Stochastic Programming - Index of

256 STOCHASTIC PROGRAMMING
Taking the expectation on both sides yields for the binomial moments S k,n
[( )] ˜ν ∑
=
k
E˜ξ
(˜χ i1 ˜χ i2 ···˜χ ik )
E˜ξ
=
1≤i 1≤···≤i k ≤n
∑
1≤i 1≤···≤i k ≤n
P (B i1 ∩···∩B ik ) .
This formulation indicates the possibility of estimating the binomial moments
from large samples through the relation
∑
S k,n =
E˜ξ
(˜χ i1 ˜χ i2 ···˜χ ik )
1≤i 1≤···≤i k ≤n
if they are difficult to compute directly.
Consider now the following example. Assume that we have a fourdimensional
random vector ˜ξ with mutually independent components. Let
z ∈ IR 4 be chosen such that with p i = P (A i ), i =1, 2, 3, 4, we have
p T =(0.9, 0.95, 0.99, 0.92).
Consequently, for q i = P (B i )=1− p i we get
q T =(0.1, 0.05, 0.01, 0.08).
Obviously we get F˜ξ(z) = ∏ 4
i=1 p i =0.778734. From the above representation
of the binomial moments, we have
S 1,n =
S 2,n =
4∑
q i =0.24
i=1
3∑
4∑
i=1 j=i+1
q i q j =0.0193
such that we get from (3.7) for P (˜ν ≥ 1) the upper bound
P U =0.24 − 2 × 0.0193 = 0.23035.
4
According to (3.6), we find i−1 = ⌊ ⌋
2×0.0193
0.24 = 0 and hence i = 1, so that (3.6)
yields the lower bound
P L = 2 2 × 0.24 − 2 × 0.0193 = 0.1757.
2
In conclusion, we get for F˜ξ(z) =0.778734 the bounds 1 − P U ≤ F˜ξ(z) ≤
1 − P L , and hence
0.76965 ≤ F˜ξ(z) ≤ 0.8243.