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In particular, it seems that the gamma random variable, with pdf (from wikipedia):

enter image description here

becomes $\infty$ when $x=0$ and $\alpha<1$

Does it make sense for a pdf to take the value $\infty$ at some value? Wouldn't that mean that value would have occur with probability 1? And if that is the case, the pdf should not be able to take on any other value. However, that isn't the case for the gamma pdf given above (when $\alpha<1$).

Or maybe I am missing something? Any insight is appreciated!

casandra
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    I would prefer to call it undefined. Here is an example. Let $f(x)=\frac{1}{2}x^{-1/2}$ on $(0,1)$, and $0$ elsewhere. The improper integral $\int_0^1 f(x),dx$ exists, and is in fact equal to $1$. Density function is not probability. And you can change density at any particular point you like and not affect probabilities. – André Nicolas Feb 25 '14 at 04:42
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    The density is only defined for $x>0$, this should be a typo (cf. top right of the Wikipedia page to see the support of the distribution). – Clement C. Feb 25 '14 at 04:42

1 Answers1

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It might be worth familiarizing yourself with the Dirac Delta distribution, as it's the canonical example of this sort of behavior. For instance, imagine a random variable $f$ (over $\mathbb{R}$, not just $\mathbb{Z}$ or $\{0,1\}$) that's $0$ with probability $\frac12$ and $1$ with probability $\frac12$. Then the CDF $F(x)$ will be $0$ for $x\lt 0$, $\frac12$ for $0\leq x\lt 1$, and $1$ for $1\leq x$. Looking at the distribution, there will be infinite spikes at both $0$ and $1$; the pdf will be a sum $\frac12\delta(x,0)+\frac12\delta(x,1)$ (note that the normalization here is necessary to make sure that the cumulative is a proper CDF!).

The key is that a probability density function will 'be infinite' anywhere the cumulative distribution has a jump discontinuity - but that jump discontinuity doesn't need to go from $0$ to $1$; it doesn't even need to be discrete, as the variable could be continuous on either side of the jump cut. It does imply that there's a finite probability that the variable takes on whatever value the jump cut occurs at, though.

Steven Stadnicki
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