Q.6.15 - A First Course in Probability

Question

Let X and Y be independent uniform (0, 1) random variables. 

(a) Find the joint density of U = X, V = X + Y. 

(b) Use the result obtained in part (a) to compute the density function of V 

Step-by-Step Solution

Answer

1. The Joint  Density Function is $f_{uv}\left(uv\right)=1\text{ for }\mathrm{Max}(v-1,0)<u<\mathrm{Min}(v,1)$
2. The density function of  V is $f_v\left(v\right)=\left\{\begin{array}{ccc}v& \text{ for }& 0<v<1\\ 2-v& \text{ for }& 1\le v\le 2\end{array}\right.$

Step 1: Given Information (Part-a)

Given in the question that let $x$and $y$ be independent uniform $(0,1)$ random variables.

Step 2: Find the Joint Density Function (Part-a)

Find the joint density of $u=x, v=x+y.$ 

The Jacobean transformation$\left(J\right)$ is

$J=\left|\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right|$

$=1\times 1-1\times 0$

$=1-0$

$=1.$

The available information $U=x$and$V=x+y$

$X=U$ and $Y=V-U$

$f_{uv}\left(uv\right)=f_{xy}(x,y)$

$=f_{x,y}(U,v-u)$

1 for$(0<u<1,0<(v-u)<1)$

Step 3: Final Answer (Part-a)

The required Joint  Density Function is$f_{uv}\left(uv\right)=1\text{ for }\mathrm{Max}(v-1,0)<u<\mathrm{Min}(v,1)$

Step 4: Given Information (Part-b)

Given in the question that let $x$and $y$ be independent uniform $(0,1)$ random variables.

Step 5: Computation of the Density Function (Part-b)

Use the outcomes acquired in part -a to compute the density function of$V$

Find the density function of $V$ for $0<v<1$

$f_v\left(v\right)={\int }_0^v du$

$=[u{]}_0^x$

$v-0=v$

For $1\le V\le 2$

$f_v\left(v\right)={\int }_{v-1}^1 du$

$=[u{]}_{v-1}^1$

$=1-v+1$

$=2-V$$f_v\left(v\right)={\int }_0^v du$

Step 6: Final Answer (Part-b)

The required density function is $f_v\left(v\right)=\left\{\begin{array}{ccc}v& \text{ for }& 0<v<1\\ 2-v& \text{ for }& 1\le v\le 2\end{array}\right.$