What does this distribution look like?

I have data with the following empirical distribution . What would be a good model for it?

Edit: here is a closer view: • There is some good general advice at stats.stackexchange.com/questions/10517/….
– whuber
Aug 1 '11 at 20:44
• @whuber: what I really wanted here was to get names of possible matches to read up on them and see if any of them make sense in my application. Aug 1 '11 at 21:53
• Usually one does the opposite: the application suggests likely distributions. Not only that, it also determines how to check whether a distribution is a reasonable fit. For example, in some cases it's important to get a good fit to the right tail (the largest values), whereas in others it's important to get a fit that doesn't deviate too much at any percentage point. Thus, some amplification on your part concerning the application would be helpful.
– whuber
Aug 1 '11 at 21:57
• Any idea what the curve would look like if continued in the negative direction? (It obviously looks exponential to the right.) Aug 2 '11 at 12:07

You have strictly positive data that is clearly skewed, so you need a distribution that includes the possibility for skew. The Gamma distribution, which has density

$$p(x) = x^{k-1} \frac{e^{-x/\theta}}{\theta^k \, \Gamma(k)}\text{ for } x \geq 0\text{ and }k, \theta > 0$$

is probably the default choice in the situation like this. There are other choices for skewed data (e.g. Skew-normal, log-normal, skew-logistic, weibull) but the gamma is more commonly used, and is directly related to some of the other choices (skew-logistic, weibull).

• It's awfully hard to eyeball a distribution, but one can usually say what it's not. In this case, the right tail is so heavy that the Gamma shape parameter $k$ has to exceed 1, but the nonzero mode shows it has to be less than 1; ergo, no Gamma is going to be a reasonable approximation for most purposes.
– whuber
Aug 1 '11 at 20:45
• The tail length is also a function of $\theta$. You could have $k$ just barely larger than 1, say $1/.75$, while having $\theta$ just barely larger than 0, say, $1/25$, would create the long tail and the non-zero mode. In fact, a histogram of x=rgamma(10000,1/.8,1/25) looks pretty close to the one seen above. Aug 1 '11 at 21:27
• typo. look at the histogram of x=rgamma(10000,1/.75,1/25) Aug 1 '11 at 21:34
• Not even close. The histogram in the OP decays roughly like $1/x$ (not exponentially). Its tail simply won't match any Gamma accurately. Histograms are notoriously bad tools for judging distributions. Constructing the q-q plot will show you what's wrong.
– whuber
Aug 8 '11 at 17:44
• I didn't have the data set so I had nothing to compare to, which is why I used the histogram. Aug 8 '11 at 17:48