\frac{0}{\frac12}=0$
사건 "Jane got one head in two tosses"를 $\displaystyle J_H_1$ 로.
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
$\displaystyle E(X|J_H_1)=0*0+1*(3/4)+2*(1/4)=5/4.$
$\displaystyle VAR(X|J_H_1)=E(X^2|J_H_1)-E(X|J_H_1)^2$
$\displaystyle =(1^2\times\frac34+2^2\times\frac14)-(\frac54)^2=\frac3{16}$
X | 0 | 1 | 2 |
$\displaystyle P(X|J_H_1)$ | 0 | 3/4 | 1/4 |
$\displaystyle VAR(X|J_H_1)=E(X^2|J_H_1)-E(X|J_H_1)^2$
$\displaystyle =(1^2\times\frac34+2^2\times\frac14)-(\frac54)^2=\frac3{16}$
이상 이산, 이후 연속
3. 누적분포함수,cdf ¶
누적분포함수,cumulative_distribution_function,CDF
def. For r.v. X, the CDF of X:
right-continuous, staircase function with jumps
$\displaystyle F_X(x)=\sum_{x_k\le x}p_X(x_k)=\sum_k p_X(x_k) u(x-x_k)$
$\displaystyle F_X(x)=P(X\le x),\quad\quad -\infty
이산확률변수,discrete_RV의 CDFright-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
$\displaystyle F_X(x)=\int_{-\infty}^x f(t)dt$
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?
$\displaystyle F_X(x)=pF_1(x)+(1-p)F_2(x)$
$\displaystyle F_X(x)=pF_1(x)+(1-p)F_2(x)$
4. 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):$$\displaystyle u(x)=\int_{t=-\infty}^x \delta(t)dt$
see 디랙_델타함수,Dirac_delta_functionProperties of PDF (확률밀도함수의 성질)
The pdf of the uniform r.v. is given by
The transmission time X of messages in a communication system has an 지수분포,exponential_distribution:
$\displaystyle \bullet\, f_X(x)\ge 0$ (since CDF is nondecreasing)
$\displaystyle \bullet\, P[a\le X\le b]=\int_a^b f_X(x)dx$
$\displaystyle \bullet\, F_X(x)=\int_{-\infty}^x f_X(t)dt$
$\displaystyle \bullet\, \int_{-\infty}^{+\infty}f_X(t)dt=1$
A valid pdf can be formed by any nonnegative, piecewise continuous function $\displaystyle g(x)$ that has a finite integral$\displaystyle \bullet\, P[a\le X\le b]=\int_a^b f_X(x)dx$
$\displaystyle \bullet\, F_X(x)=\int_{-\infty}^x f_X(t)dt$
$\displaystyle \bullet\, \int_{-\infty}^{+\infty}f_X(t)dt=1$
$\displaystyle \int_{-\infty}^{+\infty} g(x)dx=c<\infty \Rightarrow f_X(x)=g(x)/c$
example: 균등확률변수 uniform r.v.The pdf of the uniform r.v. is given by
$\displaystyle f_X(x)=\begin{cases}1/(b-a),&a\le x\le b\\0,&{\rm otherwise}\end{cases}$
$\displaystyle \Rightarrow$$\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_RVThe transmission time X of messages in a communication system has an 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)$\displaystyle f_X(x)=\frac{d}{dx}F_X(x)=\frac{d}{dx}(1-e^{-\lambda x})=\lambda e^{-\lambda x}$