I am looking for two continuous probability distributions $X,Y$ with respective densities $f_X(\cdot) \text{ and } f_Y(\cdot)$, one of them with support on $[\epsilon,a-\epsilon]$, the other with support on $[-\epsilon, \epsilon]$ (with $\epsilon,a>0$, $\epsilon<\frac{a}{2}$) with properties defined below. The convolution $(f_X*f_Y)(\cdot)$ of the probability density functions has support on [0,a].


The required properties are not very rigorous but rather soft. All densities $f_X(\cdot), f_Y(\cdot) \text{ and } (f_X*f_Y)(\cdot)$ should NOT be piecewise defined on their respective supports, i.e. I don't want to use $$f(z)=\begin{cases}...& z\in [...]\\ ... & z\in[...] \end{cases}$$ to write the functions. I do not care whether intervals are open, closed or semi-closed. I need this example for teaching/simple models where I dont want to many cases.


An example which doesn't work is $X\sim\text{Unif}([\epsilon,a-\epsilon])$ and $Y\sim\text{Unif}([-\epsilon,\epsilon])$ as the density of $X+Y$ is $$ f_{X+Y}=(f_Y*f_Y)(y)= \begin{cases} \frac{y}{2\epsilon(a -2\epsilon)} & 0<y<2\epsilon \\ \frac{1}{a-2 \epsilon} & 2\epsilon\leq y \leq a-2\epsilon \\ \frac{a-y}{2\epsilon(a -2\epsilon)} & a-2\epsilon< y <a. \end{cases} $$

Thanks in advance for suggestions or proofs that such a thing is not possible.


I don't see how you can be satisfied. The range of integration in the convolution integral depends on how the endpoints of the supports of the two densities interlace. As you translate one support against the other, the interlacing changes. You can see this in the example you supply, and I think it will work out pretty much the same way in general. Replace your uniform densities with more interesting functions, and the same break points will appear but with more complicated formulas on the pieces.

  • $\begingroup$ Thanks for your suggestion. I was wondering whether there was a distriution with surprising properties where distributions add up nicely. I found it hard to prove that this wasn't possible. Don't you see any chance for this happening? $\endgroup$ – MathProb Dec 11 '17 at 9:04
  • $\begingroup$ You can make the join as smooth as you like, by convolving smooth bump functions, but the piecewise formula aspect will survive, together with the teacher-constructed artificial counterexample air to the whole thing. $\endgroup$ – kimchi lover Dec 11 '17 at 12:22

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