# Sufficient conditions for inequality with integral of reliability functions

Let $$Y$$ and $$W$$ be two random variables with support $$(y_1,y_2)$$ and $$(w_1,w_2)$$ and distributions $$F_Y$$ and $$F_W$$, both twice continuously differentiable (densities $$f_Y$$ and $$f_W$$). Assume that both have (finite) mean $$\bar{y}$$ and $$\bar{w}$$. Assume also that $$f_Y(y)>0$$ for all $$y\in (y_1,y_2)$$ and $$f_W(w)>0$$ for all $$w\in (w_1,w_2)$$.

Define $$g(x)=\mathrm{E}[Y\,|\,Y>x]$$. Let $$A:\mathbb{R}\to \mathbb{R}$$ be a (strictly) decreasing function and $$\tilde{w}$$ denote the unique solution $$w$$ to $$A(w)=g^{-1}(A(\bar{w}))$$. Assume that $$\bar{y}< A(\bar{w})$$.

Consider:

$$P_n = \int_{w_1}^{w_2}\overline{F}_Y(g^{-1}(A(w)))f_W(w)\mathrm{d} w$$

$$P_m = \frac{1}{2}\overline{F}_Y(g^{-1}(A(\bar{w}))) + \int_{\tilde{w}}^{w_2}\bigl[\overline{F}_Y(A(w))-\overline{F}_Y(g^{-1}(A(\bar{w}))) \bigr]f_W(w)\mathrm{d} w$$

Are there sufficient conditions under which $$P_n\geq P_m$$ or $$P_m\geq P_n$$?

MOTIVATION

A machine is labeled "reliable" if the probability that failure does not occur (random variable $$Y$$) is above a threshold $$A(W)$$ (the threshold is stochastic).

The machine designer will eventually know the realization of $$Y$$. In fact, he will know it before he knows the realization of $$W$$. He can commit to a disclosure policy regarding $$Y$$. An authority who only observes the realization of $$W$$ and the designer's disclosure policy will decide on granting the "reliable" certification. Since $$\bar{y}< A(\bar{w})$$, committing to not disclosing the realization of $$Y$$ irrespective of its value precludes the machine from being granted the "reliable" certificate.

The machine designer can choose between two disclosure policies:

• Option 1

Wait until he observes the realization of $$W$$ to disclose. This option prescribes the following. Let $$y^n$$ denote the unique solution to $$\mathrm{E}[Y|Y>y^n]=A(W)$$.

If $$y< y^n$$, discloses $$y$$; otherwise, discloses nothing.

Thus, the ex-ante probability that the machine being granted the "reliable" certification is given by $$P_n$$.

• Option 2

Disclosure of realization of $$Y$$ is made as soon as possible (if at all). This option prescribes the following. Let $$y^m=A(\bar{w})$$.

If $$y< y^m$$, discloses $$y$$; otherwise, discloses nothing.

Thus, the ex-ante probability that the machine being granted the "reliable" certification is given by $$P_m$$.

So, under what conditions would the designer prefer option 1 over option 2? Under what conditions would he prefer option 2 over option 1?

• @PaceNielsen Done. – capadocia Aug 13 '19 at 1:12
• Hopefully that will get more interest in your question! – Pace Nielsen Aug 13 '19 at 2:09
• @capadocia u from turkey? – kawa Aug 16 '19 at 3:57

Rewriting $$P_m$$
\begin{align} P_m &= \int_{\bar{w}}^{w_2}\overline{F}_Y(g^{-1}(A(\bar{w})))f_W(w)\mathrm{d} w + \int_{\tilde{w}}^{w_2}\bigl[\overline{F}_Y(A(w))-\overline{F}_Y(g^{-1}(A(\bar{w}))) \bigr]f_W(w)\mathrm{d} w & \\ &= \int_{\bar{w}}^{\tilde{w}}\overline{F}_Y(g^{-1}(A(\bar{w})))f_W(w)\mathrm{d} w + \int_{\tilde{w}}^{w_2}\overline{F}_Y(A(w))f_W(w)\mathrm{d} w & \end{align}
Thus \begin{align} P_n - P_m &= \int_{w_1}^{\bar{w}}\overline{F}_Y(g^{-1}(A(w)))f_W(w)\mathrm{d} w +\int_{\bar{w}}^{\tilde{w}}\bigl[\overline{F}_Y(g^{-1}(A(w)))-\overline{F}_Y(g^{-1}(A(\bar{w}))) \bigr]f_W(w)\mathrm{d} w + \int_{\tilde{w}}^{w_2}\bigl[\overline{F}_Y(g^{-1}(A(w))) - \overline{F}_Y(A(w)) \bigr]f_W(w)\mathrm{d} w & \end{align}
Since $$w\geq \bar{w} \Rightarrow g^{-1}(A(w))\leq g^{-1}(A(\bar{w}))\Rightarrow \overline{F}_Y((g^{-1}(A(w)))\geq \overline{F}_Y(g^{-1}(A(\bar{w})))$$, the second integral is non-negative.
Also, $$f_Y(y)>0 \text{ for all } y \Rightarrow g^{-1}(A(w))\overline{F}_Y(A(w))$$. Hence the third integral is positive.
It follows that $$P_n>P_m$$.