ch02 JWBK035-Vince February 12, 2007 6:50 Char Count= 0
56 HANDBOOK OF PORTFOLIO MATHEMATICS
variance=1. Now, N′(Z) will give us the Y axis value (the height of the
curve) for any value of Z:N′(Z)= 1 /
√
2 ∗ 3. 1415926536 ∗EXP(−(Z^2 /2))
=. 398942 ∗EXP(−(Z^2 /2)) (2.15a)where: Z=(X−U)/S
and U=The mean of the data.
S=The standard deviation of the data.
X=The observed data point.
EXP ( )=The exponential function.Equation (2.16) gives us the number ofstandard unitsthat the data
point corresponds to—in other words, how many standard deviations away
from the mean the data point is. When Equation (2.16) equals 1, it is called
thestandard normal deviate.A standard deviation or a standard unit
is sometimes referred to as asigma.Thus, when someone speaks of an
event’s being a “five sigma event,” they are referring to an event whose
probability of occurrence is the probability of being beyond 5 standard de-
viations.
Consider Figure 2.7, which shows this equation for the Normal curve.
Notice that the height of the standard Normal curve is .39894. From Equa-
tion (2.15a), the height is:N′(Z)=. 398942 ∗EXP(−(Z^2 /2))
N′(0)=. 398942 ∗EXP(−(0^2 /2))
N′(0)=. 398942Notice that the curve iscontinuous—that is, there are no “breaks” in
the curve as it runs from minus infinity on the left to positive infinity on the
right. Notice also that the curve is symmetrical, the side to the right of the
peak being the mirror image of the side to the left of the peak.
Suppose we had a group of data where the mean of the data was 11
and the standard deviation of the group of data was 20. To see where a data
point in that set would be located on the curve, we could first calculate it
as a standard unit. Suppose the data point in question had a value of−9. To
calculate how many standard units this is, we first must subtract the mean
from this data point:− 9 − 11 =− 20