Question

The joint density function of $X$and $Y$is

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

(a) Are $X$and $Y$independent?

(b) Find the density function of $X$.

(c) Find$P\{X+Y<1\}$.

Short answer

1. The variables X and Y are dependent.
2. The density function of random variable X is:

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

$c.P(X+Y<1)=\frac{1}{3}$

Step-by-step solution

Step 1. Given information:

$f(x,y)=\left\{\begin{array}{c}x+y\text{;}0<x<1,0<y<1\\ 0\text{;otherwise}\end{array}\right\}$

Step 2. Prove the independence of X and Y.

In order to test the independence of X and Y, we need to find the marginal distribution function of X and Y. Therefore,

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

Since the variables $X$and $Y$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, $X$ and $Y$ are dependent random variables.

Density function of X

From the above step, we get-

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

Calculation of P(X+Y<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.