Let X and Y be independent, uniform r.v's in (-1, 1). Compute the pdf of V=$(X+Y)^2 $

Mumbai University > Electronics and Telecommunication > Sem5 > Random Signal Analysis

Marks: 5M

Year: May 2015

Given V=${(X+Y)}^2$

∴ $\sqrt v=$x+y and u=y

∴ $x=\sqrt v-u$

Also X, Y is independent uniform random variables

$$ i.e f_X (x)=\frac{1}2 .... -1≤x≤1$$

$$ =0 $$

$$ f_Y (y)=\frac{1}2 ... -1≤y≤1$$

$$∴f_{XY} (x,y )=f_X (x)×f_Y (y)=1/4 .... -1≤x,y≤1$$

$$ ∴ J=\begin{vmatrix} \frac{δx}{δu} & \frac{δx}{δv} \\\\ \ \frac{δy}{δu} & {δy}{δv} \\\\ \end{vmatrix} $$

$$ ∴ J=\begin{vmatrix} {-1} & {\frac{1}{2\sqrt v}} \\\\ \ {1} & {}0 \\\\ \end{vmatrix} $$

$$|J|=\frac{1}{2\sqrt v}$$

$$f_{UV} (u,v)=|J| f_{XY} (x,y)$$

$$=\frac{1}{(2\sqrt v)}×\frac{1}{2}$$

$$ f_{UV} (u,v)=\frac{1}{4\sqrt v}$$

The range space of (U, V) is obtained as follows:

-1≤x≤1 and -1≤y≤1

-2 ≤ x+y ≤ 2

i.e |x+y| ≤ 2

i.e |x+y|^2 ≤ 4

∴ |v| ≤ 4

Now -1 ≤ y ≤ 1 ∴ - 1 ≤ u ≤ 1

Pdf of V is given by $f_{V} (v)=∫_{-1}^1\frac{1}{(4\sqrt v)} du= \frac{1}{(2\sqrt v)}$ |v| ≤ 4