How to Make Scores Consistent

As it turns out to be, this is straightforward. For any scoring matrix E for which S_E(d) is strictly monotonic we can invert S_E(d). Our new score is computed using the inverse, i.e.

 

d^* = S_E^-1(w/n) (6)

where w is the actual score obtained from a pair of aligned sequences $\langle a, b \rangle$.

$$w = \sum_{i=1}^n E_{a_ib_i}$$ (7)

For n sufficiently large, w/n converges to its expected value S_E(d) and d^* converges to d. Since S_E(d) is a sum of exponentials in d, it is generally not possible to invert it algebraically. The computation of the inverse will be done numerically. Let $S_E^k(d) = \sum_{B=1}^{20} f_B \sum_{A=1}^{20} M_{AB}^d E_{AB}^k$ be the k^th moment of the scoring matrix E. The central moments of w/n are:

E[w/n] = S_E(d)

$$\sigma^2(w/n) = E[(w/n - S_E(d))^2] = \frac{S_E^2(d) - S_E(d)^2}{n} = \frac{\mu_2}{n}$$ (8)

$$E[(w/n - S_E(d))^3] = \frac{S_E^3(d) - 3S_E^2(d) S_E(d) + 2 S_E(d)^3}{n^2} = \frac{\mu_3}{n^2}$$ (9)

$$E[(w/n - S_E(d))^k] = O(n^{\lfloor -k/2 \rfloor }), \;\;\;\; (k \ge 1)$$ (10)