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We are given two Independent Identically Distributed random variables $X$ and $Y$ where $X,Y$~$U(0,1)$. Letting $Z=X+Y$ , we need to find the distribution of $Z$.

The text I am reading goes as follows :

$f_Z(z)=\int_{- \infty}^{\infty}f_X(x)f_Y(z-x)dx = \int_{- \infty}^{\infty}I_{(0,1)}(x)I_{(0,1)}(z-x)dx$

=> $\int_{- \infty}^{\infty}(I_{(0,z)}(x)I_{(0,1)}(z)+I_{(z-1,1)}(x)I_{[1,2)}(z))dx$

I have no idea what happened there in the third step. Also , if $X$~$U(0,1)$ , then why the integration limits of $X$ are taken from $- \infty$ to $\infty$ ? Can anyone explain ?

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2 Answers 2

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Your remark is exactly the thing they do… The density of the sum of two independent rv is the convolution, that's the first equation. The the definition the integral is from $-\infty$ to $\infty$. In the second the definition of the density of X and Y is plugged in. Remember: $f_X(x) = f_Y(x) = I_{(0,1)}(x)$ So the densities have only mass on $(0,1)$ so indeed it holds

$$\int_{- \infty}^{\infty}f_X(x)f_Y(z-x)dx= \int_{0}^{1}f_X(x)f_Y(z-x)dx$$

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Here $ I_{(0,1)}(x)$ represents indicator function which means it takes value one one when $x \in (0,1) $ else where it is zero. This makes perfect sense in the limits of integration being $(-\infty,\infty )$. The tricky part is the third step where it is decomposed as sum of 2 indicator functions. This needs some works but when you write $I_{(0,1)}(z-x) =1 \ when \ 0<z-x<1 \ or \ z>x>z-1 \implies I_{(0,1)}(z-x)=I_{(z-1,z)}(x)$. The third step becomes trivial.

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