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torch/distributions/binomial.py
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| def param_shape(self): | ||
| return self._param.size() | ||
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| def _log1pmprobs(self): |
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Since it is a function for internal use, I think this can be moved to the top, like in MVN. Something like:
def _log1pmtensor(tensor):
# Do the same thingUses of the function in kl.py can be done via importing this function along with Binomial.
torch/distributions/binomial.py
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| values = values.expand((-1,) + self._batch_shape) | ||
| return values | ||
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| def _Elnchoosek(self): |
torch/distributions/binomial.py
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| s = self.enumerate_support() | ||
| s[0] = 1 # 0! = 1 | ||
| # x is factorial matrix i.e. x[k,...] = k! | ||
| x = torch.cumsum(s.log(), dim=0) |
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x is the log of factorial matrix right?
torch/distributions/binomial.py
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| indices[0] = torch.arange(x.size(0) - 1, -1, -1, | ||
| dtype=torch.long, device=x.device) | ||
| # x[tuple(indices)] is x reversed on first axis | ||
| lnchoosek = x[-1] - x - x[tuple(indices)] |
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I think x.flip(dim=0) will exhibit same behaviour.
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weird, I tried using flip and it didn't work before- maybe I messed with arguments...
torch/distributions/binomial.py
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| elognfac = x[-1] | ||
| elogkfac = ((lnchoosek + s * self.logits + self.total_count * self._log1pmprobs()).exp() * | ||
| x).sum(dim=0) | ||
| elognmkfac = ((lnchoosek + s * self.logits + self.total_count * self._log1pmprobs()).exp() * |
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E[log(n-k)!] = E[log k!] but for Bin(n, (1 - p)). Can we use this fact here?
torch/distributions/kl.py
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| return kl | ||
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| @register_kl(Binomial, Poisson) |
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Heterogeneous combinations were placed below. This section was for homogeneous combinations.
torch/distributions/kl.py
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| @register_kl(Binomial, Geometric) |
torch/distributions/kl.py
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| return -p.entropy() - torch.log1p(-q.probs) / p.probs - q.logits | ||
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| @register_kl(Geometric, Binomial) |
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@vishwakftw thanks for the comments! One thing I want us to work out before wrapping this up is an approximation to E[logk!] for large n, I tried Stirling's but couldn't come up with a closed form. Any ideas? |
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I think we have to make use of Stirling's inequality and the Taylor series to compute this. I guess the reason you are unable to come up with a closed form is because of the log (k) term. I tried using them, and got about 0.5% relative error.
This might help after the expansion of log k! <= 1 + klog k + 0.5 log k - k <source: wikipedia: https://en.wikipedia.org/wiki/Taylor_expansions_for_the_moments_of_functions_of_random_variables> |
vishwakftw
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vishwakftw
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Looks good to me!! @fritzo what do you think?
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Also, are you going to try the large |
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@vishwakftw btw I tried it with 0.01 precision as well. 2 things on my wishlist:
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@alicanb I have a closed form solution for E[log x!], E[log (n - x)!] and E[log n!] (this is simply log n!) for large |
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Great, have you experimented with any large |
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This is the gist for the approximations. I ran some tests: n = {10, 20, 50, 75, 100} and p = {0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9} |
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btw |
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This is a larger PR than I intended but basically it adds binomial entropy and binomial-poisson and binomial-geometric KL with some helper functions:
binomial._log1pmprobs: I used this a lot so I made it a separate function. it calculates(-probs).log1p()safely.binomial._Elnchoosek(): for x~Bin(n, p), this calculates E[log(nchoosek)], E[log(n!)], E[log(x!)], E[log((n-x)!)]