# Class of density functions that f(x-c)/f(x) is non-increasing

I am looking for a class of probability density functions $f_X(x)$, with support $\subset [0,\infty)$, for which there exist a non-empty subset of $\mathbb{R}^+$ denoted by $S_X$, such that $\forall c\in S_X$, $\frac{f_X(x-c)}{f_X(x)}$ is a non-increasing function of $x$ over the intersection of supports of $f_X (x)$ and $f_X (x-c)$. Find each $f_X(x)$ with its associated $S_X$.

Clearly, the uniform distribution is one of them since $\frac{f_X(x-c)}{f_X(x)}=1$ for all $c >0$ and therefore, $S_X=\mathbb{R}^{+}$. The other one is exponential distribution $f_X(x)=\lambda e^{-\lambda x}$, since $\frac{f_X(x-c)}{f_X(x)}=e^{\lambda c}$ which is a constant with respect to x for all $c>0$. I have tried the generalized gamma function but it looks like it doesn't work for all cases. Any idea?

• A uniform distribution does not have support $[0,\infty)$ – Juho Kokkala Sep 9 '16 at 17:18
• @juhoKokkala You are right. I edited the problem statement. – Sus20200 Sep 9 '16 at 17:24

## 1 Answer

Your condition basically concerns the heaviness of the tail and so one example would be a power law distribution with density

$$f(x) = c x^{-\alpha}$$

for some $c > 0$, $\alpha > 1$ and $x > 0$. Now if we look at

\begin{align} \frac{f(x - c)}{f(x)} &= \frac{(x - c)^{-\alpha}}{x^{-\alpha}} \\ &= \left ( \frac{x}{x - c} \right )^\alpha \end{align}

we see that this is decreasing in $x$. I think you'll find that this holds for other heavy-tailed distributions but not for those with light tails.

• Thanks for your answer. This is a subset of the class of pdfs that work. Can we find the whole subset? – Sus20200 Sep 10 '16 at 15:41
• Well, your specification itself does define the whole set. Are you looking then for an equivalent definition? – dsaxton Sep 10 '16 at 19:21
• My specification defines a set, but does not give us the explicit probability density function. I need a complete set of explicit probability density functions that satisfy the condition (specification) I have written in the problem statement. Your answer, is a subset of that set. – Sus20200 Sep 10 '16 at 19:27