Probability matrix Watch

abcd123
Badges: 0
#1
Report Thread starter 10 years ago
#1
A group of families in a country have either income group low medium or high, and the families change income group from one generation to the next according to a Markov chain with transitition probabability matrix

P= \begin{pmatrix} 0.6 & 0.3 & 0.1 \\0.2 & 0.7 & 0.1 \\0.1 & 0.3 & 0.6 \end{pmatrix}

(e.g. chance of moving from low income to medium income = 0.3)

could do the 1st part which asked about probabilities of low income having its next few generations in low/medium or high. (finding P^2 and P^3)

Next part asked:
Find the proportion of families in the country that are in the 3 income groups? Give your reasoning.

How would you do this?
Thanks
0
quote
reply
Zhen Lin
Badges: 10
Rep:
?
#2
Report 10 years ago
#2
Hrm. If we assume that the income groups in the country have stabilised, then what you have to find is the eigenvector of your transition matrix corresponding to eigenvalue 1.

But, uh, are you sure you have the matrix correct? Neither the columns nor the rows add up to 1 consistently.
0
quote
reply
abcd123
Badges: 0
#3
Report Thread starter 10 years ago
#3
(Original post by Zhen Lin)
Hrm. If we assume that the income groups in the country have stabilised, then what you have to find is the eigenvector of your transition matrix corresponding to eigenvalue 1.

But, uh, are you sure you have the matrix correct? Neither the columns nor the rows add up to 1 consistently.
Just had the high to high value wrong, corrected that now. The columns don't need to add up to 1, just the rows. Not sure that's the right approach, as seems bit hard, and also the part in the question saying ' explain your reasoning', which suggests it isn't a direct calculation.
0
quote
reply
DFranklin
Badges: 18
Rep:
?
#4
Report 10 years ago
#4
No, he's right. (Note that finding the eigenvector corresponding to 1 is the same as solving Ax = x; it's not very difficult).

The "explain your reasoning" is because you need to justify why the current proportions are the same as those given by the steady state vector. (Roughly: if you assume the matrix hasn't changed over time, then over hundreds of generations it will evolve to the steady state).
0
quote
reply
abcd123
Badges: 0
#5
Report Thread starter 10 years ago
#5
I tried working it out, but don't think I did it right. (Haven't done eigenvectors for ages)

SO have to solve Ax = x (where x is a vector)
Ax - x = 0
Ax - Ix = 0
(A-I)x = 0

so
\begin{pmatrix} 0.6 & 0.3 & 0.1 \\0.2 & 0.7 & 0.1 \\0.1 & 0.3 & 0.6 \end{pmatrix} \begin{pmatrix} L \\M \\H \end{pmatrix} = \begin{pmatrix} 0 \\0 \\0 \end{pmatrix}

becomes

\begin{pmatrix} -0.4 & 0.3 & 0.1 \\0.2 & -0.3 & 0.1 \\0.1 & 0.3 & -0.4 \end{pmatrix} \begin{pmatrix} L \\M \\H \end{pmatrix} = \begin{pmatrix} 0 \\0 \\0 \end{pmatrix}

Then after row reduction I got 000 on the bottom row, so H is an arbitrary solution k, and then I also found that L and H = k. Does this sound ok?
0
quote
reply
Zhen Lin
Badges: 10
Rep:
?
#6
Report 10 years ago
#6
Er, well, you have found a correct column eigenvector, but since you're using a right stochastic matrix, aren't you supposed to find row eigenvectors, i.e. \mathbf{xA} = \mathbf{x}?
0
quote
reply
abcd123
Badges: 0
#7
Report Thread starter 10 years ago
#7
oh, so need to solve

\begin{pmatrix} L & M & H \end{pmatrix}\begin{pmatrix} 0.6 & 0.3 & 0.1 \\0.2 & 0.7 & 0.1 \\0.1 & 0.3 & 0.6 \end{pmatrix} = \begin{pmatrix} L & M & H \end{pmatrix}?

so xA = x
xA-Ix = 0
x(A-I) = 0

\begin{pmatrix} L & M & H \end{pmatrix}\begin{pmatrix} -0.4& 0.3 & 0.1 \\0.2 & -0.3 & 0.1 \\0.1 & 0.3 & -0.4 \end{pmatrix} = \begin{pmatrix} L & M & H \end{pmatrix}

Then

\begin{pmatrix} L & M & H \end{pmatrix}\begin{pmatrix} -0.4 & 0.3 & 0.1 \\0 & -0.3 & 0.3 \\0 & 1.5 & -1.5 \end{pmatrix} = \begin{pmatrix} L & M & H \end{pmatrix}

Then

\begin{pmatrix} L & M & H \end{pmatrix}\begin{pmatrix} -0.4 & 0.3 & 0.1 \\0 & -0.3 & 0.3 \\0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} L & M & H \end{pmatrix}

Hmm doing this wrong I think?
0
quote
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of Lincoln
    Mini Open Day at the Brayford Campus Undergraduate
    Wed, 19 Dec '18
  • University of East Anglia
    UEA Mini Open Day Undergraduate
    Fri, 4 Jan '19
  • Bournemouth University
    Undergraduate Mini Open Day Undergraduate
    Wed, 9 Jan '19

Were you ever put in isolation at school?

Yes (261)
26.74%
No (715)
73.26%

Watched Threads

View All