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Could someone please clarify for me the support of the gamma distribution?

According to Wikipedia (and other sources), the gamma distribution is only supported for $x>0$. However, according to Wikipedia again, the exponential distribution is a special case of the gamma distribution with the parameter $k=1$, although the exponential distribution is supported for $x \ge 0$. Does this imply that if the parameter $k$ of the gamma distribution has a certain value, say 1, that its support extends from $x>0$ to $x\ge 0$?

I would be glad if anyone could clarify this for me.

http://en.wikipedia.org/wiki/Exponential_distribution http://en.wikipedia.org/wiki/Gamma_distribution

kjetil b halvorsen
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Fridtjolfo Di Certnizie
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    What difference would it make to include $0$ in or exclude $0$ from the support, given that this has zero probability in any case? – whuber Aug 22 '13 at 19:44
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    It may be worthwhile noticing that the Wikipedia article on the Gamma distribution specifies a support (to wit, $(0,\infty)$) that *disagrees* with the definition of [support of a distribution](http://en.wikipedia.org/wiki/Support_(mathematics)#Support_of_a_distribution) elsewhere on Wikipedia: according to that definition, the support is always a closed set. I suspect the authors of the Gamma article just didn't want to keep specifying that certain functions, like the PDF, are not defined at $0$ for certain values of the shape parameter. – whuber Aug 22 '13 at 21:36
  • thank you very much for your helpful answer! However, I think it would make a difference if 0 is included in the support, as for x=0 the value for the PDF of the exponential distribution is lambda, and not 0. – Fridtjolfo Di Certnizie Aug 26 '13 at 09:38
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    What difference do you think this makes? After all, the PDF is a *density*. Its value at any given $x$ means absolutely nothing by itself and acquires any meaning only in using it to measure *areas* under the curve. Please see http://stats.stackexchange.com/questions/4220/a-probability-distribution-value-exceeding-1-is-ok/4223#4223 for a more elaborate explanation. – whuber Aug 26 '13 at 15:11

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Usually, the support of a distribution is defined to always be a closed set, so in the case of the gamma distribution, $[0,\infty)$. Even if the density function defined by some formula, for some parameter values, then is undefined, that is not a problem. The reason is that density functions are not really functions! Their values at any given point do not give meaning, they have meaning only through being integrated.

So, in one point of view, densities are equivalence classes of functions, two functions $f, g$ being equivalent if they give the same probabilities (that is, integrals) for all events $A$: $\int_A f(x)\; dx = \int_A g(x)\; dx$ for all events $A$. So the value you assign to the density function at zero do not matter (This is the $L^1$ view.) See also Can a probability distribution value exceeding 1 be OK? for discussion.

Another point of view (leading to the same conclusions) is that densities are differential forms, see Intuitive explanation for density of transformed variable?

kjetil b halvorsen
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