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In my research project is useful to classify cumulative distributions functions of random variables with support in $[a,b]$ with $0\le a<b\le+\infty$ depending on whether the ratio, $$\dfrac{F(x)}{x}$$ is increasing (or decreasing or constant) everywhere in [a,b].

Of course several random variables do not fit in this classification. Also if $0<a$ then the ratio can never be everywhere constant nor decreasing. My question is:

  1. Does the property of $F(x)/x$ being increasing has a known name?
  2. As we can interpret this ratio as a ratio of the CDF of our original random variable and the CDF of a Uniform $[0,b]$ if $b<\infty$, is there any known property/name regarding the ratio of two CDFs being increasing?
  3. If yes to 1 or 2, any references?
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    $\begingroup$ I wonder whether you have accurately described the problem, because when $0\lt a$ certainly the ratio can decrease. Take, for instance, $F(x)=1-\exp(-(x-1))$ for $1\le x\lt\infty$: $F(x)/x$ decreases for $x\gt 2.14619$ (approximately). This contradiction with one of your assertions suggests there may be some typographical errors that need to be fixed or at least some clarification of what you mean by "increasing" and "decreasing" would be in order. $\endgroup$
    – whuber
    Commented Oct 9, 2014 at 17:20
  • $\begingroup$ @whuber: I (wrongly) assumed it was clear that it was decreasing everywhere in the support. Your example, fails to be decreasing near 1. $\endgroup$ Commented Oct 10, 2014 at 14:35
  • $\begingroup$ Thanks. I am still puzzled, because the distinction between $0\le a$ and $0\lt a$ seems to be immaterial. You cannot have $F(0)\ne 0$ because $F(0)/0$ would be undefined, whence necessarily $0 \lt a$. That would seem to eliminate your "classification" altogether because it reduces to the single class of everywhere increasing! $\endgroup$
    – whuber
    Commented Oct 10, 2014 at 14:39
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    $\begingroup$ @whuber At zero one should use limits and also we could also allow for $+\infty$ as a value. For example, if $X$ is uniform $[0,b]$ then the ratio $F(x)/x$ is constant; if $X$ has CDF $F(x)=\sqrt{x}$ so $a=0$ and $b=1$ then the ratio is $F(x)/x=\dfrac{1}{\sqrt{x}}$ which is decreasing in $(0,1]$ and if use the extended-real numbers (allowing $F(0)/0=+\infty$) then it is decreasing in $[0,1]$. $\endgroup$ Commented Oct 10, 2014 at 14:44
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    $\begingroup$ That helps; thank you. It suggests that perhaps you might prefer to analyze $Y=\log(X)$ instead of $X$. If $Y$ has a PDF $g$ and CDF $G$, your decreasing criterion amounts to $G(y)\gt g(y)$ for all $y\lt \log(b)$. I doubt there has ever been interest in this condition holding for all $y$, but certainly it is of interest in studying the behavior of the left tail and is relevant to the behavior of extreme (minimal) values. $\endgroup$
    – whuber
    Commented Oct 10, 2014 at 15:08

2 Answers 2

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First, a few comments:

  1. As shown in the comment by whuber, the assertion that if $a<0$, then $F(x)/x$ cannot decrease on $[a,b]$ is false. I have no problem with the assertion that $F(x)/x$ cannot be constant.
  2. The CDF of a uniform random variable on [0,b] (with $b<\infty$) is given by $F(x)=\frac xb$. Thus, if $b$ is finite, then $b\cdot \frac{F(x)}x$ is the ratio of the CDFs you mention. Note, however, that $\frac{F(x)}x$ will be increasing/decreasing if and only if $b\cdot \frac{F(x)}x$ is increasing/decreasing. If $b=\infty$, then there is no such thing as a uniform random variable on $[0,b)$, hence your intuitive interpretation of the ratio can only make sense when $F$ is supported on a finite interval.

As for some sort of classification, if you assume the variables you're considering have a density (i.e., $F$ is differentiable), then coming up with a nice algebraic test to see if $F(x)/x$ increases/decreases shouldn't be too difficult:

Suppose that the density $f(x)=F'(x)$ exists. According to the quotient rule for differentiation, one has $$\left(\frac{F(x)}x\right)'=\frac{F'(x)x-F(x)}{x^2}=\frac{f(x)x-F(x)}{x^2}$$

Given that a differentiable function $g$ is increasing on $[a,b]$ if and only if $g'(x)\geq0$ for all $x\in[a,b]$ and decreasing if and only if $g'(x)\leq0$ for all $x\in [a,b]$, then we obtain

On $[a,b]$, $F(x)/x$ is

  1. increasing if and only if $f(x)x\geq F(x)$ for every $x\in[a,b]$;

  2. decreasing if and only if $f(x)x\leq F(x)$ for every $x\in[a,b]$; and

  3. constant if and only if $f(x)x=F(x)$ for every $x\in[a,b]$.

Maybe this does not offer an entirely satisfying classification, but it seems like a good place to start.

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    $\begingroup$ If $a>0$ then $F(a)/a=0$ and $F(b)/b=1/b>0$ so clearly this ratio can not be everywhere constant. It also can not be everywhere decreasing. $\endgroup$ Commented Oct 10, 2014 at 14:38
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Reading the book "Stochastic Orders" I found the answer. Two random variables $X$ and $Y$ are ranked by the reverse hazard rate, $X\stackrel{rhr}{\le} Y$, if and only if: $$ \dfrac{F_X(x)}{F_Y(x)}\text{ is non-decreasing in } [\min(a_X,a_Y),+\infty)$$ where $a_X$ and $a_Y$ are the lower bound of the supports of the random variables.

This is equivalent to $\dfrac{f_Y}{F_Y}\ge \dfrac{f_X}{F_X}$ (which are the respective reverse hazard rates).

The classification above is similar (but not entirely identical) to ranking the the uniform $[0,b]$ and the original random variable according to the reverse hazard rate order.

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    $\begingroup$ It is unclear how this is related to your $F(x)/x$ condition in any more than a rough, qualitative way. Could you be more explicit about that? $\endgroup$
    – whuber
    Commented Oct 10, 2014 at 15:14
  • $\begingroup$ @whuber If Y is uniform $[0,b]$ and $F=F_X$ is the CDF of X then $F(x)/x$ is proportional to $F_X(x)/F_Y(x)$. In the case $a=0$: $F(x)/x$ increasing means $X$ is dominated by Y in the reverse hazard rate; $F(x)/x$ decreasing means $Y$ is dominated by $X$ in the reversed hazard rate. $\endgroup$ Commented Oct 10, 2014 at 23:52

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