Incremental Risk Vulnerability
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to one, and E(∆(y)) = 0. Equivalently, this can be reformulated as a parametric linear
program where the non-linearity is eliminated by writing r as a parameter
min f (w,y,s) = ∑ qi [∆(yi) {-u'"(Wi) - u"(Wi)f}] (11)
{qi} i
s.t.
∑ q i∆( y i) = 0 (12)
i
∑ q i = 1, (13)
i
the definitional constraint for the parameter r
r ∑ qiu'(Wi) = - ∑ qiu"(Wi) (14)
ii
and the non-negativity constraints
qi ≥ 0, ∀i. (15)
Consider the optimal solution. Since this optimization problem has three constraints, there
are three variables in the basis. Number these as i =1, 2,a, with ∆(y1) < 0 < ∆(y2)
and y1 + s∆(y1) <y2 + s∆(y2). The associated probabilities are q1,q2,qa, such that
q1∆(y1)+qa∆(ya)+q2∆(y2)=0. There are two possibilities with respect to the state
a.
Either:
a = 0. Then ∆(ya) = ∆(y0) = 0. Hence, we immediately obtain equation (16).
or:
a = 0. In this case we drop the constraint on q0 ≥ 0 (with all the other qis staying
non-negative). Hence the probability associated with y0 can be negative. Dropping this
constraint will lead to a condition that is too demanding. However, since we are searching
for a sufficient condition, this is fine. In the original optimisation, all the non-basis variables
had nonnegative coefficients in the objective function in the final simplex tableau. Allowing
q0 < 0 must result therefore in q0 replacing either q1, q2 or qa in the optimal basis. Also,
the new f -value is either lower or the same as before.
Suppose, first, that q0 replaces qa in the optimal basis. Then the new basis variables
are q1, q2 and q0. Since ∆(y0) = 0, we can write the objective function (11) as