Question

The joint density function of X and Y is$f(x,y)=\left\{\begin{array}{l}x+y0<x<1,0<y<1\\ 00otherwise\end{array}\right.$

(a) Are X and Y independent?

(b) Find the density function of X.

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

Short 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-by-step solution

Content Introduction

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

Explanation (Part a)

The joint density function of X and Y is

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

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

Explanation (Part b)

The marginal density of X is

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

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

Explanation (Part c)

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