Q. 6.22.P - A First Course in Probability

Question

The joint density function of $\mathbf{X}$ and $\mathbf{Y}$ is

$\mathit{f}\left(x,y\right)\mathbf{=}\{\begin{array}{l}x+y 0<x<1,0<y<1\\ 0 \mathrm{otherwise}\end{array}$

1. Are $\mathbf{X}$ and $\mathbf{Y}$ independent?
2. Find the density function of $\mathbf{X}$.
3. Find $\mathit{P}\left\{X+Y<1\right\}$.

Step-by-Step Solution

Answer

Short answer:

1. The variables $\mathrm{X}$ and $\mathrm{Y}$ are dependent.
2. The density function of random variable $\mathrm{X}$ is
   $f_X\left(x\right)=x+\frac{1}{2}\text{ ; }0<x<1$
   
3. $P(X+Y<1)=\frac{1}{3}$
   

Part a. Step 1. Prove the independence of X and Y .

In order to test the independence of $\mathrm{X}$ and $\mathrm{Y}$, we need to find the marginal distribution function of $\mathrm{X}$ and $\mathrm{Y}$. Therefore,

$\begin{array}{l}{f}_X\left(x\right)=\underset{y}{\int }f(x,y)dy\\ =\underset{0}{\overset{1}{\int }}(x+y)dy\\ ={\left(xy+\frac{y^2}{2}\right)}_0^1\\ =x+\frac{1}{2}\\ f_X\left(x\right)=x+\frac{1}{2}\text{ ; }0<x<1\end{array}$

Since the variables, $\mathrm{X}$ and $\mathrm{Y}$ are interchangeable/symmetric, therefore,

$f_Y\left(y\right)=y+\frac{1}{2}\text{ ; }0<y<1$

Since $f(x,y)\ne f_X\left(x\right)f_Y\left(y\right)$

Therefore, $\mathrm{X}$ and $\mathrm{Y}$ are dependent on random variables.

Part b. Step 2. The density function of X .

From the above step, we get-

$f_X\left(x\right)=x+\frac{1}{2}\text{ ; }0<x<1$

Part c. Step 3: Calculation of P ( X + Y &#60; 1 ) .

$\begin{array}{l}P(X+Y<1)=\underset{x}{\int }\underset{y}{\int }f(x,y)dxdy\\ =\underset{x=0}{\overset{1}{\int }}\underset{y=0}{\overset{1-x}{\int }}(x+y)dxdy\\ =\underset{0}{\overset{1}{\int }}{\left|xy+\frac{y^2}{2}\right|}_0^{1-x}dx\\ =\underset{0}{\overset{1}{\int }}\left[x(1-x)+\frac{{\left(1-x\right)}^2}{2}\right]dx\\ =\underset{0}{\overset{1}{\int }}\left(\frac{1-x^2}{2}\right)dx\\ ={\left|\frac{3x-x^3}{6}\right|}_0^1\\ \Rightarrow P(X+Y<1)=\frac{1}{3}\end{array}$

which is the required solution.