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Is it possible to define a generic probability distribution function (pdf) so Mathematica can use its properties to simplify expressions?

For example:

Simplify[Integrate[f[x],{x,-Infinity,Infinity}]]

How to tell to Mathematica that f[x] is a PDF so it can simplify (in this case) to 1?

This is easy for a PDF from a given distribution, but my question is about the case of a generic PDF f[x]

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2 Answers 2

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ProbabilityDistribution does precisely that.

https://reference.wolfram.com/language/ref/ProbabilityDistribution.html

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  • $\begingroup$ Yes, it's true that you can use something like ProbabilityDistribution[f[y], {y, -Infinity, Infinity}] for a generic f[x], and it works. I didn't see that on the examples, so I assumed that was not possible. Do you know if is it possible to assume further conditions, for example, that f[x] is a continuous distribution? So it can simplify even further (for example, Simplify[ Integrate[f[y], {y, -Infinity, 10}] + Integrate[f[y], {y, 10, Infinity}]]) $\endgroup$
    – Borelian
    Nov 24, 2022 at 12:17
  • $\begingroup$ Just define your f(x) to be continuous. $\endgroup$ Nov 24, 2022 at 18:04
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$Version

(* "13.1.0 for Mac OS X x86 (64-bit) (June 16, 2022)" *)

Clear["Global`*"]

Use TagSet or TagSetDelayed to associate the integral to the symbol f, e.g.,

f /: Integrate[f[x_], {x_, -Infinity, Infinity}] = 1;

Then

Integrate[f[y], {y, -Infinity, Infinity}]

(* 1 *)

EDIT:

Mathematica doesn't combine integrals unless explicitly told to do so. Define a rule

intRule = Integrate[expr_, {t_, a_, b_}] +
    Integrate[expr_, {t_, b_, c_}] :>
   Integrate[expr, {t, a, c}];

Applying the rule,

Integrate[f[y], {y, -Infinity, 10}] + 
  Integrate[f[y], {y, 10, Infinity}] /. intRule

(* 1 *)
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  • $\begingroup$ It works for that integration, but not for other. For example, Simplify[ Integrate[f[y], {y, -Infinity, 10}] + Integrate[f[y], {y, 10, Infinity}]] does not simplify to 1. $\endgroup$
    – Borelian
    Nov 24, 2022 at 12:08

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