I have the minimum value of N iid random variables that are gamma-distributed. The parameters of the gamma distribution are known. What would be the distribution of the distance of the remaining N - 1 values from the minimum in terms of the parameters and the minimum value?

For the problems at hand, N is small (<10). If the general problem doesn't have a solution, how could it be solved for N = 3 or 4?


1 Answer 1


I think I just figured it out. For $f(x) = Gamma(\alpha, \beta, x)$, if the minimum of the $N$ values is $x_m$, then the remaining values will still follow the same gamma distribution, but with the left tail trimmed at $x_m$. The resulting distribution will have to be normalized by area of the trimmed gamma, which is $1- F(x_m)$. The resulting PDF is then:

$f(x)/(1- F(x_m))$, $x\in[x_m,\infty)$

The distribution of the distance ($\delta_n=x_n-x_m$) of these points from the minimum will be:

$f(\delta+x_m)/(1- F(x_m))$, $\delta\in[0,\infty)$

  • $\begingroup$ Could you explain how you drew this conclusion? It's not even clear what your notation means, because it appears to try to define an $N-1$-variate distribution using a function of just one real variable! $\endgroup$
    – whuber
    Jul 30, 2018 at 14:01
  • $\begingroup$ @whuber Sure. Maybe you can help with my notation. The remaining values came from the same gamma distribution, but given the minimum, they will follow kind of a left-trimmed gamma (hence the $x\in[x_m,\infty)$), which has to be normalized with the compliment of the CDF of the original gamma at $x_m$. $\endgroup$
    – jblood94
    Jul 30, 2018 at 14:20
  • $\begingroup$ This doesn't look like a mere notational problem: indeed, your solution doesn't even integrate to unity, as any PDF must. Most likely you meant to divide by the complementary CDF rather than multiply by it. But what is the rationale for proposing this solution? $\endgroup$
    – whuber
    Jul 30, 2018 at 14:49
  • 1
    $\begingroup$ @whuber Yes. Of course. I got myself confused. I reversed the edit. It's back to being divided now. I updated the answer with the reasoning. $\endgroup$
    – jblood94
    Jul 30, 2018 at 17:10

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