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How to Make Scores Consistent

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

 
d* = SE-1(w/n) (6)

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

 \begin{displaymath}
w = \sum_{i=1}^n E_{a_ib_i} \end{displaymath} (7)

For n sufficiently large, w/n converges to its expected value SE(d) and d* converges to d. Since SE(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 kth moment of the scoring matrix E. The central moments of w/n are:

E[w/n] = SE(d)


 \begin{displaymath}
\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} \end{displaymath} (8)


 \begin{displaymath}
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} \end{displaymath} (9)


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


next up previous
Next: How general are these Up: Dayhoff Scores and Evolutionary Previous: Scores are inconsistent to
Chantal Korostensky
1999-07-14