Sum of 2 independent uniform r.v.

**noob mathematician** · April 20th 2009, 08:57 AM

If X and Y are independent r.v, both uniformly distributed on (0, 1).

Given that $f_{X+Y}(a)=\int_{-\infty}^{\infty}f_X(a-y)f_Y(y)dy$

we obtain $f_{X+Y}(a)=\int_{0}^{1}f_X(a-y)dy$

For $0\leq a\leq 1$ , we have $f_{X+Y}=\int_{0}^{a}dy=a$

For $1<a<2$ , we have $f_{X+Y}=\int_{a-1}^{1}dy=2-a$

How do I determine the two upper and lower bounds?
Such as for this case we are using: $\int_{0}^{a}$ and $\int_{a-1}^{1}$ ??

Another question:

Given joint density of X and Y:

$f(x,y)=\frac{1}{4}(y-x)e^{-y}$
$-y<x<y, 0<y<\infty$
In order to find the density function of X, I meet the same problem:

for x>0, $f_X(x)=\frac{1}{4}\int_{x}^\infty (y-x)e^{-y} dy$

for x<0, $f_X(x)=\frac{1}{4}\int_{-x}^\infty (y-x)e^{-y} dy$
Why the two different lower bounds?

**The Second Solution** · April 20th 2009, 10:53 PM

Originally Posted by noob mathematician

If X and Y are independent r.v, both uniformly distributed on (0, 1).

Given that $f_{X+Y}(a)=\int_{-\infty}^{\infty}f_X(a-y)f_Y(y)dy$

we obtain $f_{X+Y}(a)=\int_{0}^{1}f_X(a-y)dy$

For $0\leq a\leq 1$ , we have $f_{X+Y}=\int_{0}^{a}dy=a$

For $1<a<2$ , we have $f_{X+Y}=\int_{a-1}^{1}dy=2-a$

How do I determine the two upper and lower bounds?
Such as for this case we are using: $\int_{0}^{a}$ and $\int_{a-1}^{1}$ ??

[snip]

Read example 7.3 p292 here: http://www.dartmouth.edu/~chance/tea...k/Chapter7.pdf

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