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If the weather can either be sunny or rainy on a particular day, and the probability of sun on any particular day depends on the weather in the previous two days, assume the following is true: $$P[\text{sun today}] = \begin{cases} \frac{1}{6} \text{ if }r = 2 \\ \frac{1}{3} \text{ if }r = 1 \\ \frac{1}{2} \text{ if }r = 0 \\ \end{cases}$$ where $r$ is the number of rainy days in the previous two days.

What is the PTM of this Markov chain?

I'm not sure how to enumerate the states of this chain.

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  • $\begingroup$ Will be easier both for you to understand and for others to answer if you include the definitions in the question. $\endgroup$ – mathreadler Oct 3 '16 at 21:56
  • $\begingroup$ A hint is that you can represent the space of one yes/no event on two consecutive days as two-bit binary number. $\endgroup$ – mathreadler Oct 3 '16 at 22:11
  • $\begingroup$ Following @mathreadler's suggestion, you have a 4-state chain and a $4 \times 4$ transition matrix P. State space might be $\{00, 01, 10, 11\}.$ where 1=sun and 0=rain, and 00 means two rainy days in a row. $\endgroup$ – BruceET Oct 3 '16 at 23:08
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You consider two recent days and each day has two possibilities. Therefore, you have four states. Let R represent rainy and S sunny. So the states are: RR, RS, SR, SS.

So for instance if you are in the state RR, you go to state RS with probability $\frac{1}{6}$ and remain in that state with probability $1-\frac{1}{6}$. Hence the transtion matrix becomes

$\hspace{6.75cm}$RR$\hspace{5mm}$ RS$\hspace{4mm}$ SR$\hspace{4mm}$ SS $$\begin{bmatrix} 1-\frac{1}{6}&\frac{1}{6} & 0 & 0\\ 0&0 & 1-\frac{1}{3} & \frac{1}{3}\\ 1-\frac{1}{3}&\frac{1}{3} & 0 & 0\\ 0&0 & 1-\frac{1}{2} & \frac{1}{2} \end{bmatrix}$$

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  • $\begingroup$ Hi I got $\begin{bmatrix} \frac{5}{6} & \frac{1}{6} & 0 & 0 \\ 0 & 0 & \frac{4}{6} & \frac {2}{6} \\ \frac{4}{6} & \frac{2}{6} & 0 & 0 \\ 0 & 0 & \frac{1}{2} & \frac{1}{2} \end{bmatrix}$ $\endgroup$ – MoronicHero Oct 3 '16 at 23:45
  • $\begingroup$ I'm not sure how you got the first row. I got 5/6 and 1/6. $\endgroup$ – MoronicHero Oct 3 '16 at 23:49
  • $\begingroup$ I thought that only if there were no rainy days in the last two days you would have probability rain today 1/2 and probability sun today 1/2 (as given in the conditions). please let me know if I'm misunderstanding something $\endgroup$ – MoronicHero Oct 3 '16 at 23:52
  • $\begingroup$ Correct. The case you mentioned is the last row of the transition matrix. Another example is the first row which is explained in the answer. But notice that we can have rain today even if there had been one or even two rainy days before. $\endgroup$ – msm Oct 4 '16 at 0:02

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