Expectation, Variance example

d4r6j·2023년 11월 23일

Reference

Random Variable $X, Y$ 의 joint probability density function
$p_{XY}(x, y) = \left\{ \begin{array}{cc} \begin{aligned} 8xy, \quad\quad &0 < x \leq 1, \; 0 \leq y \leq x \\ \\ 0, \quad\quad &{\rm other} \end{aligned} \end{array} \right.$
일 때,
Area of joint probability density function.

\mathbb{E}[X] = \int^{\infty}_{\infty}xp_X(x)dx

를 적용하면

\begin{aligned} \mathbb{E}[X] &= \int^{\infty}_{-\infty}\int^{\infty}_{-\infty} xp_{XY}(x, y)dydx \\ &= \int^1_0 \int^{x}_0 x(8xy)dydx \\ &= \int^1_0[4x^2y^2]^{x}_{0}dx \\ &= \int^1_0 4x^4dx = \frac{4}{5} \end{aligned}

\begin{aligned} \mathbb{E}[Y] &= \int^{\infty}_{\infty}\int^{\infty}_{\infty}yp_{XY}(x,y)dydx \\ &= \int^1_0\int^x_0 y(8xy)dydx \\ &= \int^1_0\left[ \frac{8}{3}xy^3 \right]^x_0dx \\ &= \int^1_0\frac{8}{3}x^4dx = \frac{8}{15} \end{aligned}

\begin{aligned} \mathbb{E}[X^2] &= \int^{\infty}_{-\infty}\int^{\infty}_{-\infty} x^2p_{XY}(x, y)dydx \\ &= \int^{\infty}_{-\infty}\int^{\infty}_{-\infty}x^2(8xy)dydx \\ &= \int^{1}_{0}[4x^3y^2]^{x}_{0}dx \\ &= \int^{1}_{0}4x^5dx = \frac{2}{3} \end{aligned}

\begin{aligned} Var(X) &= \mathbb{E}[(X-\mathbb{E}[X])^2] \\ &= \int^{\infty}_{-\infty}(x-\mathbb{E}[X])^2p_{X}(x)dx \\ &= \int^{\infty}_{-\infty}(x^2 - 2x\mathbb{E}[X] + \mathbb{E}[X]^2)p_{X}(x)dx \\ &= \int^{\infty}_{-\infty}x^2p_{X}(x)dx - \int^{\infty}_{\infty}2x\mathbb{E}[X]p_{X}(x)dx + \int^{\infty}_{-\infty}\mathbb{E}[X]^2p_{X}(x)dx \\ &= \mathbb{E}[X^2] - 2\mathbb{E}[X]\mathbb{E}[X] + \mathbb{E}[X]^2 \cdot 1 \\ &= \mathbb{E}[X^2] - (\mathbb{E}[X])^2 \\ &= \frac{2}{3} - \left( \frac{4}{5} \right)^2 = \frac{2}{75} \end{aligned}

\begin{aligned} Cor(X,Y) &= \mathbb{E}[XY] \\ &= \int^{\infty}_{-\infty}\int^{\infty}_{-\infty}xyp_{XY}(x, y)dydx \\ &= \int^1_0\int^{x}_0xy(8xy)dydx \\ &= \int^{1}_{0}\left[\frac{8}{3}x^2y^3\right]^{x}_{0}dx \\ &= \int^{1}_{0}\frac{8}{3}x^5dx = \frac{4}{9} \end{aligned}

$X$ 의 standard deviation $\sigma_X = \sqrt{Var(X)}$

\begin{aligned} Cov(X,Y) &= \mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] \\ &= \int^{\infty}_{-\infty} \int^{\infty}_{-\infty}(x-\mathbb{E}[X])(y-\mathbb{E}[Y])p_{XY}(x,y)dydx \\ &= \int^{\infty}_{-\infty}\int^{\infty}_{-\infty}(xy-\mathbb{E}[X]y - x\mathbb{E}[Y] + \mathbb{E}[X]\mathbb{E}[Y])p_{XY}(x,y)dydx \\ &= \int^{\infty}_{-\infty}\int^{\infty}_{-\infty}xyp_{XY}(x,y)dydx - \mathbb{E}[X]\int^{\infty}_{-\infty}yp_{Y}(y)dy - \mathbb{E}[Y]\int^{\infty}_{-\infty}xp_{X}(x)dx + \mathbb{E}[X]\mathbb{E}[Y] \\ &= \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y] \\ &= \frac{4}{9} - \frac{4}{5}\left(\frac{8}{15}\right) = \frac{4}{225} \end{aligned}

$Y$ of conditional probability density function
$p_{X}(x) = \int^{x}_{0}8xydy =8x\int^{x}_{0}ydy = 8x\left[\frac{1}{2}y^2 \right]^{x}_{0} = 4x^3$ $p_{Y|X}(y|x) = \frac{p_{XY}(x,y)}{p_{X}(x)} = \left\{ \begin{array}{cc} \begin{aligned} \frac{2y}{x^2}, \quad\quad &0 \leq y \leq x, \;\; &0 < x \leq 1 \\ \\ 0, \quad\quad &{\rm other} \end{aligned} \end{array} \right.$
$X$ 의 conditional expectation when given $Y=y$ ,
$\mathbb{E}[X|Y=y] = \int^{\infty}_{\infty}xp_{x|y}(x|y)dx$
Random Variable $X=x$ 인 $Y$ 의 conditional expectation.
$\begin{aligned} \mathbb{E}[Y|X=x] &= \int^{\infin}_{-\infin}yp_{Y|X}(y|x)dy \\ &= \int^{x}_{0}y\frac{2y}{x^2}dy =\left[ \frac{2}{3}x^3\right]^x_0 \\ &= \frac{2}{3}x, \quad 0<x \leq 1 \end{aligned}$
$\mathbb{E}[Y|X=x]$ 는 실수 $x$ 의 함수로서, 실수 함수
Reference (Continuous random variable)
https://online.stat.psu.edu/stat414/lesson/20/20.2