Accuracy of stats.rv_continuous integrating methods

[denkorzh]

I changed a bit code from scipy.stats.rv_continuous documentation to get a custom exponential distribution with parameter 0.5:

class CustomExpon(rv_continuous):
    """"custom exponential distribution with parameter 0.5"""
    def _pdf(self, x, *args):
        if x < 0.:
            return 0.
        else:
            return .5 * np.exp(- .5 * x)
custom_expon = CustomExpon(name='custom_expon')

But when I evaluate print custom_expon.mean() I get a number of InterationWarning and the result is 2.0576933609. (The correct answer is 2.)

I have read the issue #6185 and the advice to set range of distributon a=0. is helpful. But are there any other methods to increase the accuracy of integrating?

[ev-br]

This is really an issue of communicating the intent rather then accuracy.

On the technical level, if you do not define a=0, then your distribution is supposed to have the support of (-inf, inf). Then the numerical integration routine tries to evaluate the integral over the this interval, and finds it difficult (which is not too surprising in retrospect,cf #5428).

Now, the framework has a standard way of specifying the support of a distribution, by setting a and b. What you are doing, you're tricking it into believing that you want the support of (-inf, inf), and normalize the PDF for [0, inf].

As an aside, as soon as you do if x < 0, you lose vectorization (try custom_expon.pdf([1, 2, 3])).

Conclusion: setting a=0 is the correct thing to do. I do not believe there is a bug, so I'm closing this ticket. Feel free to keep discussing though (even though usage questions are better asked on StackOverflow or scipy-user mailing list)