hw9 110590049

tags probability

2023PB_HW9.pdf

1.a

$$\begin{gathered} \sum _{x\in 1,2,3}\sum _{y\in 1,2}P(x,y)=1 \\ \sum _{x\in 1,2,3}\sum _{y\in 1,2}c(x+y)=1 \\ 21c=1 \\ c=\frac{1}{21} \end{gathered}$$

1.b

$$\begin{gathered} P_x(x)=\frac{1}{21}(2x+3) \\ {P}_y(y)=\frac{1}{21}(3y+6) \end{gathered}$$

1.c

$$\begin{gathered} P(x\ge 2|y=1)=\frac{P(x\ge 2\cap y=1)}{P_y(y=1)}= \\ \frac{\frac{1}{3}}{\frac{3}{7}}=\frac{7}{9} \end{gathered}$$

1.d

$$\begin{gathered} E(x)=\sum _{x\in 1,2,3}x\frac{1}{21}(2x+3)=2.1904 \\ E(y)=\sum _{2\in 1,2}y\frac{1}{21}(3y+6)=1.5714 \end{gathered}$$

2

$$\begin{gathered} P(x,y)=\begin{array}{ccccccc}x\setminus y& 1& 1& 2& 3& 4& 5\\ 2& \frac{1}{36}& 0& 0& 0& 0& 0\\ 3& 0& \frac{2}{36}& 0& 0& 0& 0\\ 4& \frac{1}{36}& 0& \frac{2}{36}& 0& 0& 0\\ 5& 0& \frac{2}{36}& 0& \frac{2}{36}& 0& 0\\ 6& \frac{1}{36}& 0& \frac{2}{36}& 0& \frac{2}{36}& 0\\ 7& 0& \frac{2}{36}& 0& \frac{2}{36}& 0& \frac{2}{36}\\ 8& \frac{1}{36}& 0& \frac{2}{36}& 0& \frac{2}{36}& 0\\ 9& 0& \frac{2}{36}& 0& \frac{2}{36}& 0& 0\\ 10& \frac{1}{36}& 0& \frac{2}{36}& 0& 0& 0\\ 11& 0& \frac{2}{36}& 0& 0& 0& 0\\ 12& \frac{1}{36}& 0& 0& 0& 0& 0\end{array} \\ {p}_x(x)=\frac{1}{36}(6-|7-x|) \\ {p}_y(y)=\{\begin{array}{ll}\frac{6}{36}& y=0\\ \frac{2}{36}(6-y)& 0<y\le 5\\ 0& else\end{array} \end{gathered}$$

3.a

$$\begin{gathered} f_x(x)={\int }_0^{x}2dy=2y{|}_0^x=2x \\ {f}_y(y)={\int }_y^{1}2dx=2x{|}_y^1=2-2y \end{gathered}$$

3.b

$$\begin{gathered} E(X)={\int }_0^1{\int }_0^{1}x\times f_x(x)dydx=\frac{2}{3} \\ E(Y)={\int }_0^1{\int }_0^{1}y\times f_y(y)dydx=\frac{1}{3} \end{gathered}$$

3.c

$$\begin{gathered} P(x<\frac{1}{2})={\int }_0^{\frac{1}{2}}2xdx=\frac{1}{4} \\ P(x<2y)={\int }_0^1{\int }_{\frac{v}{2}}^{v}2dudv=\frac{1}{2} \\ P(x=y)={\int }_0^1{\int }_v^{v}2dudv=0 \end{gathered}$$

3.d

$$\begin{gathered} f_{xy}\ne f_x(x)f_y(y) \\ \text{not independent} \end{gathered}$$

3.e

$$f_{x|y}(x|y)=\frac{f(x,y)}{f_y(y)}=\frac{2}{2-2y}$$

4

$$\begin{gathered} p_x(x)=\{\begin{array}{ll}\frac{3}{7}& x=1\\ \frac{4}{7}& x=2\\ 0& else\end{array} \\ {p}_y(y)=\{\begin{array}{ll}\frac{5}{7}& y=1\\ \frac{2}{7}& y=2\\ 0& else\end{array} \\ {p}_{xy}(1,1)=\frac{1}{7} \\ {p}_x(1)p_y(1)=\frac{15}{49} \\ p(x,y)\ne p_x(x)p_y(y) \\ \text{not independent} \end{gathered}$$

5

$$E(x^{2}y)={\int }_0^{\infty }{\int }_0^{\infty }{x}^{2}y\times 2e^{-(x+2y)}dxdy=1$$

6

$$\begin{gathered} P_{x|y}(x|y)=\frac{f(x,y)}{f_y(y)} \\ {P}_y(y)=\{\begin{array}{ll}\frac{5}{25}& y=0\\ \frac{7}{25}& y=1\\ \frac{13}{25}& y=2\\ 0& else\end{array} \\ P(x=2|y=1)=\frac{f_{x,y}(x=2,y=1)}{p_y(y=1)}=\frac{\frac{5}{25}}{\frac{7}{25}}=\frac{5}{7} \\ E(x|y=1)=\sum _{x}x\times f_{x,y}(x,y=1)=\frac{12}{25} \end{gathered}$$

7.a

$$\begin{gathered} {\int }_0^1{\int }_0^{1-x}\lambda dydx=1 \\ \lambda =2 \end{gathered}$$

7.b

$$\begin{gathered} f_{x|y}(x|y)=\frac{f(x,y)}{f_y(y)} \\ {f}_y(y)={\int }_0^{1-y}2dx=2-2y \\ {f}_{x|y}(x|y)=\frac{1}{1-y} \end{gathered}$$

7.c

$$\begin{gathered} E(X|Y=y)={\int }_0^{1-y}x\times f_{x|y}(x|y)dx= \\ {\int }_0^{1-y}x\times \frac{1}{1-y}dx=\frac{x^2}{2(1-y)}{|}_0^{1-y}=\frac{1-y}{2} \end{gathered}$$