[5분 고등수학] 확률변수 aX+b의 평균, 분산, 표준편차

 

 

확률변수 aX+b에 대한 세가지 등식을 유도해보겠습니다. 

세가지 등식은 아래와 같습니다. 

1) $ E\left(aX+b\right)=aE\left(X\right)+b$
2) $ V\left(aX+b\right)={a}^2V\left(X\right)$
3) $ \sigma \left(aX+b\right)=\left|{\sigma }\right|V\left(X\right)$

 

하나씩 유도해봅시다. 

 

1) $ E\left(aX+b\right)=aE\left(X\right)+b$

유도하기 위해서 아래와 같은 표를 그리겠습니다. 

 

X	$x_{1}$	...	$x_{n}$	합계
P(X=x)	$p_{1}$	...	$p_{n}$	1

확률변수 X의 기댓값은 아래와 같이 계산됩니다. 

$E\left(X\right)=\sum _{i=1}^n{x}_i{p}_i$

확률변수 aX+b의 기댓값은 아래와 같이 계산됩니다. 

$E\left(aX+b\right)=\sum _{i=1}^n\left(a{x}_i+b\right){p}_i$

전개해봅시다. 

$E\left(aX+b\right)=\sum _{i=1}^n\left(a{x}_i{p}_i+b{p}_i\right)$

분리해서 씁시다. 

$E\left(aX+b\right)=\sum _{i=1}^na{x}_i{p}_i+\sum _{i=1}^nb{p}_i$

a와 b는 상수이므로 앞으로 꺼냅시다. 

$E\left(aX+b\right)=a\sum _{i=1}^n{x}_i{p}_i+b\sum _{i=1}^n{p}_i$

모든 확률의 합은 1이므로 아래와 같습니다. 

$E\left(aX+b\right)=a\textcolor{#ff0010}{\sum _{i=1}^n{x}_i{p}_i}+b$

위 식의 빨강부분은 E(X)이므로 아래와 같습니다. 

$E\left(aX+b\right)=a\textcolor{#ff0010}{E\left(X\right)}+b$E

이렇게 첫번째 등식을 유도했습니다. 

 

 

2) $ V\left(aX+b\right)={a}^2V\left(X\right)$

확률변수 X의 분산은 아래와 같이 계산됩니다. 

$V\left(X\right)=\sum _{i=1}^n{x}_i^2{p}_i-{\left(\sum _{i=1}^n{x}_i{p}_i\right)}^2$

확률변수 aX+b의 기댓값은 아래와 같이 계산됩니다. 

$V\left(aX+b\right)=\sum _{i=1}^n{\left(a{x}_i+b\right)}^2{p}_i-{\left(\sum _{i=1}^n\left(a{x}_i+b\right){p}_i\right)}^2$

첫항의 제곱을 전개하겠습니다. 

$V\left(aX+b\right)=\sum _{i=1}^n\left({a}^2{{x}_i^2}+2ab{x}_i+{b}^2\right){p}_i-{\left(\sum _{i=1}^na{x}_i{p}_i+\sum _{i=1}^nb{p}_i\right)}^2$

첫항은 시그마를 나눠쓰고, 두번째항은 전개합시다. 

$V\left(aX+b\right)=\sum _{i=1}^n{a}^2{{x}_i^2}{p}_i+\sum _{i=1}^n2ab{x}_i{p}_i+\sum _{i=1}^n{b}^2{p}_i$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -{\left(\sum _{i=1}^na{x}_i{p}_i\right)}^2-{\left(\sum _{i=1}^nb{p}_i\right)}^2-2\sum _{i=1}^na{x}_i{p}_i\sum _{i=1}^nb{p}_i$ 

상수들은 밖으로 꺼냅시다. 

$V\left(aX+b\right)={a}^2\sum _{i=1}^n{{x}_i^2}{p}_i+2ab\sum _{i=1}^n{x}_i{p}_i+{b}^2\textcolor{#ff0010}{\sum _{i=1}^n{p}_i}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -{a}^2{\left(\sum _{i=1}^n{x}_i{p}_i\right)}^2-{b}^2{\left(\textcolor{#ff0010}{\sum _{i=1}^n{p}_i}\right)}^2-2ab\sum _{i=1}^n{x}_i{p}_i\textcolor{#ff0010}{\sum _{i=1}^n{p}_i}$ 

위 식에 빨간 항들은 모든 확률의 합이므로 1입니다. 

$V\left(aX+b\right)={a}^2\sum _{i=1}^n{{x}_i^2}{p}_i+\textcolor{#00b3f2}{2ab\sum _{i=1}^n{x}_i{p}_i}+{b}^2$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -{a}^2{\left(\sum _{i=1}^n{x}_i{p}_i\right)}^2-{b}^2\textcolor{#00b3f2}{-2ab\sum _{i=1}^n{x}_i{p}_i}$

파란 항끼리 소거되고, b제곱끼리도 소거됩니다. 

$V\left(aX+b\right)={a}^2\sum _{i=1}^n{{x}_i^2}{p}_i-{a}^2{\left(\sum _{i=1}^n{x}_i{p}_i\right)}^2$

a제곱으로 묶겠습니다. 

$V\left(aX+b\right)={a}^2\left(\textcolor{#ff0010}{\sum _{i=1}^n{{x}_i^2}{p}_i-{\left(\sum _{i=1}^n{x}_i{p}_i\right)}^2}\right)$

오른쪽 항의 빨간부분은 V(X)이므로, 아래 등식이 성립함을 보일 수 있습니다. 

$V\left(aX+b\right)={a}^2V\left(X\right)$​

 

3) $ \sigma \left(aX+b\right)=\left|{\sigma }\right|V\left(X\right)$

아래 식의 양변에 루트를 씌웁시다. 

$V\left(aX+b\right)={a}^2V\left(X\right)$

루트를 씌우면 분산은 표준편차가 됩니다. 

$\sigma \left(aX+b\right)=\sqrt{{a}^2}\sigma \left(X\right)$

아래 등식이 성립하므로, 절댓값으로 변형할 수 있습니다. 

$\sqrt{{a}^2}=\left|{a}\right|$

$\sigma \left(aX+b\right)=\left|{a}\right|\sigma \left(X\right)$