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If I have the identity

$$f(a)f(b)=f(a')f(b')$$

for a given normalized probability distribution $f$, and additionally the constraint

$$a+b = a'+b' = const.$$

for any suitable pairs of real numbers $a,b$ and $a',b'$, would this be sufficient to conclude that

$$f(a)f(b)=f(a+b)$$

or are there further conditions required?

I know that the latter relationship implies that $f$ is an exponential function, but I would like to know which conditions are mathematically required for this, and I can't quite see how this would follow from the two conditions alone I mentioned.

Thomas
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    this is very vague. What is $f$? What does "suitable" mean? What does this have to do with probability? – lulu Mar 17 '24 at 16:07
  • @lulu Well, $f$ is a probability distribution, and "suitable" pairs $a,b$ and $a′,b$′ means those that satisfy the constraint $a+b=a′+b′$ – Thomas Mar 17 '24 at 16:13
  • I suggest editing your post for clarity. You don't, for instance, mention that $f$ is a probability distribution in the post. If you have an example of a probability distribution with this property, you should include it. – lulu Mar 17 '24 at 16:17
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    If you have the condition $f(0)=1$ then $f(a)f(b)=f(a+b)f(0)=f(a+b)$. – Shean Mar 17 '24 at 16:26
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    @Shean But by the same token, the desired relation is false unless $f(0)=1$ – lulu Mar 17 '24 at 16:27
  • @lulu I am trying to find some simple mathematical conditions sufficient for a function to be a normal (exponential) distribution. I know that $f(0)=1$ does it, but this is not something I can assume from the start, it should rather come out as a result. – Thomas Mar 17 '24 at 16:41

1 Answers1

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$$f(a)f(b)=f(a+b)f(0)=f(a+b)\iff f(0)=1$$

Hence, $f(a)=c^a$ for some constant $c$ (see e.g. A function with a property $f(x+y)=f(x)f(y)$).

Shean
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  • Sorry, but how do you get to this equation "$f(a)f(b)=f(a+b)f(0)$"? – Thomas Mar 17 '24 at 16:57
  • Let $a'=a+b$ and $b'=0$. Then $a+b=a'+b'$ satisfying your constraint. – Shean Mar 17 '24 at 16:59
  • and the second equal sign $=f(a+b)$? – Thomas Mar 17 '24 at 17:13
  • But there are no functions of the form $x\mapsto c^x$ which integrate to $1$ over the entire real line. – lulu Mar 17 '24 at 18:44
  • @Thomas $f(0)=1$. – Shean Mar 17 '24 at 20:52
  • So $f(0)=1$ would be a further condition to have $f(a)f(b)=f(a+b)$, not something that follows from the other conditions mentioned? This is not really what I was looking for, but it would also be sufficient for me if I can rule out $f(0)=0$. From your above equation $f(0)=0$ would mean $f(a)f(b)=0$ so either $f(a)=0$ or $f(b)=0$ or both, which however would be impossible unless $f$ is identically zero. Would you agree? – Thomas Mar 17 '24 at 21:43
  • @lulu $e^{-x^2}$ integrates to 1 over the real line (if suitably normalized), as does $e^{-x}$ for $x>0$. – Thomas Mar 17 '24 at 21:54
  • $f(0)=1$ is a sufficient and necessary condition, thats the $\iff$ in my answer. If $f(0)$ you can show $f$ is identically equal to 0, yes. Also, $f(x)=e^{-x^2}$ wouldn't work. Do you see why? – Shean Mar 18 '24 at 07:26
  • The equal sign in the equation $f(a+b)f(0)=f(a+b)$ I don't understand. You probably meant $f(a+b)f(0)\overset!=f(a+b)$, but this would mean enforcing the equation. I would like to derive it by means of the conditions/ constraints mentioned.

    I know what you mean regarding $f(x)=e^{-x^2}$ . In this case just replace the $a$ and $b$ with $a^2$ and $b^2$ in the above equations.

     – Thomas Mar 18 '24 at 18:49
  • $f(a+b)f(0)=f(a+b)$ if and only if $f(0)=1$. This condition ($f(0)=1$) must be satisfied, otherwise you will not have your desired conclusion $f(a)f(b)=f(a+b)$. – Shean Mar 19 '24 at 14:55
  • If $f(x)=e^{-x^2}$ then $f(a)f(b)=e^{-a^2}e^{-b^2}=e^{-a^2-b^2}$, but $f(a+b)=e^{-(a+b)^2}=e^{-a^2-2ab-b^2}\neq f(a)f(b)$. – Shean Mar 19 '24 at 14:57