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I am trying to show that the following function is not increasing in $\widetilde{k}$ ($\widetilde{k} > 0$):

\begin{equation} \gamma(\widetilde{k}) = \frac{\sum_{k \geq \widetilde{k}} f(k) \cdot G\left[\frac{k}{\widetilde{k}}-1\right]}{\sum_{k \geq \widetilde{k}} f(k) \cdot k \cdot G\left[\frac{k}{\widetilde{k}}-1\right]} \end{equation} Here I used the following notation:

  • $k$ is a discrete variable taking values from finite support $\{1,\ldots, K_m\}$
  • $f(k)$ - some discrete probability mass function, hence $f(k)\in[0,1], \forall k$
  • $G(\cdot)$ - complementary cdf of some continuous distribution with positive support, thus it is decreasing in its argument

If there was no $\widetilde{k}$ inside $G(\cdot)$ function, function $\gamma(\widetilde{k})$ would be decreasing. Indeed, say there is only dependence on $k$ under $G(\cdot)$: for instance some function $\beta(k)$, then with some normalization $f(k) \cdot G(\beta(k))$ is a pmf of some distribution (call it $A$). In the same way, under different normalization $f(k) \cdot k \cdot G(\beta(k))$ is a pmf of another distribution (call it $B$). Clearly their ratio is decreasing in $k$ and hence there is a likelihood ratio order $A \leq_{lr} B$, which implies the same hazard rate order which implies $\gamma(\widetilde{k})$ decreasing (again if there is no dependency on $\widetilde{k}$ inside $G$.

Unfortunately, I cannot come up with any way to show that my original function $\gamma(\widetilde{k})$ is not increasing. Could someone help me?

Thank you in advance.

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    Where $\langle k \rangle$ is defined, it looks like it means the infinite series $f(1)+2f(2)+3f(3)+\ldots$. But then it would be a constant, not depending on $k$, so that would be a weird notation, and anyway you'd have to worry about convergence. Then I thought you might mean something like $\langle k\rangle=f(1)+2f(2)+\ldots+kf(k)$. But then $\langle k\rangle $ might equal $0$ for some values of $k$, if $f(k)$ is allowed to be zero sometimes, and that would be a problem since $\langle k\rangle$ appears as a denominator. Can you please clarify the definition of $\langle k\rangle$? – 2017-02-27
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    @mathmandan, let's say $k \in \{1, \ldots, K_m\}$, so that the support is finite (I made a correction in the initial question). Thus, $\langle k \rangle = f(1) + 2 f(2) + \ldots + f(K_m) K_m$ - which is just expectation of $k$ and doesn't depend on $k$. – 2017-02-27
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    OK. Then $\langle k \rangle $ is constant (and positive), so you can just factor it out of the summation, and it won't ultimately affect monotonicity. Does that help you? – 2017-02-27
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    @mathmandan sure, it can be factored out, but this doesn't make things easier. I left it in the initial formula in order to show similarity between the two summations - i.e. the fact that the denumerator is just weighted version of the enumerator. But, I think I will remove it along with $\alpha$, not to confuse people. – 2017-02-27

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