The presumption of analytical liberty signifies that the joint likelihood thickness function of the elements of ?X? is:

(B1) f X; ? 1, ? 2, … ? m = ? i = 1 m ? i ag ag e – ? i = 1 m ? i x i (B1)

## The likelihood that the arbitrary breakthrough will have attributes satisfying the conditions imposed by the minimal set is:

(B2) P r o b x 1 x 1, min, …. X n x n, min = 1 – F X min = e – ? i = 1 n ? i x i, min (B2)

The numerator of this expected energy in Equation (4) when you look at the unique situation is:

(B3) ? x n, min ? ? ? x 1, min ? U X f X d x 1 ? d x n = ? x n, min ? ? ? x 1, min ? ? i = 1 n w i x i ? i = 1 n ? i ag ag e – ? i i ? i = 1 letter d x i (B3)

The analytical liberty for the random factors in X let the phrase in the right-hand part of Equation (B3) to solve as:

(B4) ? x n, min ? ? ? x 1, min ? U X f X d x 1 ? d x n = ? i = 1 n ? x i, min ? w i x i ? i e – ? i x i d x i (B4)

It really is a simple calculation to show that the definite integral when you look at the summand in B4 is:

(B5) ? x i, min ? x i ? i e – ? i x i d x i = x i, min + 1 ? i e – ? i x i, min (B5)

Holding out of the integration from the side that is right-hand of and summing, we now have: