Q.6.6 - A First Course in Probability

Question

6. Let X and Y be continuous random variables with joint density function $f(x,y)=\left\{\begin{array}{l}\frac{x}{5}+cy\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0<x<1,1<y<5\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ otherwise }\end{array}\right.$

where c is a constant.

 (a) What is the value of c? 

(b) Are X and Y independent? 

(c) Find $P[X+Y>3]$

Step-by-Step Solution

Answer

a) The value of c is $\frac{1}{20}$

b) No X and Y are not independent.

c) The probability is  $\frac{11}{15}$

Part (a) - Step 1 : To find

The value of c

Part(a) - Step 2: Explanation

The given function is $f(x,y)=\left\{\begin{array}{l}\frac{x}{5}+cy\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0<x<1,1<y<5\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ otherwise }\end{array}\right.$

The value of c can be calculated as

$\begin{array}{r}{\int }_0^1 {\int }_1^5 \left(\frac{x}{5}+Cy\right)dydx=1\\ {\int }_0^1 \left(\frac{4x}{5}+12C\right)=1\\ 12C+\frac{2}{5}=1\\ C=\frac{1}{20}\end{array}$

Hence the value of c is $\frac{1}{20}$

Part (b) - Step 3: To find

Are X and Y are independent

Part(b) - Step 4: Explanation

From the part(a) we know that c is $\frac{1}{20}$

Thus $\begin{array}{r}f(x,y)=\left\{\begin{array}{c}\frac{x}{5}+\frac{1}{20}y0<x<1,1<y<5\\ 0\text{ Otherwise }\end{array}\right.\\ f(x,y)=\left\{\begin{array}{ll}\frac{4x+y}{20}& 0<x<1,1<y<5\\ 0\text{ Otherwise }& \end{array}\right.\end{array}$

The marginal density X could be calculated as:

$\begin{array}{r}{f}_x\left(x\right)={\int }_1^5 \left(\frac{4x+y}{20}\right)dy\\ =\frac{1}{20}{\left[4xy+\frac{y^2}{2}\right]}_1^5\\ =\frac{1}{20}[16x+12]\\ =\frac{4x+3}{5}\end{array}$

The marginal density Y could be calculated as:

$\begin{array}{r}{f}_Y\left(y\right)={\int }_0^1 \left(\frac{4x+y}{20}\right)dy\\ =\frac{1}{20}{\left[\frac{4x^2}{2}+xy\right]}_0^1\\ =\frac{1}{20}[2+y]\\ =\frac{y+2}{20}\end{array}$

Now,

$\begin{array}{c}{f}_x\left(x\right)\times f_y\left(y\right)=\left(\frac{4x+3}{5}\right)\left(\frac{y+2}{20}\right)\\ \frac{8x+4xy+6+3y}{100}\ne \frac{4x+y}{20}\\ f_x\left(x\right)\times f_y\left(y\right)\ne f_{xy}(x,y)\end{array}$

Hence X and Y are not independent function.

Part(c) - Step 5: To find

$P[X+Y>3]$

Part (c) - Step 6: Explanation

The probability is
$\begin{array}{c}P[X+Y>3]={\int }_0^1 {\int }_{3-x}^5 \left(\frac{x}{5}+\frac{y}{20}\right)dydx\\ ={\int }_0^1 \left[(x+2)\frac{x}{5}+\frac{25}{40}-\frac{(3-x{)}^2}{40}\right]dx\\ =\frac{1}{5}+\frac{1}{15}+\frac{5}{8}-\frac{19}{20}\\ =\frac{11}{15}\end{array}$

Hence the probability is $\frac{11}{15}$