# Finding a distribution with a particular invariance property: F(x/b) - F(x/a) independent of x

Suppose $$F$$ is a cdf for some random variable on some support, and that $$a,b$$ are constants with $$a<1. I'm hoping to find a distribution such that: $$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!

• 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??
– whuber
Jan 13, 2021 at 17:09

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.
The distribution function will be

$$F(x)=\frac{\log(x/A)}{\log(B/A)}$$ for $$A;
$$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
$$F\left(\frac{x}b\right)-F\left(\frac{x}a\right)=\frac{\log((x/b)/A)}{\log(B/A)}-\frac{\log((x/a)/A)}{\log(B/A)}=\frac{\log(a/b)}{\log(B/A)}$$

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 $$Ab. Then, for all $$Ab 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 $$[Ab,Ba]$$.

• 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! Jan 14, 2021 at 18:32
• It was the only function I could think of that would have that property. Jan 14, 2021 at 19:25
• 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. Jan 14, 2021 at 20:27