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.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
누적분포함수, cdf ¶
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
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