Sum of Two Independent Random Variables

[Shannon Young]

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
f_X(x) = f_Y(x) = 1/2 if -1<x<1 and 0 otherwise

The density of Z will be given by
f_Z= $$\int$$f_X(z-y)f_Y(y)dy

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

f_Z= $$\int$$(1/2)f_X(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

 

[mathman]

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
f_X(x) = f_Y(x) = 1/2 if -1<x<1 and 0 otherwise

The density of Z will be given by
f_Z= $$\int$$f_X(z-y)f_Y(y)dy

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

f_Z= $$\int$$(1/2)f_X(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.

 

[Shannon Young]

This is incorrect, it should read =1/2 within the interval and =0 outside.

I figured the problem out, thanks