Q.6.7 - A First Course in Probability

Question

The joint density function of X and Y is $f(x,y)=\left\{\begin{array}{l}xy\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0<x<1,0<y<2\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ otherwise }\end{array}\right.$ 

(a) Are X and Y independent?

(b) Find the density function of X. 

(c) Find the density function of Y. 

(d) Find the joint distribution function.

(e) Find$E\left[Y\right].$

(f) Find$P[X+Y<1]$

Step-by-Step Solution

Answer

a) Yes they are independent to each other.

b) The density function of$X$ is $2x.$

c) The density function of Y is $\frac{y}{2}$.

d) The joint distribution function $F(x,y)=\left(\frac{x^2{y}^2}{4}\right)$

e) $E\left[Y\right] = \frac{4}{3}$

f) $P[X+Y<1] = \frac{1}{24}$

Part (a) - Step 1: To determine

 Are X and Y independent.

Part(a) - Step 2: Explanation

To show : X and Y are independent.

To given: $f(x,y)=\left\{\begin{array}{l}xy\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0<x<1,0<y<2\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ otherwise }\end{array}\right.$

Formula: $f(x,y)=f\left(x\right)f\left(y\right)$

proof:

 $\begin{array}{c}f(x,y)=\left\{\begin{array}{ll}xy,& 0<x<1,0<y<2\\ 0,& \text{ otherwise }\end{array}\right\}\\ f\left(x\right)={\int }_y f(x,y)dx\\ ={\int }_0^2 xydx\\ =2x\\ f\left(y\right)={\int }_x f(x,y)dx\\ ={\int }_0^1 xydx\\ =\frac{y}{2}\end{array}$

Therefore $f(x,y) = f\left(x\right)f\left(y\right)$

Hence proved that $X$ and $Y$ are independent.

Part(b) - Step 3: To determine

The density function of X.

Part(b) - Step 4: Explanation

To find: the density function of $X .$

        As per the above part (a)

$\begin{array}{c}f\left(x\right)={\int }_y f(x,y)dx\\ ={\int }_0^2 xydx\\ =2x\end{array}$

Hence the density function of X is $f\left(x\right) = \{2x,0<x<1\}$

Part(c) - Step 5: To determine

The density function of Y

Part (c) - Step 6: Explanation

To find: density function of Y

$\begin{array}{r}f\left(y\right)={\int }_x f(x,y)dx\\ ={\int }_0^1 xydx\\ =\frac{y}{2}\end{array}$

Hence the density function of$f\left(Y\right) = \{\frac{y}{2}, 0<y<2\}$

Part d) - Step 7: To determine

The joint distribution function

Part (d) - Step 8: Explanation

To find : The joint distribution function

Formula used: $F(x,y) = {\int }_x{\int }_{y}f(u,v)dudv$

Calculation: The joint distributions 

                     $\begin{array}{c}F(x,y)={\int }_x {\int }_y f(u,v)dudv\\ =\left({\int }_0^x udu\right)\left({\int }_0^y vdv\right)\\ ={\left(\frac{u^2}{2}\right)}_0^x{\left(\frac{u^2}{2}\right)}_0^y\\ F(x,y)=\left(\frac{x^2}{2}\right)\left(\frac{y^2}{2}\right)\\ =\left(\frac{x^2{y}^2}{4}\right)\end{array}$

Hence $F(x,y)=\left\{\left(\frac{x^2{y}^2}{4}\right),0<x<1,0<y<2\right\}$

Part (e) - Step 9: To determine

The expectation of $Y$

Part (e) - Step 10: Explanation

Formula to use to find the expectation value of Y is: $E\left(Y\right)={\int }_{y}yf\left(y\right)dy$

Calculation:

 $\begin{array}{c}f\left(y\right)=\left\{\frac{y}{2},0<y<2\right\}\\ E\left(Y\right)={\int }_y y\left(\frac{y}{2}\right)dy\\ ={\int }_0^2 \left(\frac{y^2}{2}\right)dy\\ ={\left[\frac{y^3}{6}\right]}^{2}0\end{array}$

Further $\begin{array}{r}E\left(Y\right)=\left[\frac{2^3-0^3}{6}\right]\\ =\frac{8}{6}\\ =\frac{4}{3}\end{array}$

Therefore $E\left[Y\right] = \frac{4}{3}$

Part (f) - Step 11: To find

probability of $X+Y<1$

Part(f) - Step 12: Explanation

The probability of X+Y<1 is 

$\begin{array}{r}P(X+Y<1)={\int }_x {\int }_y xydxdy\\ ={\int }_0^1 {\int }_0^{1-x} xydxdy\\ ={\int }_0^1 x{\left[\frac{y^2}{2}\right]}_0^{1-x}dx\\ ={\int }_0^1 x\left[\frac{(1-x{)}^2}{2}\right]dx\end{array}$

Now,

$\begin{array}{r}P(X+Y<1)={\int }_0^1 x\left[\frac{1+x^2-2x}{2}\right]dx\\ ={\int }_0^1 \frac{x+x^3-2x^2}{2}dy\\ =\frac{1}{2}{\left(\frac{x^2}{2}+\frac{x^4}{4}-2\frac{x^3}{3}\right)}_0^1\end{array}$

$\begin{array}{r}P(X+Y<1)=\frac{1}{2}\left(\frac{1^2}{2}+\frac{1^4}{4}-2\frac{1^3}{3}\right)\\ =\frac{1}{2}\left(\frac{1}{12}\right)\\ =\frac{1}{24}\end{array}$

Therefore $P(X+Y<1)=\frac{1}{24}$