Markov Chain Question, need help!

Question

The weather on any given day in a particular city can be sunny, cloudy, or rainy. It has been observed to be predictable largely on the basis of the weather on the previous day. Specfically:

if it is sunny on one day, it will be sunny the next day 1/5 of the time, and be cloudy the next day 1/5 of the time

if it is cloudy on one day, it will be sunny the next day 3/5 of the time, and be cloudy the next day 1/5 of the time

if it is rainy on one day, it will be sunny the next day 2/5 of the time, and be cloudy the next day 3/5 of the time

Using 'sunny', 'cloudy', and 'rainy' (in that order) as the states in a system, set up the transition matrix for a Markov chain to describe this system.

Find the proportion of days that have each type of weather in the long run. I need a transformative 3x3 matrix and proportions of each weather(sunny, cloudy rainy)

Answer 1

Represent the three stable weather conditions as s, c, r. The table below shows the transition probabilities from yesterday's state (1st column in the table) to today's state (1st row in the table).

	sunny	cloudy	rainy
sunny	0.2	0.2	0.6
cloudy	0.6	0.2	0.2
rainy	0.4	0.6	0

The figures are derived from the given info in the question. The figures in red are deduced from the simple fact that each row must sum to 1, since the three conditions are the only allowable or measurable conditions applicable to the situation, therefore, knowing two of the probabilities automatically determines the third. (For example, no provision has been made for snowy or windy.)

This table becomes the transition matrix, A:

⎛ 0.2 0.2 0.6 ⎞

⎜0.6 0.2 0.2 ⎟

⎝ 0.4 0.6  0  ⎠

Let's now start with an arbitrary initial condition: today is sunny, represented by the matrix π₀=

( 1 0 0 ). What is the probability for tomorrow's weather? We carry out the matrix multiplication π₁=π₀A=( 0.2 0.2 0.6 ) which, of course, matches expectation.

π₂=π₁A=

( 0.2×0.2+0.2×0.6+0.6×0.4  0.2×0.2+0.2×0.2+0.6×0.6  0.2×0.6+0.2×0.2+0.6×0 ),

π₂=( 0.04+0.12+0.24 0.04+0.04+0.36, 0.12+0.04 )=( 0.40 0.44 0.16 ), corresponding to sunny 40%; cloudy 44%; rainy 16%.

π₃=( 0.408 0.264 0.328 ); π₄=( 0.3712 0.3312 0.2976);

π₅=( 0.385408 0.315584 0.299008 ); π₆=( 0.3860352 0.3196032 0.2943616 );

π₇=( 0.3867136 0.31774464 0.29554176 ); π₈=( 0.386206208 0.318216704 0.295577088 );

π₉=(  0.3864020992 0.3182308352 0.2953670656 ).

So we approach stability at approximately sunny 38.6%; cloudy 31.8%; rainy 29.5%. Note that these figures sum to 100% given that there are rounding errors of 0.1%.

A more accurate assessment is sunny 38.64%; cloudy 31.82%; rainy 29.55%. Rounding error about 0.01%.

Expressed exactly as fractions: sunny 17/44; cloudy 7/22; rainy 13/44 (see proof below).

To get this solution we have the stable weather forecast π_n=( s c r ).

π_n+1=π_n=π_nA when we have stability, so ( s c r )=( ⅕s+⅗c+⅖r ⅕s+⅕c+⅗r ⅗s+⅕c ).

Therefore, s=⅕s+⅗c+⅖r, -⅘s+⅗c+⅖r=0, -4s+3c+2r=0;

c=⅕s+⅕c+⅗r, ⅕s-⅘c+⅗r=0, s-4c+3r=0;

r=⅗s+⅕c, ⅗s+⅕c-r=0, 3s+c-5r=0.

We also know that s+c+r=1, so r=1-s-c.

-4s+3c+2r=-4s+3c+2-2s-2c=0, -6s+c=-2, c=6s-2.

s-4c+3r=s-24s+8+3-3s-18s+6=0, -44s+17=0, s=17/44, c=6s-2=7/22, r=1-17/44-7/22=13/44.