# Thread: Concept of a markov chain

1. ## Problem of a markov chain

Suppose that whether or not it rains today depends on previous weather
conditions through the last two days.

Specifically , suppose that
if it has rained for the past two days,then it will rain tommorow with probability 0.7;
if it rained today but not yesterday, then i will rain tommorow with probability o.5;
if it rained yesterday but not today, then it will rain tommorow with probability 0.4;
if it has not rained in the past two days, then it will rain tommorow with probability 0.2.

=====================================
the transition matrix

P = $\begin{bmatrix}0.7&0&0.3&0\\0.5&0&0.5&0\\0&0.4&0&0 .6\\0&0.2&0&0.8 \end{bmatrix}$
======================

i dont understand the mechanism to obtain the transition matrix in such a pattern............
why there are zeros? what are the meanings behind them?

2. If it rained yesterday and it rained today, the probability it rained today and will rain tomorrow is 0.7.

If it did not rain yesterday but it rained today, the probability it rained today and will rain tomorrow is 0.5.

If it rained yesterday but it didn't rain today, the probability it rained today and will rain rain tomorrow is 0.

If it did not rain yesterday and did not rain today, the probability it rained today and will rain tomorrow is 0.

If it rained yesterday and it rained today, the probability it didn't rain today and will rain tomorrow is 0.

If it did not rain yesterday but it rained today, the probability it didn't rain today and will rain tomorrow is 0.

If it rained yesterday but it didn't rain today, the probability it didn't rain today and will rain tomorrow is 0.4.

If it did not rain yesterday and did not rain today, the probability it didn't rain today and will rain tomorrow is 0.2.

If it rained yesterday and it rained today, the probability it rained today and will not rain tomorrow is 1-0.7= 0.3.

If it did not rain yesterday but it rained today, the probability it rained today but will not rain tomorrow is 1-0.5=0.5.

and so on