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Probability of rain today given rain yesterday is $0.6$, $0.2$ otherwise.

(a) What's the P(rain on the day after tomorrow if it rained today)?

(b) What's mean number of days of a rainy period?

(c) What fraction of days does rain fall?

(b) Really not sure how to do this one. Any help will be appreciated.

(c) For this problem, am I supposed to do pi = pi * P? Then I can compute the stationary distribution and find the proportion in the state for rain? Is that right?

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    The transition matrix is $$\begin{bmatrix}0.6 & 0.2\0.4 & 0.8\end{bmatrix}$$

    Here, the left column represents the probabilities of rain or no rain on the next day, given that it rained today. The right column represents the probabilities of rain or no rain on the next day, given that it did not rain today.

     – Zubin Mukerjee Oct 15 '19 at 20:11
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    If we square the transition matrix, we get

    $$\begin{bmatrix}0.6 & 0.2\0.4 & 0.8\end{bmatrix}^2 =\begin{bmatrix}0.44 & 0.28\0.56 & 0.72\end{bmatrix}$$

    Here, the left column represents the probabilities of rain or no rain two days from now, given that it rained today. The right column represents the probabilities of rain or no rain two days from now, given that it did not rain today.

    The probability you are interested in is the top left corner, which is $0.44$.

     – Zubin Mukerjee Oct 15 '19 at 20:11
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    For part $(b)$, these are Bernoulli Trials as soon as the first rainy day starts, you may compute a geometric series: If $R$ is rain and $N$ is no rain, then find the probability of $RN$, $RRN$, $RRRN$, $RRRRN$, and so on up to infinity . Then multiply each term by the length of the rainy-day run ($1$, $2$, $3$, etc.). This is an expected value calculation. – Zubin Mukerjee Oct 15 '19 at 20:14
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    For part $(c)$, you are finding the steady state(s) of the process, you want to find a matrix $A$ so that $$A \begin{bmatrix}0.6 & 0.2\0.4 & 0.8\end{bmatrix} = A$$ This will give you the long-term probabilities of rain or no rain on an arbitrarily chosen day. – Zubin Mukerjee Oct 15 '19 at 20:16
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    you could write this as an answer @ZubinMukerjee This is hard to read in as a comments – nonuser Oct 15 '19 at 20:16
  • @Aqua I didn't want to put it as an answer, since I'm only answering part $(a)$, and giving a method to solve the other two parts, maybe it can be three separate questions :D – Zubin Mukerjee Oct 15 '19 at 20:19

1 Answers1

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We have the transition matrix $$P = \begin{pmatrix}3/5&2/5\\1/5&4/5\end{pmatrix}. $$ Denote the columns in order by $1$ and $2$ (as according to raining and not raining). Then by the Chapman-Kolmogorov equation: \begin{align} \mathbb P(X_2 = 1 \mid X_0=1) &= \mathbb P(X_2=1\mid X_1=1)\mathbb P(X_1=1\mid X_0=1)\\ &\quad+ \mathbb PX_2=1\mid X_1=2)\mathbb P(X_1=2\mid X_0=1)\\ &= P_{11}P_{11} + P_{21}P_{12}\\ &= (3/5)^2 + 2/5\cdot1/5\\ &= 11/25\approx 0.44. \end{align}

Assume now that $X_0=1$ (it is initially rainy) and let $\tau=\inf\{n>0:X_n=2\}$ (the first day it is not rainy. Then $\tau$ is geometrically distributed with success parameter $2/5$, and therefore has expectation $(2/5)^{-1}=5/2$.

Part (c) is asking for the stationary distribution of the Markov chain, which is the (unique) vector $\pi$ that satisfies $\pi P=\pi$ and $\sum_i \pi_i=1$. From $\pi P = \pi$ we have the system of equations \begin{align} 3/5\pi_1 + 1/5\pi_2 &= \pi_1\\ 2/5\pi_1+4/5\pi_2 &= \pi_2, \end{align} which yields $\pi_1=1/3,\pi_2=2/3$.

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