NaN variance with a or b as infinity

[JJMorton]

Hi,

It seems that when a or b are defined as math.inf, the variance becomes undefined. Unless I am missing something, the variance should still be well defined for these cases

For example, comparing to the equivalent scipy function scipy.stats.truncnorm:

• scipy.stats.truncnorm(a=-0.5, b=math.inf, loc=7500, scale=15000) gives variance 1.1E+08
• TruncatedNormal(a=-0.5, b=math.inf, loc=7500, scale=15000) gives variance NaN

I believe it comes down to the line

self._lpbb_m_lpaa_d_Z = (self._little_phi_b * self.b - self._little_phi_a * self.a) / self._Z

Little phi is correctly calculated as zero for the infinite limit, but 0 * math.inf results in a NaN.

As a quick fix, I replaced infinities with zeros in the line above, for example:

self._lpbb_m_lpaa_d_Z = (self._little_phi_b * (self.b if self.b != math.inf else 0) - self._little_phi_a * (self.a if self.a != math.inf else 0)) / self._Z

[toshas]

Hi, thanks for spotting this, Is it handled similarly in scipy? If yes, could you perhaps file a pull request? Originally I wanted to maintain differentiability of moments wrt the parameters, but I'm guessing a and/or b being -inf and inf respectively is a reasonable case which should allow backprop only wrt finite parameters of the distribution. Thanks, Anton

[JJMorton]

I'm not familiar enough with scipy to be sure how it internally handles cases like this, but this fix yields the same behaviour as far as I've tested. I'll submit a pull request.

[toshas]

Fixed in #5