normal_lcdf and normal_lccdf give infinite gradients for infinite inputs. #2881

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1. ASKurz commented on Mar 6, 2023

   

   ASKurz

   on Mar 6, 2023

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   Thank you for opening this issue, @jsocolar.
2. syclik commented on Mar 7, 2023

   

   syclik

   on Mar 7, 2023

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   @jsocolar, thanks for bringing this up. I'm putting up a minimal example in C++ to walk through it.

   If you pop this into a file called test/unit/math/normal_lcdf_test.cpp, you'll be able to run it from the command line like: python runTests.py test/unit/math/normal_lcdf_test.cpp

   #include <stan/math.hpp>
   #include <gtest/gtest.h>
   
   TEST(normal_lcdf, inf_y) {
     double y = stan::math::INFTY;
     stan::math::var mu = 0.0;
     stan::math::var sigma = 1.0;
   
     stan::math::var target = stan::math::normal_lcdf(y, mu, sigma);
   
     EXPECT_FLOAT_EQ(0.0, target.val());
   
     target.grad();
   
     EXPECT_FLOAT_EQ(0.0, mu.adj());
     EXPECT_TRUE(isnan(sigma.adj()));
     
   
     stan::math::recover_memory();
   }

   When y is infinite (and given that mu and sigma are finite, here ensured by declaring them as parameters), then the gradient is zero. However, it seems that Stan yields infinite gradients here.

   Just curious... how'd you come to this conclusion? And which gradient did you think was "infinite"? (The gradient has two elements.) It'd be good to know how you're thinking about what the math library does and how it's generating gradients, especially at the boundary conditions.

   To get into the example, if you look at the test:

   • d normal_lcdf(y, mu, sigma) / dmu = 0
   • d normal_lcdf(y, mu, sigma) / dsigma = NaN

   (NaN does not equal infinity, but it doesn't mean that's good.)

   I think you're expecting d / dsigma = 0? Is that right?

   I didn't trace through to figure out why it's having trouble computing that term, but I'm sure it could be addressed. I'd lean towards throwing an exception at the boundaries, but I think that would change the behavior of Stan in a way that we'd have to have a larger discussion to implement.

   Thoughts?

   👍React with 👍1Reacted by Jacob B. Socolar
3. jsocolar commented on Mar 7, 2023

   

   jsocolar

   on Mar 7, 2023

   Last edited by jsocolar

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   @syclik Thanks for taking a look!

   Just curious... how'd you come to this conclusion? And which gradient did you think was "infinite"? (The gradient has two elements.) It'd be good to know how you're thinking about what the math library does and how it's generating gradients, especially at the boundary conditions.

   I concluded that the gradient should be zero because that's the well defined limit of the gradient with respect to both mu and sigma as y approaches infinity. I concluded (incorrectly I think) that Stan's gradient was infinite (correct conclusion: not finite) by running the Stan program from the OP via cmdstanr with

   lcdf_mod <- cmdstan_model("/Users/JacobSocolar/Desktop/lcdf.stan")
   lcdf_mod$sample(data = list(y = Inf))
   

   and getting back a bunch of

   Chain 1 Rejecting initial value:
   Chain 1   Gradient evaluated at the initial value is not finite.
   Chain 1   Stan can't start sampling from this initial value.
   

   Based on your follow-up, I assume that the problem is the partial wrt sigma, that the NaN is the problem, and that when I said "infinite" I really meant "not finite".

   The use case for returning zero here is given in the discourse thread linked from the OP. It shouldn't be particularly burdensome to code around the current behavior, which I agree isn't necessarily wrong, but if there's appetite on the Stan side for returning zero for the partial wrt sigma then that'd also solve the original discourse issue.
4. avehtari commented on Oct 7, 2026

   

   avehtari

   on Oct 7, 2026

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   This is fixed by #3406 (not yet merged)

   👍React with 👍1Reacted by Jacob B. Socolar

   😄React with 😄1Reacted by A. Solomon Kurz