8
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Bug introduced in 8.0 or earlier and persisting through 13.2.0


I've noticed this strange behavior and I'm wondering if it's a bug.

I define a Cauchy distribution:

c = CauchyDistribution[0, 1];

If I evaluate Mean[c], I get Indeterminate, as expected.

If I evaluate Expectation[x, x \[Distributed] c], I get Expectation[x, x \[Distributed] CauchyDistribution[0, 1]]. I would have expected Indeterminate, too, but that's ok.

However, if I evaluate Expectation[x + y, {x \[Distributed] c, y \[Distributed] c}], I get 0.

The sum of two independent Cauchy distributed random variables should be another Cauchy distributed random variable, right? Why is the expectation 0?

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    $\begingroup$ Another example of why I think it is best to avoid using prepackaged written up results , and go for general symbolic methods. Indeed, if you ask Mma to find the general solution, it works perfectly. Here: f = (1/(Pi (1 + x^2))) (1/(Pi (1 + y^2))); and Integrate[(x + y)*f, {x, -Infinity, Infinity}, {y, -Infinity, Infinity}] returns the appropriate 'does not converge' message. $\endgroup$
    – wolfies
    Dec 10, 2016 at 6:19

1 Answer 1

8
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You can check that :

TransformedDistribution[ x + y, {x \[Distributed] CauchyDistribution[0, 1], 
                                y \[Distributed] CauchyDistribution[0, 1]}]

(* CauchyDistribution[0, 2] *)  

Expectation[z, z \[Distributed] CauchyDistribution[0, 2]]
(* Expectation[z, z \[Distributed] CauchyDistribution[0, 2]] *)

Mean[TransformedDistribution[ x + y, {x \[Distributed] c, y \[Distributed] c}]]
(* Indeterminate *)
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    $\begingroup$ So, the conclusion is it's a bug? $\endgroup$ Jan 11, 2013 at 12:19
  • $\begingroup$ @SjoerdC.deVries Yes it looks like it. $\endgroup$ Jan 11, 2013 at 12:42
  • $\begingroup$ @AndyRoss can you confirm this diagnosis? $\endgroup$
    – rcollyer
    Jan 11, 2013 at 15:48
  • $\begingroup$ @rcollyer I was hoping for news from him too. $\endgroup$ Jan 11, 2013 at 16:31

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