Thread: Check if everything is OK in the way I have solved this problem?

Can you please just look this over like peer review? I think I am fairly confident about my general procedure, but I am scared that I have committed minor errors. Thank you for your help.


Suppose a gene in a chromosome is of type A or type B. Assume that the probability that a gene of type A will mutate to type B in one generation is 10^-4 and that a gene of type B will mutate to type A is 10^-6.

(1) What is the transition matrix?

10^-4 = 1/10^4 = 0.0001

10^-6 = 1/10^6 = 0.000001

	A	B
A	0.9999	0.0001
B	0.000001	0.999999

(2) In the long run, what is the probability that the gene will be of type A? Of type B? (Find the stationary matrix.)

[s1 s2] [0.9999 0.0001] = [s1 s2]
[0.000001 0.999999]

[0.9999 s1 + 0.000001 s2 0.0001 s1 + 0.999999 s2] = [s1 s2]

Linear system

0.9999 s1 + 0.000001 s2 = s1
0.0001 s1 + 0.999999 s2 = s2
s1 + s2 = 1

-0.0001 s1 + 0.000001 s2 = 0
0.0001 s1 - 0.000001 s2 = 0
s1 + s2 = 1

Algebraically solving for s1 and s2 using the equations from the linear system

0.0001 s1 - 0.000001 s2 = 0
0.0001 s1 = 0 + 0.000001 s2
0.0001 s1 = 0.000001 s2
s1 = (0.000001 s2 / 0.0001)

s1 + s2 = 1
(0.000001 s2 / 0.0001) + s2 = 1
0.000001 s2 + 0.0001 s2 = 0.0001
0.000101 s2 = 0.0001
s2 = 0.0001/0.000101
s2 = 0.990099009

s1 + s2 = 1
s1 = 1 - 0.990099009 = 0.00990099

S = [0.00990099 0.9900099009]

In the long run, the probability that a gene will be type A is 0.0099009 and the probability that it will be of type B is 0.9900099009.

Hey Yoodle15.

Using Octave we get the answers:

>> A
A =

1.00000 1.00000 1.00000
-0.00010 0.00000 0.00000
0.00010 -0.00000 0.00000

>> rref(A)
ans =

1.00000 0.00000 0.00990
0.00000 1.00000 0.99010
0.00000 0.00000 0.00000

which give s1 = 0.00990 and s2 = 0.99010 which is extremely close to your answer (I'm guessing its got to do with round-off errors).