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In this question density of sum of two uniform random variables $[0,1]$ two different methods (convolution and cdf) are explained in order to do the sum of two independent random variables. In both cases the key seems to be in the geometric properties. For the case of $Z = X + Y$ and with $X$ and $Y$ uniform distributions it is relatively easy to solve the problem.

But I was wondering how one could proceed when one has to sum for instance 3 or 4 random variables and you just can't draw the hypersurface to know the limits of integration. That seems a very intimidating problem to me. Does anyone know a general method to proceed in such cases?

David
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    Use indicator functions, for example, write the PDF of $X$ uniformly distributed on $(0,1)$ as the function $f_X$ defined on the whole real line by $$f_X(x)=\mathbf 1_{(0,1)}(x)$$ and carry these indicator terms in the computations. – Did May 11 '17 at 12:44
  • Thanks. Do you have an example? – David May 11 '17 at 13:47
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    In the case $Z=X+Y$ you consider, note that $$f_Z(z)=\int_\mathbb Rf_X(x)f_Y(z-x)dx=\int_\mathbb R\mathbf 1_{(0,1)}(x)\mathbf 1_{(0,1)}(z-x)dx$$ and $$0<x<1,0<z-x<1\iff0<x<1,z-1<x<z\iff\max{0,z-1}<x<\min{1,z}$$ hence $$f_Z(z)=(\min{1,z}-\max{0,z-1}),\mathbf 1_{\max{0,z-1}<\min{1,z}}$$ and the last indicator function is obviously zero unless $z-1<1$ and $0<z$, that is, unless $0<z<2$. It remains to identify $\min{1,z}-\max{0,z-1}$ when $0<z<2$, considering separately the case $z<1$, and then $$f_Z(z)=\min{1,z}-\max{0,z-1}=z-0=z$$ – Did May 11 '17 at 15:17
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    ... and the case $z>1$, and then $$f_Z(z)=\min{1,z}-\max{0,z-1}=1-(z-1)=2-z$$ as desired. – Did May 11 '17 at 15:17
  • Really thanks. I have a couple of questions: a) you write min${1,z} - $max${0,z-1}$ for $f_{Z}(z)$ because the integral can be performed immediately, but I suppose that if the pdf weren't uniform, the integral would be not immediate, but the indicator function and all the previous work would be the same but considering other intervals, right? b) when you are considering $z - 1 < 0$ and $0 < z$ is because the cases $0 < 1$ and $z - 1 < z$ are immediate and you don't need to consider them but in other occasions you should check the four possible combinations of cases, right? – David May 11 '17 at 20:19
  • And finally, why do you consider separately the case $z < 1$? You have the interval $0 < z < 2$, why do you divide in the point 1 and not, for example in the point 1/2 or 3/2? – David May 11 '17 at 20:25
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    a) Yes. b) Yes. c) One splits the interval at $z=1$ because this is where the minimum and the maximum switch from $z$ and $0$ to $1$ and $z-1$ respectively (only one split is necessary since they switch both at the same value). – Did May 11 '17 at 20:47

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