Markov Models for Allele Frequencies

In the lecture, you saw that the leading eigenvalue of the Markov matrix \(M\), whose R output is:

      [,1]  [,2]  [,3]  [,4]
[1,] 0.980 0.005 0.005 0.010
[2,] 0.005 0.980 0.010 0.005
[3,] 0.005 0.010 0.980 0.005
[4,] 0.010 0.005 0.005 0.980

produced an eigenvector modeling a situation where the alleles are represented equally (each with probability 0.25).

In this exercise, we use a for-loop to iterate the process of mutation from an initial allele distribution of:

[1] 1 0 0 0

and show that this is indeed what happens - that the eigenvector yields the right information in lieu of the for-loop.

For more on Markov Processes, see this link.

This exercise is part of the course

Linear Algebra for Data Science in R

View Course

Exercise instructions

Print x, the allele distribution after the 1000 mutations.
Find and scale the first eigenvector of M (which is loaded for you) so that it sums to 1. Assign to v1.
Print v1, the scaled first eigenvector of M and compare with x.

Hands-on interactive exercise

Have a go at this exercise by completing this sample code.

# This code iterates mutation 1000 times
x <- c(1, 0, 0, 0)
for (j in 1:1000) {x <- M%*%x}

# Print x
print(___)

# Print and scale the first eigenvector of M
Lambda <- eigen(M)
v1 <- Lambda$vectors[, ___]/sum(Lambda$___[, 1])

# Print v1
print(___)

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