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I am calculating the sum of two uniform random variables $X$ and $Y$, so that the sum is $X+Y = Z$. Since the two are independent, their densities are $f_X(x)=f_Y(x)=1$ if $0\leq x\leq1$ and $0$ otherwise. The density of the sum becomes $f_Z(z)=\int_{-\infty}^\infty f_X(z-y)f_Y(y)dy=\int_0^1f_X(z-y)dy$ by convolution. I am stuck at this stage. How do I proceed with my integral? I think a diagram make it easy but I dont know how to proceed.

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up vote 2 down vote accepted

Hint: Split the calculation into two cases: (i) $0\le z\le 1$ and (ii) $1\lt z\le 2$.

Added: (i) if $0\le z\le 1$, then $f_X(z-y)=1$ if $0\le y\le z$, and $f_X(z-y)=0$ if $y\gt z$. It follows that $$\int_0^1 f_X(z-y)\,dy=\int_0^z 1\cdot dy=z$$.

(ii) If $1\lt z\le 2$, then $f_X(z-y)=1$ if $z-1\le y \le 1$, and $f_X(z-y)=0$ elsewhere. It follows that $$\int_0^1 f_X(z-y)\,dy=\int_{z-1}^1 1\cdot dy=2-z.$$ Thus $f_Z(z)=z$ if $0\le z\le 1$, and $f_Z(z)=2-z$ if $1\le z\le 2$. And for completeness, $f_Z(z)=0$ if $z$ is outside the interval $[0,2]$.

Remark: I suspect that the convolution way is in this case effectively no faster than the "slow" way of finding first the cumulative distribution function $F_Z(z)$, and differentiating.

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could you elaborate on that. –  Vaolter Oct 25 '12 at 8:41
    
@Vaolter: Details filled in. –  André Nicolas Oct 25 '12 at 9:28
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hint: the integrand is zero unless $0 \le z-y \le 1$

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ok, $f_X(z-y)$ is 1 in that case how do I proceed? –  Vaolter Oct 25 '12 at 8:41
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