In Freedonia, every day is either cloudy or sunny (not both). If it's sunny on any given day, then the probability that the next day will be sunny is $\frac 34$. If it's cloudy on any given day, then the probability that the next day will be cloudy is $\frac 23$.

a. In the long run, what fraction of days are sunny?

b. Given that a consecutive Saturday and Sunday had the same weather in Freedonia, what is the probability that that weather was sunny?

I tried using weighted coins, but that didn't work. Can I get two answers, one for each problem, solution not necessary, as I need to figure out which of my methods leads to the correct answer. Thanks.

I found a congruent problem, but it didn't have answers I could comprehend.


Consider a 'cycle': Start with a sunny day; probability of changing to a rainy day is 1/4. so on average a run of sunny days will be four days long. Then (by similar logic) there will be a run of rainy days averaging four days in length. Average cycle length is 7 days of which 4 are sunny. Answer to (a) is 4/7.

Addendum: Here is a simulation of 100,000 steps of the chain using R software, where state 0 = Sun and state 1 = Rain. (Note that @GrahamKemp's excellent Answer, posted while I was working on this, uses 1 = Sun and 0 = Rain.)

For your part (a), We already know that the steady-state distribution has sum 4/7 of the time; this is consistent with simulation results. For your part (b), the required conditional probability is 3/5, again consistent with simulation results.

 m = 100000; n = 1:m;  alpha = 1/4;  beta = 1/3
 x = numeric(m); x[1] = 0
 for (i in 2:m)  {
     if (x[i-1]==0) x[i] = rbinom(1, 1, alpha)
     else x[i] = rbinom(1, 1, 1 - beta)  }
 mean(x==0); 4/7
 ## 0.56847    
 ## 0.5714286
 x1 = x[1:(m-1)];  x2 = x[2:m]
 mean(x1[x1==x2]==0);  3/5
 ## 0.5958692
 ## 0.6

A sketch of my rationale for (b) is $P(\text{SS}) = (4/7)(3/4) = 3/7,\;$ $P(\text{RR}) = 2/7.\;$ So the answer is $\frac{3/7}{3/7 + 2/7}.$

You asked for answers to check against the ones you already obtained. I have tried to give them to you in a way that may show you how to think about such simple Markov chains from points of view that may not be in your textbook.


(a) If $p$ is the long-term probability (aka equilibrium point) that it is sunny, then the probability that it is sunny on a following day is also $p$, so: $\Box p + \Box (1-p) = p$

Likewise the probability that it is not sunny on the subsequent day is: $\Box p + \Box (1-p) = (1-p)$.

Fill in the boxes from the given information and solve the simultaneous equation

(b) Use this value as the prior probability that the weather is sunny on Saturday, and construct a Bayesian case for the posterior probability for the given condition.

Let $T$ be the indicator event that it is sunny on Saturday, and $N$ the indicator event that it is sunny on Sunday.   We have from the long-term probability: $\mathsf P(T=1)=p, \mathsf P(T=0)=(1-p)$ and also from the given information: $\mathsf P(N=1\mid T=1)=3/4, \mathsf P(N=0\mid T=0)=2/3$.

Find $\mathsf P(T=1 \mid N=T)$ using conditional probability rules.

  • $\begingroup$ So in a) the first equation will be .9p+.1(1-p)=p and the second equation: .2p+.8(1-p)=1-p ? Am I right? $\endgroup$ – Aaron Martinez Apr 30 '17 at 0:24

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