Suppose $F$ is a cdf for some random variable on some support, and that $a,b$ are constants with $a<1<b$. I'm hoping to find a distribution such that: $$F \left( \frac{x}{b} \right) - F \left( \frac{x}{a} \right) = \text{constant}$$ i.e. that this difference is independent of $x$.

Intuitively, as you increase $x$, the slope of the CDF has to decrease at just the right speed to constantly outweigh the shrinking size of the interval $\left( \frac{x}{b}, \frac{x}{a} \right)$.

If it helps, in my use-case: $a = \frac{1}{1+c}, b = \frac{1}{1-c}$ with $c\in(0,1)$.

I'm not sure of where to begin here. Is there a way to prove that such a distribution does/does not exist? Is there a well-known distribution that already satisfies this? Is there way to simulate this?

Thank you!

  • 1
    $\begingroup$ Let $x$ grow large: the left hand side must approach $1-1=0$ (presuming $a\gt 0$), so the constant can only be zero. Perhaps you intended to apply some qualifications to your condition, such as that $x$ should be restricted to some particular interval?? $\endgroup$
    – whuber
    Jan 13, 2021 at 17:09

1 Answer 1


Suppose $F$ has a derivative $f$ and the support is all real numbers.
Then, $f(\frac{x}b) \frac{1}b=f(\frac{x}a) \frac{1}a$ and with $x=a y$, $f(\frac{a y}b) \frac{a}b=f(y)$.
Let $r=\frac{a}b$.
Then, $r f(r x)=f(x)$ for all $x$.
In particular $r^2 f(r^2 x)=r (r f(r (r x)))=r f(r x)=f(x)$ for all $x$.
By induction, $r^n f(r^n x)=f(x)$ for all $x$ and for any positive integer $n$.
Thus, I would try $f(x)=\gamma x^{-1}$ for some constant $\gamma$.
That will have the property that $r^n f(r^n x)=r^n \gamma (r^n x)^{-1}=f(x)$.
But, a density has to be non-negative and it has to integrate to 1. So, the support will have to be restricted to $[A,B]$ where $0<A<B$.
The distribution function will be

$F(x)=\frac{\log(x/A)}{\log(B/A)}$ for $A<x<B$;
$F(x)=0$ for $x\le A$;
$F(x)=1$ for $x\ge B$.

If $x/b$ and $x/a$ are between $A$ and $B$, then

This is constant. However if either $x/b$ or $x/a$ are outside $[A,B]$, then the invariance property will not hold.
To summarize, choose any $A>0$ close to $0$ and choose $B>0$ very large such that $Ab<Ba$. Then, for all $Ab<x<Ba$ the invariance property will hold with the distribution function defined above. The invariance property will not hold for points in the support $[A,B]$ that are not within the interval $[Ab,Ba]$.

  • $\begingroup$ This is brilliant, thank you so much. How did you go from the line "By induction..." to "Thus, I would try..." ? I see that the property in the first line does hold in the suggestion in the second; but I can't reconstruct your thought process for making that jump. Did you just rely on your knowledge of the properties of the function f(x)=1/x ? Thanks again! $\endgroup$
    – bhalperin
    Jan 14, 2021 at 18:32
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    $\begingroup$ It was the only function I could think of that would have that property. $\endgroup$
    – John L
    Jan 14, 2021 at 19:25
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    $\begingroup$ I think $f(x)$ could be defined arbitrarily first for $1 \le x<r$. Then, the induction formula would define $f(x)$ for all other $x$. Then, you have to multiply by the right constant to make it integrate to 1. $\endgroup$
    – John L
    Jan 14, 2021 at 20:27

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