Stochastic Matrix/Steady-State vector

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November 6th 2006, 05:28 PM
fifthrapiers

Stochastic Matrix/Steady-State vector

Suppose that we have a nucloetide string consisting of A, C, G, and T. We are interested in the regions which are "CG rich". In one of the "CG rich" regions of a few thousand letters of DNA we have that the chance that an A will follow an A is .18, the chance that an A will follow a C is .274, the chance that an A will follow a G is .426 and the chance that an A will follow a T is .12. Also, we have that the chance that a C will follow an A is .17, the chance that a C will follow a C is .368, the chance that a C will follow a G is .274 and the chance that a C will follow a T is .188. Further, we have that the chance a G will follow an A is .161, the chance that a G will follow a C is .339, the chance that a G will follow a G is .385 and the chance that a G will follow a T is .115. Finally, we have that the chance a T will follow an A is .079, the chance that a T will follow a C is .355, the chance that a T will follow a G is .384 and the chance that a T will follow a T is .182.

a.) From the information above, show the 1-step stochastic matrix for moving from 1 letter in the code above to the next letter. What are the eigenvalues and corresponding eigenvectors of this matrix?

b.) Using the information in a, state whether a steady-state (stationary) vector exists for the code. If one does exist, find it and interpret it in the context of the propoortion of A's, C's, G's, and T's we expect to find further along the code. If no steady-state vector exists, say whether we'll see a significant rise in the proportion of C's and the proportion of G's, or whether we will see a loss (that is, an extinction) of C's & G's further along the code.
November 6th 2006, 09:46 PM
fifthrapiers

a.) Is the stochastic matrix:

A,C,G,T along the side and A,C,G,T along the top. Then im not quite sure where to place the values. For example, "an A will follow a C is .274" so where does the .274 go? We have 16 values so this is going to be a 4 x 4 matrix. Then I guess the .274 would go in the slot A,C ??

The eigenvals then I can figure out as well as the eigenvects once I find out what the matrix looks like.

But for part b, I am clueless. Steady state vector determines when it is "stable"?
November 6th 2006, 11:26 PM
CaptainBlack

Quote:

Originally Posted by fifthrapiers

a.) Is the stochastic matrix:

A,C,G,T along the side and A,C,G,T along the top. Then im not quite sure where to place the values. For example, "an A will follow a C is .274" so where does the .274 go? We have 16 values so this is going to be a 4 x 4 matrix. Then I guess the .274 would go in the slot A,C ??

The eigenvals then I can figure out as well as the eigenvects once I find out what the matrix looks like.

But for part b, I am clueless. Steady state vector determines when it is "stable"?

Think of it this way, you have a col. vector $X$ with the probabilities of the
current nucloetide being A,C,G,T down the columns.

Then the stochastic $S$ matrix will give you the probabilities for the nucloetide
at the next position as $SX$ .

So if the current nucloetide is A, the first column of $S$ gives the probability
that next nucloetide is A, C, G or T, similarly if the current nucloetide is C
the second column gives the probability that next nucloetide is A, C, G or T,
and so on.

Now the numbers given for the transition probabilities in the question do
not run down the columns as described above but accross the rows, so:

$ S=\left[ \begin{array}{cccc} 0.18&0.17&0.161&0.079\\ 0.274&0.368&0.339&0.355\\ 0.426&0.274&0.385&0.384\\ 0.12&0.188&0.115&0.182 \end{array} \right] $

You say that you can work out the eigen values and vectors yourself so
I will leave that to you.

A steady state is a vector $X$ such that:

$ SX=X $

that is it is an eigen vector corresponding to an eigen value of $1$

RonL
November 7th 2006, 05:04 AM
fifthrapiers

Thanks.

Am I correct when I do the following then, to find a steady-state vector.

Let the steady-state vector be S;

We want SX = X, as you said, for some eigenvalue of 1 corresponding to the eigenvector X.

SX - X = 0

(S - I)X = 0

Subtract 1 from each of the diagonals in S;

Then, row reduce S and that is your answer. But when I row reduce S, I get the identity matrix.
November 7th 2006, 06:56 AM
CaptainBlack

Quote:

Originally Posted by fifthrapiers

Thanks.

Am I correct when I do the following then, to find a steady-state vector.

Let the steady-state vector be S;

We want SX = X, as you said, for some eigenvalue of 1 corresponding to the eigenvector X.

SX - X = 0

(S - I)X = 0

Subtract 1 from each of the diagonals in S;

Then, row reduce S and that is your answer. But when I row reduce S, I get the identity matrix.

Do you have the eigen values of S? If you do is 1 one of them, if not there
is no steady state. If there is you need to solve

(S-I)X=0

for X, which is a standard "find the eigen vector corresponding to eigen
value 1", (subject to the condition that sum(X_i)=1 as these are probabilities).

RonL
November 7th 2006, 09:08 AM
AfterShock

Quote:

Originally Posted by CaptainBlack

Think of it this way, you have a col. vector $X$ with the probabilities of the
current nucloetide being A,C,G,T down the columns.

Then the stochastic $S$ matrix will give you the probabilities for the nucloetide
at the next position as $SX$ .

So if the current nucloetide is A, the first column of $S$ gives the probability
that next nucloetide is A, C, G or T, similarly if the current nucloetide is C
the second column gives the probability that next nucloetide is A, C, G or T,
and so on.

Now the numbers given for the transition probabilities in the question do
not run down the columns as described above but accross the rows, so:

$ S=\left[ \begin{array}{cccc} 0.18&0.274&0.426&0.12\\ 0.17&0.368&0.274&0.188\\ 0.161&0.339&0.385&0.115\\ 0.079&0.355&0.384&0.182 \end{array} \right] $

You say that you can work out the eigen values and vectors yourself so
I will leave that to you.

A steady state is a vector $X$ such that:

$ SX=X $

that is it is an eigen vector corresponding to an eigen value of $1$

RonL

I agree with everything CaptainBlack has said, however, his stochastic matrix is wrong. Take the transpose of that and it is correct. Recall that the columns of the stochastic matrix always equal 1, since they are "probability vectors". That should help.
November 7th 2006, 09:16 AM
CaptainBlack

Quote:

Originally Posted by AfterShock

I agree with everything CaptainBlack has said, however, his stochastic matrix is wrong. Take the transpose of that and it is correct. Recall that the columns of the stochastic matrix always equal 1, since they are "probability vectors". That should help.

Oppps I had read the descriptions the wrong way around (without thinking about what the result meant):mad:

Matrix now transposed (but worth checking anyway - its fiddly typing matrices in LaTeX

RonL