早稲田大学 創造理工学研究科 経営システム工学専攻 2023年8月実施 概率统计

标签：

• Waseda-University
• Joint-Distribution
• Independence-of-Random-Variables
• Cumulative-Distribution-Function-and-Probability-Density-Function

Author

思齐塾, 祭音Myyura

Description

確率変数(random variable) $x$ と $y$ の同時確率密度関数 (joint probability density function)が次式であるとき、 $x$ と $y$ が独立(independent)になるように定数 $a$ と $b$ の値を定めよ。ただし, $0<x<1$ , $0 <y<1$ とする。

$$f(x, y) = a(2xy + x + by + 3)$$

题目描述

随机变量 $x,y$ 的联合概率密度函数在

$$0<x<1,\qquad0<y<1$$

上为

$$f(x,y)=a(2xy+x+by+3).$$

确定常数 $a,b$，使 $x$ 与 $y$ 相互独立。

Kai

Since $x$ and $y$ are independent, we can write $f(x, y) = g(x)h(y)$ , where $g(x)$ is a function of $x$ only and $h(y)$ is a function of $y$ only. Thus, we need to factor $2xy + x + by + 3$ into a product of a function of $x$ and a function of $y$ .

$2xy + x + by + 3 = x(2y + 1) + by + 3$

If $x$ and $y$ are independent, then $f(x, y)$ must be separable as $g(x)h(y)$ for some functions $g$ and $h$ . $f(x, y) = a(2xy + x + by + 3) = a(x(2y+1) + by + 3)$ $= a(2xy+x+by+3) = a(2x+\frac{x}{y} +b + \frac{3}{y})y$ Let $f_x(x) = \int_0^1 f(x, y) dy$ and $f_y(y) = \int_0^1 f(x, y) dx$ .

For $x$ and $y$ to be independent, we must have $f(x, y) = f_x(x) f_y(y)$ .

For $f(x,y)$ to be the product of $g(x)$ and $h(y)$ , we must have

$$2xy + x + by + 3 = (Ax + B)(Cy + D) = ACxy + ADx + BCy + BD$$

Comparing the coefficients: $AC = 2$ , $AD = 1$ , $BC = b$ , $BD = 3$

Then $\frac{AD}{BD} = \frac{1}{3} = \frac{A}{B}$ and $\frac{AC}{BC} = \frac{2}{b} = \frac{A}{B}$ . Therefore, $\frac{1}{3} = \frac{2}{b} \implies b = 6$ .

So $2xy+x+6y+3=(x+3)(2y+1)$ . Then $2xy+x+by+3 = (Ax+B)(Cy+D)$ , $AC=2$ , $AD=1$ , $BC=6$ , $BD=3$ If $D=1$ , then $B=3$ , $A=1$ , and $C=2$ . $f(x, y) = a(x+3)(2y+1)$ , $\int_0^1 \int_0^1 f(x, y) dxdy = 1$ $\int_0^1 \int_0^1 a(x+3)(2y+1) dxdy = a \int_0^1 (x+3)dx \int_0^1 (2y+1) dy$ $= a [\frac{x^2}{2} + 3x]_0^1 [y^2+y]_0^1 = a(\frac{1}{2} + 3)(1+1) = a(\frac{7}{2})(2) = 7a = 1$ So $a = \frac{1}{7}$ . Therefore, $a = \frac{1}{7}$ and $b = 6$ .