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There are three random variables, x,y,z distributed uniformly on $[0,1]$. They are Independently distributed.

The question I was trying to answer is the following:

What is the probability that $max(x,y,z)-min(x,y,z) \leq \frac{2}{3} $.

I am interested in solving this by first calculating the joint PDF of the LHS. However, I can't seem to replicate the results I've found online. Here is what I did:

I see that to get the PDF of the max, and min we simply follow the procedure in this thread and the expected value is here.

The PDF of the max is $f_{max}=3a^2$ and pdf of the min is $f_{min}=3a^2-6a+3$, where $a \in [0,1]$

The source at the bottom says that the convolution of the two is:

$f_{minmax} = 6(a-a^2)$

To be clear, I want to know how the convolution is attained because everybody seems to just do it very quickly and I don't see the approach. What I am doing is $\int_0^1 f_{max}(z-t)f_{min}(-t) dt $, but i'm pretty sure this isn't the right expression to get the answer.

To see the solutions and problem see here

Dio
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1 Answers1

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Here's the volume in $x$, $y$, $z$ space:

Max[x,y,z]-Min[x,y,z] < 2/3

Integrate the region's volume to find that the probability is $20/27$.

David G. Stork
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  • This is fun but I was asking how we get the expression for the convolution! – Dio Jul 06 '18 at 13:28
  • What does convolution have to do with anything here? – David G. Stork Jul 11 '18 at 18:45
  • I can't figure how to @ you. Quite simply, the max and the min follow a distribution each, and then their difference is a convolution. The last link I have in my post shows an answer that claims the convolution is the $f_{minmax}$ – Dio Jul 12 '18 at 13:31