The Leverage Space Model 309
We can rewrite Equation (9.02) as:HPRk=(1+C)xC=∑ni= 1(fi∗(−PLk,i/BLi))x=(n− 1
∏k= 1( n
∏j=i+ 1P(ik|jk)))(1/(n−1))Notice that the exponent in Equation (9.02), which we must keep track of,
is expressed as the variablexin Equation (9.02a). This is also expressed in
Equation (9.03).
So, to obtainC, we simply go through each scenario spectrum, taking the
outcome of the scenario currently being used in that spectrum as dictated
byk, dividing its negative by the scenario in that spectrum with the worst
outcome, and multiplying this quotient by thefvalue being used with that
scenario spectrum. As we go through all of the scenario spectrums, we total
these values.
The variableiis the scenario spectrum we are looking at. The biggest
loss in scenario spectrum 1 is tails, which sees a loss of one dollar (i.e.,−1).
Thus, BL 1 is−1 (as will be BL 2 and BL 3 since the biggest loss in each of the
other two scenario spectrums—the other two coins—is−1). The associated
PL, that is, the outcome of the scenario in spectrumicorresponding to the
scenario in that spectrum thatkpoints to, is−1 in scenario spectrum 1 (as
it is in the other two spectrums). Thefvalue is currently .1 (as it also is now
in the other two spectrums). Thus:
C=
∑ni= 1(
fi*(
−PLk,i
BLi))
C=
(
. (^1) *
(
−− 1
− 1
))
+
(
. (^1) *
(
−− 1
− 1
))
+
(
− (^1) *
(
−− 1
− 1
))
C=(. (^1) −1)+(. (^1) −1)+(. (^1) *−1)
C=−. 1 +−. 1 +−. 1 =−. 3
Notice that the PLs are negative and, since PL has a minus sign in front
of it, that makes them positive.
Now we take the value forCin Equation (9.02) above and add 1 to it,
obtaining .7 (since 1+−.3=.7). Now we must figure the exponent, the
variablexin Equation (9.02) above.
P(ik|jk) means, simply, the joint probability of the scenario in spectrum
ipointed to byk, and the scenario in spectrumjpointed to byk. Sincekis
presently 1, it points to tails in all three scenario spectrums. To findx,we