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I am interested in finding an analytic expression for the below cdf of an order distribution, whose form I now describe.

Suppose that the time until a particular process occurs can be modelled as the sum of two gamma random variables: $\tau=t_1+t_2$, where $t_1\sim\text{gamma}(a,b)$ and $t_2\sim\text{gamma}(c,d)$ and $a\neq c$ and $b\neq d$.

In Mathematica, I can get a distribution that describes $\tau$ using:

bDist = TransformedDistribution[ u + v, {u \[Distributed] GammaDistribution[a, b], v \[Distributed] GammaDistribution[c, d]}]

Now suppose I have $n$ independent random variables of the above composite gamma, each of which is $\tau_i$ (that is, in Mathematica, each distribution is described by bDist). Further, suppose that I am interested in $T=\text{min}(\tau_1,...,\tau_n)$, that is, the time taken for the first such process to occur.

$T$ is the 1st order statistic of the composite distribution; that is, in Mathematica:

cDist = OrderDistribution[{bDist, n}, 1];

For a particular set of parameters, I can use simulation to estimate what the cdf for the above distribution looks like:

temp = RandomVariate[ cDist /. {a -> 8, b -> 0.5, c -> 3, d -> 2, n -> 5}, 10000];

approxDist = SmoothKernelDistribution[temp];

Plot[CDF[approxDist, x], {x, 0, 20}]

which produces a curve like:

enter image description here

My question is, is there a way to derive an analytic expression for the above curve for any set of parameters?

I have tried:

Assuming[n >= 1 && n \[Element] Integers && a > 0 && b > 0 && c > 0 && d > 0, CDF[cDist, x]]

But Mathematica just returns this unevaluated (after much thought).

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    $\begingroup$ The order distribution needs the CDF for bDist but it is unlikely that there is an analytic expression for that. $\endgroup$ – JimB Jan 26 '20 at 0:08
  • $\begingroup$ I know you want $b\neq d$ but if you allow $b=d$, then there is an analytic expression. See stats.stackexchange.com/questions/252191/…. From that link I think the answer is $1-\left(\frac{\Gamma (a+c,x/b)}{\Gamma (a+c)}\right)^n$. $\endgroup$ – JimB Feb 16 '20 at 22:05
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Well maybe one can find a general form for the cdf and pdf in terms of n given specified values of $a$, $b$, $c$, and $d$. Using your example the PDF's for n=2 to 6 are the following:

bDist = TransformedDistribution[u + v, 
  {u \[Distributed] GammaDistribution[a, b], 
   v \[Distributed] GammaDistribution[c, d]}];

The pdf for a single observation from bDist is given by

Simplify[PDF[bDist, z] // FunctionExpand, Assumptions -> z > 0]
(* (b^-a d^-c E^(-(z/d)) z^(-1 + a + c)*
   Hypergeometric1F1[a, a + c, (-(1/b) + 1/d) z])/Gamma[a + c] *)

So using the definition of the pdf and cdf for the minimum of a sample size of n we can write the following function that gives both the pdf and cdf:

pdfcdf[n_, a_, b_, c_, d_] := Module[{pdf0, cdf0, bDist},
  pdf0 = (b^-a d^-c E^(-(z/d)) z^(-1 + a + c)*
      Hypergeometric1F1[a, a + c, (-(1/b) + 1/d) z])/Gamma[a + c] //  FullSimplify;
  cdf0 = Integrate[pdf0, {z, 0, z0}, Assumptions -> z0 > 0] /. z0 -> z;
  {n pdf0 (1 - cdf0)^(n - 1) // FullSimplify, 1 - (1 - cdf0)^n}]

Using your example:

pc = pdfcdf[n, 8, 1/2, 3, 2]

[![PDF and CDF in terms of sample size n][1]][1]

Plot[pc[[2]]/.n->5, {z, 0, 20}, Frame -> True]

CDF of minimum order statistic for sample size of 5

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