0.16 Markov chains

We obtain the following transition matrix by properly placing the row and column entries. Note that if, for example, Professor Symons bicycles one day, then the probability that he will walk the next day is $1/4$ , and therefore, the probability that he will bicycle the next day is $3/4$ .

Let $W$ denote that Professor Symons walks and $B$ denote that he rides his bicycle.

We use the following tree diagram to compute the probabilities.

The probability that Professor Symons walked on Wednesday given that he walked on Monday can be found from the tree diagram, as listed below.

$P\left(\text{Walked Wednesday}\mid \text{Walked Monday}\right)=P\left(\text{WWW}\right)+P\left(\text{WBW}\right)=1/4+1/8=3/8$ .

$P\left(\text{Bicycled Wednesday}\mid \text{Walked Monday}\right)=P\left(\text{WWB}\right)+P\left(\text{WBB}\right)=1/4+3/8=5/8$ .

$P\left(\text{Walked Wednesday}\mid \text{Bicycled Monday}\right)=P\left(\text{BWW}\right)+P\left(\text{BBW}\right)=1/8+3/\text{16}=3/5/\text{16}$ .

$P\left(\text{Bicycled Wednesday}\mid \text{Bicycleed Monday}\right)=P\left(\text{BWB}\right)+P\left(\text{BBB}\right)=1/8+9/\text{16}=\text{11}/\text{16}$ .

We represent the results in the following matrix.

Alternately, this result can be obtained by squaring the original transition matrix.

We list both the original transition matrix $T$ and $T^2$ as follows:

The reader should compare this result with the probabilities obtained from the tree diagram.

Consider the following case, for example,

It makes sense because to find the probability that Professor Symons will walk on Wednesday given that he bicycled on Monday, we sum the probabilities of all paths that begin with $B$ and end in $W$ . There are two such paths, and they are $\text{BWW}$ and $\text{BBW}$ .