Class 9
The probabilities of outcomes of each experiment depend only on the outcome of the previous experiment.
\[ \operatorname{P}(X_{n+1} = j \mid X_n = i, X_{n-1} = i_{n-1}, X_{n-2} = i_{n-2}, \dots, X_0 = i_0) = \operatorname{P}(X_{n+1} = j \mid X_n = i) = P_{ij} \]
Transition probabilities
\[P_{ij} = \operatorname{P}(X_{n+1} = j \mid X_n = i)\]
Transition Matrix
\[ \mathbf{P} = \begin{bmatrix} P_{00} & P_{01} & P_{02} & \cdots \\ P_{10} & P_{11} & P_{12} & \cdots \\ \vdots & \vdots & \vdots & \\ P_{i0} & P_{i1} & P_{i2} & \cdots \\ \vdots & \vdots & \vdots & \\ \end{bmatrix} \]
Transition over \(k\) steps:
\[P^k_{ij} = \operatorname{P}(X_{n+k} = j \mid X_n = i)\]
Chapman-Kolmogorov Equations:
\[P_{ij}^{m+n} = \sum_{k = 0, \dots} P_{ik}^m P_{kj}^n\]
\(n\)-step transition matrix is simply \(\mathbf{P}^n\).
\[ \mathbf{P} = \begin{bmatrix} .9 & .1 \\ .4 & .6 \end{bmatrix} \]
\[ \small \mathbf{P} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & \cdots & 0 & 0 \\ 1-p & 0 & p & 0 & 0 & \cdots & 0 & 0 \\ 0 & 1-p & 0 & p & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1-p & 0 & p & \cdots & 0 & 0 \\ 0 & 0 & 0 & 1-p & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & 0 & \cdots & 0 & p \\ 0 & 0 & 0 & 0 & 0 & \cdots & 0 & 1 \end{bmatrix} \]
In a sequence of independent coin flips, let \(N\) denote the number of flips until there are 3 consecutive heads.
Given a Markov Chain with states 1, 2, 3, 4, and 5, with some transition probabilities
Starting at state 1, what is the probability that the first time you enter a state greater than 2 is on step 4?
You have 5 stations, a message sent from each station will end up at any of the stations with the following probabilities:
\[\mathbf{P} = \begin{bmatrix} .1 & .2 & .5 & 0 & .2 \\ .3 & 0 & .7 & 0 & 0 \\ .6 & .1 & .3 & 0 & 0 \\ 0 & 0 & 0 & .6 & .4 \\ 0 & 0 & 0 & .3 & .7 \end{bmatrix}\]