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Svet_ta [14]
3 years ago
12

This question is to show that we can `recode' and model a situation that depends on nitely many past states as a homogeneous Mar

kov chain. Suppose we model the daily weather as a Markov chain. The weather has just two states: cloudy and sunny. Suppose that if it is sunny today and was sunny yesterday then it will be sunny tomorrow with probability 0:6; if sunny today but cloudy yesterday then it will be sunny tomorrow with probability 0:5; if cloudy today but sunny yesterday then it will be sunny tomorrow with probability 0:4; if it was cloudy for the last two days then it will be sunny tomorrow with probability 0:2. Calculate the expected fraction of cloudy days.
Mathematics
1 answer:
Hatshy [7]3 years ago
4 0

Answer:

F=y+z=4/6.25

Step-by-step explanation:

First, we have to consider that in the problem model we have only two possible states: sunny and cloudy. Now, according to the information given in the statement, we also have the behavior of the last two days. In any case, we can have four possible transitional states:

Today-Yesterday(S=Sunny, C=cloudy)

1) ST and SY (Sunny today and sunny yesterday)

2) ST and CY.

3) CT and SY.

4) CT and CY.

Now, according to the statement, the probabilities given for the four states can be expressed by the following matrix:

\left[\begin{array}{cccc}0.6&0&(1-0.6)&0\\0.5&0&(1-0.5)&0\\0&0.4&0&(1-0.4)\\0&0.2&0&(1-0.2)\end{array}\right]

Now, making w, x, y, z as the transition probabilities for the four states mentioned, we then have that:

x=0.6w+0.5x

w=1.25x (1)

x=0.4y+0.2z (2)

y=0.4w+0.5x

y= 0.4(1.25x)+0.5x=x

y=x (3)

replacing 3 in 2:

y=0.4y+0.2x

x=3y (4)

And as w+x+y+z= 1 (no more possible combinations):

w+x+y+z=1 (5)

So, replacing the expressions obtained previously in equation 5, we have finally that:

1.25x+x+x+3x=1

x=1/6.25=y

z=3x=3/6.25

So, the fraction of sunny days is given by:

F=y+z=4/6.25

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