ch02 JWBK035-Vince February 12, 2007 6:50 Char Count= 0
60 HANDBOOK OF PORTFOLIO MATHEMATICS
N(Z)= 1 −N′(Z)∗((1. 330274429 ∗Y∧5)−(1. 821255978 ∗Y∧4)
+(1. 781477937 ∗Y∧3)−(. 356563782 ∗Y∧2)
+(. 31938153 ∗Y)) (2.21)
If Z<0, then N(Z)= 1 −N(Z). Now recall Equation (2.15a):N′(Z)=. 398942 ∗EXP(−(Z^2 /2))where: Y=1/(1+ (^2316419) *ABS(Z))
and ABS ( )=The absolute value function.
EXP ( )=The exponential function.
We will always convert our data to standard units when finding proba-
bilities under the curve. That is, we will not describe an N(X) function, but
rather we will use the N(Z) function where:
Z=(X−U)/S
and U=The mean of the data.
S=The standard deviation of the data.
X=The observed data point.
Refer now to Equation (2.21). Suppose we want to know what the prob-
ability is of an event’s not exceeding+2 standard units (Z=+2).
Y= 1 /(1+ 2316419 ∗ABS(+2))
= 1 / 1. 4632838
=. 68339443311
N′(Z)=. 398942 ∗EXP(−(Z^2 /2))
=. 398942 ∗EXP(−2)
=. 398942 ∗. 1353353
=. 05399093525
Notice that this tells us the height of the curve at−2 standard units.
Plugging these values for Y and N′(Z) into Equation (2.21) we can obtain
the probability of an event’s not exceeding+2 standard units:
N(Z)= 1 −N′(Z)∗((1. 330274429 ∗Y∧5)−(1. 821255978 ∗Y∧4)
+(1. 781477937 ∗Y∧3)−(. 356563782 ∗Y∧2)
+(. 31938153 ∗Y))