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Let $X,Y$ be nonnegative random variables with cummulative distribution functions $F$ and $G$ respectively. Then $X\le_{\text{st}}Y$ iff $1-F(x)\le 1-G(x)$ for all $x\ge 0$ annd $X\le_{\text{mrl}}Y$ iff $$\frac{1}{1-F(x)}\int_{x}^{+\infty}(1-F(u))du\le \frac{1}{1-G(x)}\int_{x}^{+\infty}(1-G(u))du$$ for all $x\ge 0$. (Assume that denominators are not zero, is not important here). Shaked and Shanthikumar write in p.83 of their 2007 "Stochastic Orders"

Neither of the orders $\le_{\text{st}}$ and $\le_{\text{mrl}}$ implies the other; counterexamples can be found in the literature.

I tried to construct counterexamples with the Beta, Gamma, Uniform (which are well known to be stochastically increasing in some of their parameters), but I did not have any success. That is, whenever the variables that I construct are stochastically increasing, then they are also increasing in the MRL order.

Can someone provide some references for the examples that Shaked and Shanthikumar suggest, or can someone provide such an example?

Thank you.

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I found counterexamples by working with a function that is elementary to integrate: the exponential (which includes the trigonometric functions sine and cosine). I thought that a wiggly survival function $1-F$ might do the trick, and it does when suitably adjusted.


Consider, then, the density

$$f(t;\omega,\kappa,\phi) = \frac{\kappa \left(\kappa ^2+\omega ^2\right) \color{red}{e^{-\kappa t} (\cos (\omega (t-\phi ))+1)}}{\kappa ^2 \cos (\omega \phi )+\kappa ^2+\kappa \omega \sin (\omega \phi )+\omega ^2}$$

defined for $t\ge 0$ with parameters $\omega \gt 0$ (the frequency), $\kappa \gt 0$ (the decay rate), and $\phi$ (the phase). (Although it might look complicated, it's really just an exponentially decaying sinusoid--shown in red--times a suitable normalizing factor.) Integration gives the survival function

$$1-F(t;\omega,\kappa,\phi) = \frac{e^{-\kappa t} \left(\kappa ^2+\kappa ^2 \cos (\omega (t-\phi ))-\kappa \omega \sin (\omega (t-\phi ))+\omega ^2\right)}{\kappa ^2 \cos (\omega \phi )+\kappa ^2+\kappa \omega \sin (\omega \phi )+\omega ^2}$$

from which we may compute the "mean residual life" function

$$\eqalign{h(t;\omega,\kappa,\phi) &= \frac{1}{1-F(t;\omega,\kappa,\phi)}\int_t^\infty (1-F(t;\omega,\kappa,\phi))dt \\ &=\frac{2 \kappa }{\kappa ^2+\omega ^2}-\frac{\kappa ^2+\kappa ^2 \cos (\omega (t-\phi ))-\omega ^2}{\kappa \left(\kappa ^2+\kappa ^2 \cos (\omega (t-\phi ))-\kappa \omega \sin (\omega (t-\phi ))+\omega ^2\right)}. }$$

I am going to compare the distribution with parameters $(\omega,\kappa,\phi)=(0,1,0)$ (shown in blue in all plots of density, survival, and MRL) to distributions with other parameters $(2\pi, \kappa,\phi)$ (shown in red).

The first one doesn't wiggle-it's exponential--while the other does. It is straightforward to show that the patterns visually established for the survival ratio and MRL continue in to zero and out to infinity, so I won't go into the details.

First up is the case $\kappa=0.96$ and $\phi=1.7$: the wiggly distribution decays a little less slowly.

Figure 1

The "survival ratio" is the ratio of the two survival functions, taking the base (blue) as numerator. In this case neither distribution dominates the other in terms of MRL, but the second always has a larger survival function: it stochastically dominates the first.

Next consider $\kappa=4/3$ and $\phi=0.79$. The wiggly density decays more rapidly, keeping its MRL low, but a change in its phase affects the survival ratio for small arguments $t$.

Figure 2

Because the survival ratio is both positive and negative, neither distribution stochastically dominates the other; however, the first one always has the greater MRL.

These two examples illustrate how neither stochastic dominance nor dominance in MLR imply the other.

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    $\begingroup$ Although not the simplest example, (e.g. can one still construct examples where the MRL is decreasing for both functions?) it is an impressive answer. It is working perfectly. Thank you very much, I will use it! $\endgroup$
    – Jimmy R.
    Feb 19, 2018 at 7:52

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