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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?

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  • 1
    $\begingroup$ @PaceNielsen Done. $\endgroup$ – capadocia Aug 13 at 1:12
  • $\begingroup$ Hopefully that will get more interest in your question! $\endgroup$ – Pace Nielsen Aug 13 at 2:09
  • $\begingroup$ @capadocia u from turkey? $\endgroup$ – kawa yesterday
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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, $f_Y(y)>0 \text{ for all } y \Rightarrow g^{-1}(A(w))<A(w)\Rightarrow \overline{F}_Y(g^{-1}(A(w)))>\overline{F}_Y(A(w))$. Hence the third integral is positive.

It follows that $P_n>P_m$.

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