9
$\begingroup$

It is well known that a random variable being Gamma distributed with integer shape parameter $k$ is equivalent to the sum of the squares of $k$ normally distributed random variables.

But what can I say about a gamma distributed random variable with non-integer $k$? Is there any other interpretation other than the Gamma distribution at all?

$\endgroup$
  • 5
    $\begingroup$ Gamma with shape parameter $k/2$ is the sum of the squares of $k$ normally distributed random variables. Gamma with shape parameter $k$ is the sum of $k$ iid exponential distributions. $\endgroup$ – Greenparker May 3 '16 at 14:19
  • 2
    $\begingroup$ One more interpretation of the gamma with integer $k$: it's the waiting time until the $k$th arrival in a one-dimensional Poisson process with intensity $1/\theta$. $\endgroup$ – S. Kolassa - Reinstate Monica May 3 '16 at 14:34
1
$\begingroup$

If $X\sim{\cal G}(\alpha,1)$ and $Y\sim{\cal G}(\beta,1)$ are independent then$$X+Y\sim{\cal G}(\alpha+\beta,1)$$ In particular, if $X\sim{\cal G}(\alpha,1)$, it is distributed with the same distribution as $$X_1+\cdots+X_n\sim{\cal G}(\alpha,1)\qquad X_i\stackrel{\text{iid}}{\sim}{\cal G}(\alpha/n,1)$$for any $n\in\mathbb N$. (This property is called infinite divisibility.) This means that, if $X\sim{\cal G}(\alpha,1)$ when $\alpha$ is not an integer, $X$ has the same distribution as $Y+Z$ with $Z$ independent from $Y$ and $$Y\sim{\cal G}(\lfloor\alpha\rfloor,1)\qquad Z\sim{\cal G}(\alpha-\lfloor\alpha\rfloor,1)$$It also implies that integer valued shapes $\alpha$ have no particular meaning for Gammas.

Conversely, if $X\sim{\cal G}(\alpha,1)$ with $\alpha<1$, it has the same distribution as $YU^{1/\alpha}$ when $Y$ is independent from $U\sim{\cal U}(0,1)$ and $$Y\sim{\cal G}(\alpha+1,1)$$And hence the distribution ${\cal G}(\alpha,1)$ is invariant in $$X \sim (X'+\xi)U^{1/\alpha}\qquad X,X'\sim{\cal G}(\alpha,1)\quad U\sim{\cal U}(0,1)\quad \xi\sim{\cal E}(1)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.