Q. 6.3 - A First Course in Probability

Question

The joint density of X and Y is given by

$f(x,y)=C(y-x)e^{-y}-y<x<y,0<y<\infty$

 (a) Find C.

 (b) Find the density function of X.

 (c) Find the density function of Y.

 (d) Find E[X]. 

(e) Find E[Y]. 

Step-by-Step Solution

Answer

(a) The value of C is $\frac{1}{4}.$

(b) The density function of X is $\frac{e^{-x}}{4}$.

(c) The density function of Y is $\frac{1}{2}{y}^2{e}^{-y}$.

(d) The value of $E\left[X\right] is -1$.

(e) The value of $E\left[Y\right] is 3$.

Step 1: Given information (part a)

The function is $f(x,y)=C(y-x)e^{-y}-y<x<y,0<y<\infty$

Step 2: Explanation (part a)

The joint density of X and Y is,

$f(x,y)=\left\{\begin{array}{cc}c(y-x)e^{-y}-y<x<y,0<y<\infty & \\ 0& \text{ Otherwise }\end{array}\right.$

The value of C is,

$C{\int }_{y=0}^{\infty }{e}^{-y}\left({\int }_{x=-y}^y(y-x)dx\right)dy=1$

$C{\int }_{y=0}^{\infty }{e}^{-y}{\left(xy-\frac{x^2}{2}\right)}_{x=-y}^{y}dy=1$

$C{\int }_{y=0}^{\infty }{e}^{-y}\left(y(y-(-y\left)\right)-\frac{1}{2}\left(y^2-(-y{)}^2\right)\right)dy=1$

$C{\int }_{y=0}^{\infty }{e}^{-y}\left(y(y+y)-\frac{1}{2}\left(y^2-y^2\right)\right)dy=1$

$C{\int }_{y=0}^{\infty }{e}^{-y}\left(2y^2\right)dy=1$

$2C{\int }_{y=0}^{\infty }{y}^2{e}^{-y}dy=1$

$$\begin{gathered} 2C\left[2\right]=1 \\ 4C=1 \\ C=\frac{1}{4} \end{gathered}$$

Step 3: Given information (part b)

The function is  $f(x,y)=C(y-x)e^{-y}-y<x<y,0<y<\infty$.

Step 4: Explanation (part b)

The density function of X is,

$$\begin{gathered} f_x\left(x\right)=\frac{1}{4}{\int }_x^{\infty }(y-x)e^{-y}dy \\ {f}_x\left(x\right)=\frac{1}{4}{\int }_x^{\infty }\left(y{e}^{-y}-x{e}^{-y}\right)dy \\ =\frac{1}{4}\left\{{\int }_x^{\infty }\left(y{e}^{-y}\right)dy-{\int }_x^{\infty }\left(x{e}^{-y}\right)dy\right\} \\ =\frac{1}{4}\left\{{\left[-y{e}^{-y}-e^{-y}\right]}_x^{\infty }+{\left[x{e}^{-y}\right]}_x^{\infty }\right\} \\ =\frac{1}{4}\left\{{\left[-y{e}^{-y}-e^{-y}+x{e}^{-y}\right]}_x^{\alpha }\right\} \\ =\frac{1}{4}\left[-0+x{e}^{-x}-\left(0-e^{-x}\right)+x\left(0-e^{-x}\right)\right] \\ =\frac{e^{-x}}{4} \end{gathered}$$

Step 5: Given information (part c)

The function is $f(x,y)=C(y-x)e^{-y}-y<x<y,0<y<\infty$

Step 6: Explanation (part c)

The density function of Y is,

$$\begin{gathered} f_y\left(y\right)=\frac{1}{4}{e}^{-y}{\int }_{x=-y}^y(y-x)dx \\ =\frac{1}{4}{e}^{-y}{\left[yx-\frac{x^2}{2}\right]}_{x=-y}^y \\ =\frac{1}{4}{e}^{-y}\left[y(y-(-y\left)\right)-\frac{1}{2}\left(y^2-(-y{)}^2\right)\right] \\ =\frac{1}{4}{e}^{-y}\left[y(y+y)-\frac{1}{2}\left(y^2-y^2\right)\right] \\ =\frac{1}{2}{y}^2{e}^{-y} \end{gathered}$$

Step 7: Given information (part d)

The function is $f(x,y)=C(y-x)e^{-y}-y<x<y,0<y<\infty$

Step 8: Explanation (part d)

The value of $E\left[X\right] is,$

$$\begin{gathered} E\left[X\right]={\int }_{x=-\infty }^{\infty }xf\left(x\right)dx \\ =\frac{1}{4}\left[{\int }_{x=-\infty }^{\infty }x{e}^{-x}dx+{\int }_{-\infty }^0\left(-2x^2{e}^x+x{e}^x\right)dx\right] \\ =\frac{1}{4}\left[{\int }_{x=-\infty }^{\infty }{x}^{2-1}{e}^{-x}dx-{\int }_0^{\infty }\left(-2y^2{e}^{-y}+y{e}^{-y}\right)dy\right] \\ =\frac{1}{4}\left[\frac{\Gamma \left(2\right)}{1}-{\int }_0^{\infty }\left(2y^2{e}^{-y}\right)dy+{\int }_0^{\infty }\left(y{e}^{-y}\right)dy\right] \\ =\frac{1}{4}\left[1-2{\int }_0^{\infty }\left(y^{3-1}{e}^{-y}\right)dy-{\int }_0^{\infty }\left(y^{2-1}{e}^{-y}\right)dy\right] \\ =\frac{1}{4}\left[1-2\left(\frac{\Gamma \left(3\right)}{1}\right)-\frac{\Gamma \left(2\right)}{1}\right] \\ =\frac{1}{4}[1-2(2!)-1!] \\ =-1 \end{gathered}$$

Step 9: Given information (part e)

The function is $f(x,y)=C(y-x)e^{-y}-y<x<y,0<y<\infty$

Step 10: Explanation (part e)

The value of $E\left[Y\right]$ is,

$$\begin{gathered} E\left[Y\right]={\int }_{y=-\infty }^{\infty }yf\left(y\right)dy \\ =\frac{1}{2}\left[{\int }_0^{\infty }y{y}^2{e}^{-y}dy\right] \\ =\frac{1}{2}\left[{\int }_0^{\infty }{y}^{4-1}{e}^{-y}dy\right] \\ =\frac{1}{2}\left[\frac{\Gamma \left(4\right)}{1^4}\right] \\ =\frac{1}{2}\left[3!\right] \\ =3 \end{gathered}$$