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Monday, 5 March 2012

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.
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)

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
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.

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