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I'm trying to find the result of the sum of two PDF that is the square of two continuous uniform distribution (rectangular distribution) that have $\mu =0$ and $\sigma =1$. The square of this continuous uniform distribution is defined from 0 to 3 and is $$p(x)= \frac{1}{2 \sqrt{3x}}$$ Now i need to find the PDF of the sum of two random variables with p(x) as distribution. The convolution integral is $$p(u)=\int_0^3 p(u-t)p(t)dt=\int_0^3 \frac{dt}{12 \sqrt{t(u-t)}}= \frac{arcsin \left( \sqrt{ \frac{3}{u}} \right)}{6}$$ This funcion is not defined on R from 0 to 6. If i extract the real part of $(p(u))$ i obtain the correct PDF from 0 to 3 but the part from 3 to 6 is wrong and p(u) is not normalized from 0 to 6. I simulate this PDF with python, here is an image

Simulation of PDF: https://imgur.com/ImqT7ya

Plot of the PDF: https://imgur.com/khg8rGM

what did I do wrong?

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  • $\begingroup$ That distribution is not uniform. Do you mean continuous? $\endgroup$
    – Graham Kemp
    Jun 26, 2018 at 1:16
  • $\begingroup$ p(x) is the pdf of the square of an uniform distribution with mean=0 and variance=1. I'm not saying that p(x) is an uniform distribution $\endgroup$
    – Piero Viscone
    Jun 26, 2018 at 1:19
  • $\begingroup$ okay sorry it is a continuous uniform distribution ( en.wikipedia.org/wiki/Uniform_distribution_(continuous) ), I got confused with the translation from italian $\endgroup$
    – Piero Viscone
    Jun 26, 2018 at 1:22

1 Answer 1

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The key is in the support interval.   Both $t$ and $u-t$ must lie within the support, or otherwise $p(t)p(u-t)=0$.   So the bounds for the integral depends on the value given for $u$.

You have $p(x)=\dfrac 1{2\sqrt{3x}}\mathbf 1_{0\leq x\leq 3}$

The the convolution is $$\begin{split}\int_\Bbb R p(t)p(u-t)~\mathsf d t &= \int_\Bbb R \tfrac{1}{12\sqrt{t(u-t)}}\mathbf 1_{0\leq t\leq 3~,~0\leq u-t\leq 3}~\mathsf d t\\&= \int_\Bbb R \tfrac{1}{12\sqrt{t(u-t)}}\mathbf 1_{0\leq t\leq 3~,~ t\leq u~,~ u-3\leq t}~\mathsf d t\\ &= \int_{\max\{0,u-3\}}^{\min\{3,u\}} \tfrac{1}{12\sqrt{t(u-t)}}~\mathbf 1_{0\leq u\leq 6}~\mathsf d t \\ &= \mathbf 1_{0\leq u\leq 3}\int_{0}^{u} \tfrac{1}{12\sqrt{t(u-t)}}~\mathsf d t+\mathbf 1_{3<u\leq 6}\int_{u-3}^{3} \tfrac{1}{12\sqrt{t(u-t)}}~\mathsf d t\end{split}$$

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  • $\begingroup$ Thanks man, i still can't define the interval of integration of the convolution integral in the right way. But this works fine $\endgroup$
    – Piero Viscone
    Jun 26, 2018 at 1:49

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