Q.6.22 - A First Course in Probability

Question

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

(a) Are X and Y independent?

(b) Find the density function of X.

(c) Find $P[X+Y<1].$

Step-by-Step Solution

Answer

a. X and Y are not independent.

b. The density function of X is 

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

c. The value of  $P[X+Y<1]=\frac{2}{3}$

Step 1: Content Introduction

X and Y are random variables with a combined density function.

Step 2: Explanation (Part a)

The joint density function of X and Y is

$f_{X,Y}(x,y) =x+y ; 0<x<1 \mathrm{and} 0<y<1$

$$\begin{gathered} \mathrm{For} \mathrm{marginal} \mathrm{density} \mathrm{of} \mathrm{X}, \mathrm{please} \mathrm{reffer} \mathrm{part} \left(\mathrm{b}\right) \mathrm{for} \mathrm{explanation} \\ {f}_X\left(x\right)=x+0.5 ; 0<x<1 \\ \mathrm{Therefore} \mathrm{by} \mathrm{symmetry}, \mathrm{for} \mathrm{Y}, \mathrm{we} \mathrm{have} \\ {f}_Y\left(y\right)=y+0.5 ; 0<y<1 \\ (\mathrm{x}+\frac{1}{2})\times (y+\frac{1}{2})\ne \mathrm{x}+\mathrm{y} \\ \mathrm{Which} \mathrm{proves} \mathrm{that} \mathrm{x} \mathrm{and} \mathrm{y} \mathrm{are} \mathrm{dependent}. \end{gathered}$$

Step 3: Explanation (Part b)

The marginal density of X is

$$\begin{gathered} f_X\left(x\right)={\int }_0^1(x+y)dy \\ ={\left[xy+\frac{y^2}{2}\right]}_0^1 \\ =\left(x+\frac{1}{2}\right) \end{gathered}$$ 

$x+\frac{1}{2} 0<x<1$

Step 4: Explanation (Part c)

$$\begin{gathered} P[X+Y<1]={\int }_{x=0}^1{\int }_{y=0}^{1-x}{f}_{X,Y}(x,y)dydx \\ ={\int }_0^1{\int }_0^{1-x}(x+y)dy \\ ={\int }_0^1 {\left[xy+{\frac{y}{2}}^2\right]}_0^{1-x}dx \\ \mathrm{Upon} \mathrm{simplification}, \mathrm{we} \mathrm{obtain} \\ ={\int }_0^1\left(\frac{x^2+1}{2}\right)dx \\ ={\left[\frac{x^3}{6}+\frac{x}{2}\right]}_0^1 \\ =\frac{2}{3} \end{gathered}$$