Sum of Two Independent Random Variables

  • #1

Main Question or Discussion Point

Suppose X and Y are Uniform(-1, 1) such that X and Y are independent and identically distributed. What is the density of Z = X + Y?

Here is what I have done so far (I am new to this forum, so, my formatting is very bad). I know that
fX(x) = fY(x) = 1/2 if -1<x<1 and 0 otherwise

The density of Z will be given by
fZ= [tex]\int[/tex]fX(z-y)fY(y)dy

fY(y) = 1/2 if -1<y<1 and 0 otherwise
So,

fZ= [tex]\int[/tex](1/2)fX(z-y)dy (bounds of integration -1 to 1)

The integrand = 0 if -1<z-y<1 or if z-1<y<z+1

That is where I get stuck, and need help to complete. Your assistance is appreciated, thanks
 

Answers and Replies

  • #2
mathman
Science Advisor
7,801
431
Suppose X and Y are Uniform(-1, 1) such that X and Y are independent and identically distributed. What is the density of Z = X + Y?

Here is what I have done so far (I am new to this forum, so, my formatting is very bad). I know that
fX(x) = fY(x) = 1/2 if -1<x<1 and 0 otherwise

The density of Z will be given by
fZ= [tex]\int[/tex]fX(z-y)fY(y)dy

fY(y) = 1/2 if -1<y<1 and 0 otherwise
So,

fZ= [tex]\int[/tex](1/2)fX(z-y)dy (bounds of integration -1 to 1)

The integrand = 0 if -1<z-y<1 or if z-1<y<z+1

That is where I get stuck, and need help to complete. Your assistance is appreciated, thanks
The integrand = 0 if -1<z-y<1 or if z-1<y<z+1
This is incorrect, it should read =1/2 within the interval and =0 outside.
 
  • #3
This is incorrect, it should read =1/2 within the interval and =0 outside.
I figured the problem out, thanks
 

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