[UnSA] The Condition for existence of minimum in Survival-Related Function.
Today, I was asked about the proof stated on this paper (theorem 2.1 page 98), (the proof is on page 110), we are dazed with the argument given there, and started to think that it's probably an errata (mistyped but not serious) or we just missed the point.
Let $S_X(t)$ be a survival function, define for each $d>0$ and $0<\alpha < S_X(0)$ the function
\[\text{VaR}_T(d,\alpha) := \left\{\begin{array}{lr} d+\delta(d)\qquad & \text{for }d \in (0,S_X^{-1}(\alpha)\\ S_X^{-1}(\alpha)+\delta(d) \qquad & \text{for } d > S_X^{-1}(\alpha)
\end{array}\right.\]
Let $S_X(t)$ be a survival function, define for each $d>0$ and $0<\alpha < S_X(0)$ the function
\[\text{VaR}_T(d,\alpha) := \left\{\begin{array}{lr} d+\delta(d)\qquad & \text{for }d \in (0,S_X^{-1}(\alpha)\\ S_X^{-1}(\alpha)+\delta(d) \qquad & \text{for } d > S_X^{-1}(\alpha)
\end{array}\right.\]
Where $\delta(d)=(p+1)\int_d^{\infty} S_X(t) \, dt$ and $p^*=\frac{1}{1+p}$.
The value $d^*>0$ such that $\text{VaR}_T(d^*,\alpha)=\min_{d>0}(\text{VaR}_T(d,\alpha))$ exists if and only if
\[S_X^{-1}(\alpha)\geq S_X^{-1}(p^*)+\delta(S_X^{-1}(p))\]
and
\[\alpha < p^* < S_X(0)\]
Recall that, the survival function is monotone decreasing function positive. Also it can be proved that $\text{VaR}_T(d,\alpha)$ is continuous. For $d>S_X^{-1} (\alpha)$, the derivative of $\text{VaR}_T(d,\alpha)$ is
Proof:
Recall that, the survival function is monotone decreasing function positive. Also it can be proved that $\text{VaR}_T(d,\alpha)$ is continuous. For $d>S_X^{-1} (\alpha)$, the derivative of $\text{VaR}_T(d,\alpha)$ is
\[\delta'(d)=-(1+p) S_X(d) < 0\]
(note that the differentiation under integral sign is verified since $S_X(d)$ is bounded, hence clearly dominated).
Therefore $\text{VaR}_T(d,\alpha)$ is decreasing on $(S_X^{-1}(\alpha),\infty)$, with limit $S_X^{-1}(\alpha)$ as $d\rightarrow \infty$. Also, for $p^*>S_X(0)$ we have the derivative of $d+\delta(d)$ is
Therefore $\text{VaR}_T(d,\alpha)$ is decreasing on $(S_X^{-1}(\alpha),\infty)$, with limit $S_X^{-1}(\alpha)$ as $d\rightarrow \infty$. Also, for $p^*>S_X(0)$ we have the derivative of $d+\delta(d)$ is
\[1-(p+1)S_X(d)<0\]
thus $d+\delta(d)$ is decreasing when $p^* > S_X(0)$ until $d=S_X^{-1}(p^*)$. Therefore the function $d+\delta(d)$ attains minimum value on $d_0=S_X^{-1}(p^*)$. If the both conditions hold then $d_0 \in (0,S_X^{-1}(\alpha))$ and the minimum point is now also the minimum point of $\text{VaR}_T(d,\alpha)$, therefore $d^*=d_0>0$.
Conversely, If $\alpha\geq p^*$ then since $S_X(t)$ is decreasing its inverse function is also decreasing, it follows that $S^{-1}_X(\alpha) \leq S_X^{-1}(p^*)=d_0$, and for all $d\leq S^{-1}_X(\alpha)$ we have $S_X(d) \geq p^*$, therefore the derivative of the function $\text{VaR}_T(d,\alpha)$ is equal to $1-(p+1)S_X(d)$ on $d\in (0,S_X^{-1}(\alpha))$ and equal to $-(p+1)S_X(d)$ when $d>S_X^{-1}(\alpha)$ which is consistently negative. Thus $\text{VaR}_T(d,\alpha)$ is decreasing on $(0,\infty)$ and as $d\rightarrow \infty$ it has limiting value $S_X^{-1}(\alpha)$, while no such $d^*$ with $\text{VaR}_T(d^*,\alpha)=S_X^{-1}(\alpha)$, so no such $d^*$ exists.
Now suppose that $p^* \geq S_X(0)$, then we have $d=S_X^{-1}(p^*) \leq 0$ thus $d_0 =0$. Since $d_0$ is the only stationery point of the function $d+\delta(d)$, it follows that this function does not have a stationery point when $d>0$, therefore on $0< d\leq S_X^{-1}(\alpha)$ the derivative of the function is either strictly positive or strictly negative (if it ever been positive in one point and then negative in other point, then since the derivative is continuous by Intermediate Value Theorem it has zero, thus it has stationery point.)
From the above argument it follows that $d+\delta(d)$ is either monotone decreasing or monotone increasing on $0< d \leq S_X^{-1}(\alpha)$.
[In the paper it is stated only the decreasing case]
It it is increasing, then $VaR_T(d,\alpha)$ is increasing on $(0,S_X^{-1}(\alpha))$ and eventually starts decreasing on $(S_X^{-1}(\alpha),\infty)$ with limited value $S_X^{-1}(\alpha)$, while it has no minimum value on $(0,S_X^{-1}(\alpha))$ the limited value $S_X^{-1}(\alpha)$ is the smallest the function $\text{VaR}_T(d,\alpha)$ can approach, but as before no such $d^*$ with $\text{VaR}_T(d^*,\alpha)=S_X^{-1}(\alpha)$. If $\delta(d)+d$ is decreasing, then $\text{VaR}_T(d,\alpha)$ is decreasing on $(0,\infty)$ with limiting value $S_X^{-1}(\alpha)$ as $d\rightarrow \infty$, thus no such $d^*$ exists.
If condition (1) does not hold while condition (2) holds then $\inf_{d>0} \text{VaR}_T(d,\alpha)=S_X^{-1}(\alpha)$, in order to have the infimum a minimum, there should be exists $d^*$ with $\text{VaR}_T(d^*,\alpha)=S_X^{-1}(\alpha)$, but this is clearly impossible, and the proof is completed.
Remark: Because I don't know much about the behavior of Survival Function, I tend to avoid the second derivative test for stationery point $d_0$, if we employ this test, we will get that this point is a minima, thus $\text{VaR}_T (d,\alpha)$ is not always decreasing when $d \in (0,\infty)$, this is where we think the argument in the paper is fishy.
Conversely, If $\alpha\geq p^*$ then since $S_X(t)$ is decreasing its inverse function is also decreasing, it follows that $S^{-1}_X(\alpha) \leq S_X^{-1}(p^*)=d_0$, and for all $d\leq S^{-1}_X(\alpha)$ we have $S_X(d) \geq p^*$, therefore the derivative of the function $\text{VaR}_T(d,\alpha)$ is equal to $1-(p+1)S_X(d)$ on $d\in (0,S_X^{-1}(\alpha))$ and equal to $-(p+1)S_X(d)$ when $d>S_X^{-1}(\alpha)$ which is consistently negative. Thus $\text{VaR}_T(d,\alpha)$ is decreasing on $(0,\infty)$ and as $d\rightarrow \infty$ it has limiting value $S_X^{-1}(\alpha)$, while no such $d^*$ with $\text{VaR}_T(d^*,\alpha)=S_X^{-1}(\alpha)$, so no such $d^*$ exists.
Now suppose that $p^* \geq S_X(0)$, then we have $d=S_X^{-1}(p^*) \leq 0$ thus $d_0 =0$. Since $d_0$ is the only stationery point of the function $d+\delta(d)$, it follows that this function does not have a stationery point when $d>0$, therefore on $0< d\leq S_X^{-1}(\alpha)$ the derivative of the function is either strictly positive or strictly negative (if it ever been positive in one point and then negative in other point, then since the derivative is continuous by Intermediate Value Theorem it has zero, thus it has stationery point.)
From the above argument it follows that $d+\delta(d)$ is either monotone decreasing or monotone increasing on $0< d \leq S_X^{-1}(\alpha)$.
[In the paper it is stated only the decreasing case]
It it is increasing, then $VaR_T(d,\alpha)$ is increasing on $(0,S_X^{-1}(\alpha))$ and eventually starts decreasing on $(S_X^{-1}(\alpha),\infty)$ with limited value $S_X^{-1}(\alpha)$, while it has no minimum value on $(0,S_X^{-1}(\alpha))$ the limited value $S_X^{-1}(\alpha)$ is the smallest the function $\text{VaR}_T(d,\alpha)$ can approach, but as before no such $d^*$ with $\text{VaR}_T(d^*,\alpha)=S_X^{-1}(\alpha)$. If $\delta(d)+d$ is decreasing, then $\text{VaR}_T(d,\alpha)$ is decreasing on $(0,\infty)$ with limiting value $S_X^{-1}(\alpha)$ as $d\rightarrow \infty$, thus no such $d^*$ exists.
If condition (1) does not hold while condition (2) holds then $\inf_{d>0} \text{VaR}_T(d,\alpha)=S_X^{-1}(\alpha)$, in order to have the infimum a minimum, there should be exists $d^*$ with $\text{VaR}_T(d^*,\alpha)=S_X^{-1}(\alpha)$, but this is clearly impossible, and the proof is completed.
Remark: Because I don't know much about the behavior of Survival Function, I tend to avoid the second derivative test for stationery point $d_0$, if we employ this test, we will get that this point is a minima, thus $\text{VaR}_T (d,\alpha)$ is not always decreasing when $d \in (0,\infty)$, this is where we think the argument in the paper is fishy.
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