Question

The pattern of dry and wet days in Cleveland is a homogeneous Markov chain with two states. Every dry day is followed by another dry day with probability 0.8. Every wet day is followed by another wet day with probability 0.6. a) Today is dry in Cleveland. What is the chance of rain (wet) the day after tomorrow? b) Compute the probability that April 1 next year is wet in Cleveland.

Answer 1

solution:

a)

Given that today is dry in Cleveland, probability that there will be a rain the day after tomorrow computed here as:

• D --> D --> R, Prob: 0.8*(1 - 0.8) = 0.16
• D --> R --> R, Prob. (1 - 0.8)*0.6 = 0.12

Therefore the required probability here is 0.16+0.12=0.28

Therefore 0.28 Is the requIred probability here.

b)

For a random data way ahead in future, we need to find the infinite time probability share for the 2 states. Let these be X and Y for dry and wet days respectively. Then, we have here :

X = 0.8X + 0.4Y

0.2X = 0.4Y

X = 2Y

Also, we know that X + Y = 1, as there are only 2 states here.

3Y = 1

Y = 1/3

Therefore 1/3 Is the requIred probability here.

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