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Let $T$(Talent Distribution) and $E$(Environment) be independently and uniformly distributed random variables over the interval [0,1] and [0,$c_1$] respectively. Considering $c_1>1$, I have now found out a distribution for $X_A=T+E$ for which I followed this blog:

$f_{X_A}(x){}={\bf{1}}_{\left\{0\le x \le 1\right\}}\dfrac{z}{c_1} +{\bf{1}}_{\left\{1 < z \le c_1\right\}} {}+{}{\bf{1}}_{\left\{c_1 < z \le c_1+1\right\}}\dfrac{c_1+1-z}{c_1}\,.$

Similarly, if we consider another distribution $X_B=T+E'$ where $E'$ is uniformly distributed between $[0,c_2]$ such that:

$f_{X_B}(x){}={\bf{1}}_{\left\{0\le x \le 1\right\}}\dfrac{z}{c_2} +{\bf{1}}_{\left\{1 < z \le c_2\right\}} {}+{}{\bf{1}}_{\left\{c_2 < z \le c_2+1\right\}}\dfrac{c_2+1-z}{c_2}\,.$

Then I am interested in finding the PDF for $X_A-X_B$.

Being new to probability distributions, I find it hard to compute it easily. Any thoughts appreciated.

Karan Deep Singh
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  • They are not "id" (identically distributed) in "iid" since $c_1\neq 1$. – user10354138 Feb 25 '21 at 04:07
  • Yes, you are right. My bad. I fixed it. Thank you. – Karan Deep Singh Feb 25 '21 at 04:21
  • $X_A-X_B=E-E'$, right? It has mean $\frac12(c_1-c_2)$, and once the mean is subtracted and $E,E'$ are symmetric around $0$, there's no difference between $E-E'$ and $E+E'$. So the distribution will be of exactly the same type as the two distributions in the OP. – Greg Martin Feb 25 '21 at 04:29
  • Actually, $X_A-X_B = (T - T) + (E-E')$, and I believe we simply cannot ignore $T-T$ since the subtraction of two uniform random variables is not $0$. Wolfram alpha explains this pretty well. https://mathworld.wolfram.com/UniformDifferenceDistribution.html – Karan Deep Singh Feb 25 '21 at 12:39

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