hw7 110590049

tags probability

2023PB_HW7.pdf

1.a

$$\begin{gathered} 1={\int }_{-\infty }^{\infty }c{e}^{-3x}dx=c\frac{e^{-3x}}{-3}{|}_0^{\infty } \\ \underset{x\to \infty }{\mathrm{limit}}\frac{1}{-3e^{3x}}=0 \\ 1=0-c\frac{1}{-3} \\ c=3 \end{gathered}$$

1.b

$$\begin{gathered} {\int }_0^{0.5}3e^{-3x}dx=-e^{-3x}{|}_0^{0.5} \\ =-e^{-1.5}+1=0.776 \end{gathered}$$

2.a

$$g(x)=\{\begin{array}{l}0,x<4\\ 32x^{-3},x\ge 4\end{array}$$

2.b

$$\begin{gathered} {\int }_0^{4}0dx+{\int }_4^{5}32x^{-3}dx=-16x^{-2}{|}_4^5=\frac{9}{25} \\ {\int }_6^{\infty }32x^{-3}dx=-16x^{-2}{|}_6^{\infty }=\frac{4}{9} \\ {\int }_5^{7}32x^{-3}dx=-16x^{-2}{|}_5^7=0.313 \\ {\int }_1^{3.5}0dx=0 \end{gathered}$$

3

$$\begin{gathered} y=x^3 \\ p(x^3\le y)=p(x\le y^{\frac{1}{3}})={\int }_{-\infty }^{y^{\frac{1}{3}}}\frac{1}{4}dx=\frac{1}{4}x{|}_{-2}^{y^{\frac{1}{3}}}=\frac{1}{4}{y}^{\frac{1}{3}}+\frac{1}{2} \\ g(y)=\{\begin{array}{l}0,y<-8\\ \frac{1}{4}{t}^{\frac{1}{3}}+\frac{1}{2},-8\le y\le 8\\ 0,y>8\end{array} \\ {g}^{\prime }(y)=\{\begin{array}{l}0,y<-8\\ \frac{1}{12}{x}^{-\frac{2}{3}},-8\le y\le 8\\ 0,y>8\end{array} \\ z=x^4 \\ p(x^4\le z)=p(z^{\frac{1}{4}}\le x\le z^{\frac{1}{4}})={\int }_{-z^{\frac{1}{4}}}^{z^{\frac{1}{4}}}\frac{1}{4}dx=\frac{1}{4}x{|}_{-z^{\frac{1}{4}}}^{z^{\frac{1}{4}}}=\frac{1}{2}{z}^{\frac{1}{4}} \\ g(z)=\{\begin{array}{l}0,z<0\\ \frac{1}{2}{z}^{\frac{1}{4}},0\le z\le 16\\ 0,z>16\end{array} \\ {g}^{\prime }(z)=\{\begin{array}{l}0,z<0\\ \frac{1}{16}{z}^{\frac{-3}{4}},0\le z\le 16\\ 0,z>16\end{array} \end{gathered}$$

4

$$\begin{gathered} y=x^{\frac{2}{3}} \\ p(0\le y\le t^{\frac{3}{2}})={\int }_0^{t^{\frac{3}{2}}}\lambda e^{-\lambda x}dx=-e^{\lambda x}{|}_0^{t^{\frac{3}{2}}}=1-e^{\lambda t^{\frac{3}{2}}} \\ g(t)=\{\begin{array}{l}0,t<0\\ 1-e^{-\lambda t^{\frac{3}{2}}},0\le t<\infty \end{array} \\ {g}^{\prime }(t)=\{\begin{array}{l}0,t<0\\ \frac{3}{2}\lambda t^{\frac{1}{2}}{e}^{-\lambda t^{\frac{3}{2}}},0\le t<\infty \end{array} \end{gathered}$$

5

$$\begin{gathered} E(e^x)={\int }_{-\infty }^{\infty }{e}^x\times 3e^{-3x}dx={\int }_0^{\infty }3e^{-2x}dx=\frac{-3}{2}{e}^{-2x}{|}_0^{\infty } \\ \underset{x\to \infty }{\lim }\frac{-3}{2}{e}^{-2x}=0 \\ E(e^x)=0+\frac{3}{2}=\frac{3}{2} \end{gathered}$$

6

$$\begin{gathered} E(x)={\int }_{-\infty }^{\infty }x\times \frac{1}{2}{e}^{-|x|}dx \\ ={\int }_0^{\infty }\frac{1}{2}x{e}^{-x}dx+{\int }_{-\infty }^0\frac{1}{2}x{e}^{x}dx \\ =-\frac{1}{2}(x+1)e^{-x}{|}_0^{\infty }+\frac{1}{2}(x-1)e^x{|}_{-\infty }^0 \\ =0+\frac{1}{2}-\frac{1}{2}+0=0 \\ E(x^2)={\int }_{-\infty }^{\infty }{x}^2\times \frac{1}{2}{e}^{-|x|}dx \\ =2{\int }_0^{\infty }\frac{1}{2}{x}^2{e}^{-x}dx=2 \\ Var(x)=E(x^2)-E(x{)}^2=2-0^2=2 \end{gathered}$$