Q.6.9 - A First Course in Probability

Question

The joint probability density function of X and Y is given by

$f(x.y)=\frac{6}{7}(x^2+\frac{xy}{2}) 0<x<1, 0<y<2$

(a) Verify that this is indeed a joint density function.

(b) Compute the density function of X.

(c) Find P{X > Y}.

(d) Find P{Y > 1 2 |X < 1 2 }. 

(e) Find E[X]. 

(f) Find E[Y].

Step-by-Step Solution

Answer

(a) The function is indeed a joint density function.

(b)  The density function of X : $fx\left(x\right)=\frac{6}{7}(2x^2+x)$

(c) Probability: $P\{X>Y\}=\frac{15}{56}$

(d) Conditional probability $P \{X,Y\}= \frac{15}{56}$

(e)For probability function of X: $E\left[X\right]=\frac{5}{6}$

(f) The probability density function of Y: $E\left[Y\right]=\frac{8}{7}$

Step 1: Introduction

The marginal density is the name given to this outcome. The marginal density for the random variable X, for example, is the random variable's marginal density. The marginal density for the random variable Y is shown below.

Step 2: Explanation (a)

The joint probability density function of $X and Y$ :

$f(x,y)=\frac{6}{7}\left(x^2+\frac{xy}{2}\right)$

Such that $0<x<1,0<y<2$

To be a probability density function,

For all possible values of the 2 random variables, it needs to be positive.

And the integration over the whole range of $X and Y$ should be $" 1 " .$Then

$f_Y\left(y\right)={\int }_0^1\frac{6}{7}\left(x^2+\frac{xy}{2}\right)dx=\frac{6}{7}\left(\frac{1}{3}+\frac{y}{4}\right)$

Integrate again

${\int }_0^2{f}_Y\left(y\right)dy={\int }_0^2\frac{6}{7}\left(\frac{1}{3}+\frac{y}{4}\right)dy=\frac{2}{7}\times 2+\frac{3}{14}\times \frac{1}{2}\times 4=1$

Thus, the function is indeed a joint density function.

Step 3: Explanation (b)

The joint probability density function of X and Y:

$f(x,y)=\frac{6}{7}\left(x^2+\frac{xy}{2}\right)$

Such that $0<x<1,0<y<2$

For density function of $X$ : $f_X\left(x\right)={\int }_0^2\frac{6}{7}\left(x^2+\frac{xy}{2}\right)dy$

Integrate the left side: $f_X\left(x\right)=\frac{6}{7}\left(2x^2+x\right)$

Thus, the density function of $X$ :

$f_X\left(x\right)=\frac{6}{7}\left(2x^2+x\right)$

Step 4: Explanation (c)

The joint probability density function of X and Y:

$f(x,y)=\frac{6}{7}\left(x^2+\frac{xy}{2}\right)$

Such that $0<x<1,0<y<2$

We have $f(x,y)=\frac{6}{7}\left(x^2+\frac{xy}{2}\right)$

Then, $P (X>Y) ={\int }_0^1{\int }_0^x\frac{6}{7}(x^2+\frac{xy}{2}) dxdy =\frac{6}{7}{\int }_0^1(x^3+\frac{x^3}{4})dx$

Also, the other way:

$P\{X,Y\} ={\int }_0^1{\int }_y^1\frac{6}{7}(x^2+\frac{xy}{2}) dxdy =\frac{15}{56}$

Step 5: Explanation (d)

The joint probability density function of X and Y :

$f(x,y)=\frac{6}{7}\left(x^2+\frac{xy}{2}\right)$

Such that $0<x<1,0<y<2$

For conditional probability:

$P\left\{Y>\frac{1}{2}\mid X<\frac{1}{2}\right\}=\frac{P\left\{Y>\frac{1}{2},X<\frac{1}{2}\right\}}{P\left\{X<\frac{1}{2}\right\}}$

Then

$P\left\{Y>\frac{1}{2},X<\frac{1}{2}\right\}={\int }_{\frac{1}{2}}^2{\int }_0^{\frac{1}{2}}\frac{6}{7}\left(x^2+\frac{xy}{2}\right)dxdy=\frac{69}{448}$

And

$P\left\{X<\frac{1}{2}\right\}={\int }_0^{\frac{1}{2}}\frac{6}{7}\left(2x^2+x\right)dx=\frac{5}{28}$

Therefore,

$P\left\{Y>\frac{1}{2}\mid X<\frac{1}{2}\right\}=\frac{P\left\{Y>\frac{1}{2},X<\frac{1}{2}\right\}}{P\left\{X<\frac{1}{2}\right\}}=\frac{69}{80}$

Step 6: Explanation (e)

The joint probability density function of $X and Y$ :

$f(x,y)=\frac{6}{7}\left(x^2+\frac{xy}{2}\right)$

Such that $0<x<1,0<y<2$

From Part (b),

We have $f_X\left(x\right)=\frac{6}{7}\left(2x^2+x\right)$

Then

$E\left[X\right]={\int }_0^{1}x·\frac{6}{7}\left(2x^2+x\right)dx=\frac{5}{6}$

Step 7: Explanation (f)

The joint probability density function of $X and Y$ :

$f(x,y)=\frac{6}{7}\left(x^2+\frac{xy}{2}\right)$

Such that $0<x<1,0<y<2$

From Part (a),

We have $f_Y\left(y\right)=\frac{6}{7}\left(\frac{1}{3}+\frac{y}{4}\right)$

Then

$E\left[Y\right]={\int }_0^{2}y·\frac{6}{7}\left(\frac{1}{3}+\frac{y}{4}\right)dy=\frac{8}{7}$