例 3.3.6 (例 3.3.1 续)

设(X, Y)的联合分布密度函数同

Transclude of 例-3.3.1-(二维随机向量的联合分布与边缘分布)#eq-def

1. 试求 $f_{X\mid Y}\left(x\mid y=\frac{1}{2}\right)$.
2. 试求 $f_{Y\mid X}\left(y\mid x=\frac{1}{3}\right)$.

解

(1) 由于

$$f_{X,Y}\left(x,y\right)=\{\begin{array}{ll}6xy,& x^2\le y,0\le x\le 1,\\ 0,& \text{ 其他 }\end{array}$$

和

$$f_Y\left(y\right)=3y^2,0\le y\le 1.$$

所以由 (3.3.4),对 $0\le x\le \sqrt{\frac{1}{2}}$ 有

$$f_{X\mid Y}\left(x\mid y=\frac{1}{2}\right)=\frac{f_{X,Y}\left(x,\frac{1}{2}\right)}{f_Y\left(\frac{1}{2}\right)}=\frac{6x\cdot \frac{1}{2}}{3\cdot {\left(\frac{1}{2}\right)}^2}=4x,$$

于是

$$f_{X\mid Y}\left(x\mid y=\frac{1}{2}\right)=\{\begin{array}{ll}4x,& x\le \frac{\sqrt{2}}{2},\\ 0,& \text{ 其他. }\end{array}$$

(2) 由于

$$f_X\left(x\right)=3x\left(1-x^4\right),0\le x\le 1.$$

所以由 (3.3.5),对 ${\left(\frac{1}{3}\right)}^2\le y\le 1$ 有

$$f_{Y\mid X}\left(y\mid x=\frac{1}{3}\right)=\frac{f_{X,Y}\left(\frac{1}{3},y\right)}{f_X\left(\frac{1}{3}\right)}=\frac{6y\cdot \frac{1}{3}}{3\cdot \left(\frac{1}{3}\right)\left(1-{\left(\frac{1}{3}\right)}^4\right)}=\frac{81}{40}y,$$

于是

$$f_{Y\mid X}\left(y\mid x=\frac{1}{3}\right)=\{\begin{array}{ll}\frac{81}{40}y,& \frac{1}{9}\le y\le 1,\\ 0,& \text{ 其他. }\end{array}$$