(very) beginner's example Markov Chain Person X is on holidays for T weeks. At the end of each week, she gets a mosquito-bite, which infects her (for the following week) with a probability of 0.2. The state-space $Y_t$ is hence either I ("infected") or N ("no disease"). $Y_1$ is "N" - the person starts out healthy. No medication is available, so once infected the person is infected until the end of the holidays.
The tasks are now to (a) specify $P(Y_1)$ as well as $P(Y_t | Y_{t-1})$ for t = 2,3,...,T (b) to draw the transition diagram and (c) to determine the probability that person X is not infected in $Y_4$, i.e. in the fourth week. 
My ideas and questions: (a)..? $P(Y_1)$ is obviously 1..? Am I missing something? For $P(Y_t | Y_{t-1})$ I would just specify the transition matrix P as follows (row 1 = infected, row 2 = not infected):
$
  \left[ {\begin{array}{cc}
   1 & 0 \\
   0.8 & 0.2 \\
  \end{array} } \right]
$
is there another way to specify $P(Y_t | Y_{t-1})$? (b) The transition diagram contains two states, from the state "I" there is only an arrow to itself with probability 1, from the state "N" an arrow to itself with probability 0.8 and an arrow to "I" with probability 0.2. Regarding (c), I would calculate the respective entry in $P^3$, calculating first the respective entry in $P^2$ we get 0.2*0+0.8*0.8 = 0.64, and then $P(Y_4=N)$ = 0.64*0.8 = 0.512, the probability that X is not infected after 4 weeks of holidays.
Are my solutions correct? Thanks for your help!
 A: You know without a doubt that at time $t=1$ the person is not infected, so $P(Y_1 = N) = 1$. Maybe you are used to seeing a Markov chain start with an initial distribution that's not just a point mass at one value but a deterministic start is perfectly reasonable.
For the transition matrix, let $\Omega = \{I, N\}$ denote the set of possible values. You have one conditional distribution per value in $\Omega$, each distribution specifies $|\Omega|$ probabilities, and the distributions do not change over time, so you'll need to have $|\Omega|^2$ probabilities specified which is what is accomplished by your $2 \times 2$ transition matrix. So there's no way that does not involve specifying 4 probabilities. This matrix is unique up to relabeling of the states.
Finally, you need to compute $P(Y_4 = N)$. Here is how I would do it: you know that $Y_t = I \implies Y_k = I$ for all $k \geq t$ so the event $\{Y_t = N\}$ is equivalent to the event $\{Y_1 = N, \dots, Y_t = N\}$. This means
$$
P(Y_4 = N) = P(Y_1 = N, Y_2 = N, Y_3 = N, Y_4 = N) \\
= P(Y_4 = N | Y_3 = N)P(Y_3 = N | Y_2 = N)P(Y_2 = N | Y_1 = N)P(Y_1 = N) \\
= (0.8)^3 \times 1 = 0.512
$$
which agrees with your computation. You got the right answer but if I can avoid computing powers of matrices by hand I try to do so, and since we have $P(Y_t = N | Y_{t-1} = I) = 0$ the computations simplify to the point that we can get away with this more direct approach. But in total generality you would indeed need to compute powers of matrices.
