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I'm trying to gain intuition for how to scale the parameters of the Gamma distribution when the desired variance is not known. I'll make up an example:

Let $\lambda$ be the number of 100-degree days in Austin, Texas per year. From historical observation, $\lambda \sim$ Gamma(18, 0.75).

If I want to estimate the probability that there will be 250 100-degree days in Austin over the next 10 years, how do I know what Gamma distribution to use? Intuitively, I have at least two reasonable choices:

  1. Set the second parameter of the Gamma distribution to be "10 years", i.e. Gamma(2400, 10)
  2. Linearly scale the first parameter to be 10 * the number of events, i.e. Gamma(180, 0.75)

If I knew what the variance of the distribution was supposed to be, then this question would be trivial. If there is no predefined variance, is there a way of identifying which of these two distributions is more reasonable?

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  • $\begingroup$ Note that "number of hundred degree days in a year" is a count (i.e. necessarily discrete), and one that has a strict upper bound (you can't have more hundred degree days than days in a year), so it cannot possibly have any Gamma distribution, despite the statement otherwise there. There has to be truncation and rounding (or similarly, some other way of dealing with the distinction between a continuous variable with semi-infinite support and a discrete variable with bounded support), leaving you with not-actually-a-gamma. $\endgroup$ – Glen_b Jul 20 '20 at 5:11
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Assuming the years are independent, the number of 100-degree days in Austin over the next 10 years will simply be $10\lambda$. The number of hot days sum up for each year over the 10 years.

Now if $\lambda\sim\mathrm{Gamma}(18,0.75)$, then $10\lambda\sim\mathrm{Gamma}(180, 0.75)$ (see here for a reference). In general, if $X_i\sim\mathrm{Gamma}(k, \theta)$, then $\sum_{i=1}^{N}{X_{i}}\sim\mathrm{Gamma}\left(Nk, \theta\right)$.

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  • $\begingroup$ Thanks - intuitively, it seems like the assumption of independence gives us a constraint on the variance, which makes this answer reasonable. Much appreciated! $\endgroup$ – m81 Jul 19 '20 at 15:43
  • $\begingroup$ If anyone else is interested in the intuition behind this answer, I found this post very helpful: stats.stackexchange.com/questions/51605/… $\endgroup$ – m81 Jul 19 '20 at 16:36

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