I don\'t really understand this class or stats in general. could you explain as
ID: 3205632 • Letter: I
Question
I don't really understand this class or stats in general. could you explain as simply as possible? Thank You! I will rate! :)
5. (10 pts) Suppose, that whether or not it rains today depends on previous weather conditions through the last three days. (a) Show how this system may be analyzed by using a Markov Chain. How many states are needed? (b) Suppose that if it has rained for all of the past three days, then it will rain today with probability 0.8; if it didn't rain for the past three days, it will rain today with probability 0.2; and in any other case the weather today will, with probability 0.6 be the same as the weather yesterday. Determine P(transition matrix) for this Markov Chain.Explanation / Answer
Define the following states. Let i = the day. For example if today is day i, then yesterday is day i-1, and the day before is day i-2. Table 1 is a summary of the states. Y denotes that it did rain on day i, where N denotes that it did not rain on day i.
Table 1: State Definition
Rain (Y/N)
0
Y
Y
Y
1
N
Y
Y
2
Y
N
Y
3
Y
Y
N
4
N
N
Y
5
Y
N
N
6
N
Y
N
7
N
N
N
Now we need to determine which transition probabilities Pij are 0. Table 2 should help.
Table 2: Definitions of Transition Probabilitities
Pij
Day j=i+1
Day i
Day i-1
Day i-2
P00
Y
Y
Y
Y
P10=0
Y
N
Y
Y
P20=0
Y
Y
N
Y
P30
Y
Y
Y
N
P40=0
Y
N
N
Y
P50=0
Y
Y
N
N
P60=0
Y
N
Y
N
P70=0
Y
N
N
N
P01
N
Y
Y
Y
P11=0
N
N
Y
Y
P21=0
N
Y
N
Y
P31
N
Y
Y
N
P41=0
N
N
N
Y
P51=0
N
Y
N
N
P61=0
N
N
Y
N
P71=0
N
N
N
N
P02=0
Y
Y
Y
Y
P12
Y
N
Y
Y
P22=0
Y
Y
N
Y
P32=0
Y
Y
Y
N
P42=0
Y
N
N
Y
P52=0
Y
Y
N
N
P62
Y
N
Y
N
P72=0
Y
N
N
N
P03=0
Y
Y
Y
Y
P13=0
Y
N
Y
Y
P23
Y
Y
N
Y
P33=0
Y
Y
Y
N
P43=0
Y
N
N
Y
P53
Y
Y
N
N
P63=0
Y
N
Y
N
P73=0
Y
N
N
N
P04=0
N
Y
Y
Y
P14
N
N
Y
Y
P24=0
N
Y
N
Y
P34=0
N
Y
Y
N
P44=0
N
N
N
Y
P54=0
N
Y
N
N
P64
N
N
Y
N
P74=0
N
N
N
N
P05=0
Y
Y
Y
Y
P15=0
Y
N
Y
Y
P25=0
Y
Y
N
Y
P35=0
Y
Y
Y
N
P45
Y
N
N
Y
P55=0
Y
Y
N
N
P65=0
Y
N
Y
N
P75
Y
N
N
N
P06=0
N
Y
Y
Y
P16=0
N
N
Y
Y
P26
N
Y
N
Y
P36=0
N
Y
Y
N
P46=0
N
N
N
Y
P56
N
Y
N
N
P66=0
N
N
Y
N
P76=0
N
N
N
N
P07=0
N
Y
Y
Y
P17=0
N
N
Y
Y
P27=0
N
Y
N
Y
P37=0
N
Y
Y
N
P47
N
N
N
Y
P57=0
N
Y
N
N
P67=0
N
N
Y
N
P77
N
N
N
N
Now we define the transition matrix P
P00
P01
P02=0
P03=0
P04=0
P05=0
P06=0
P07=0
P10=0
P11=0
P12
P13=0
P14
P15=0
P16=0
P17=0
P20=0
P21=0
P22=0
P23
P24=0
P25=0
P26
P27=0
P30
P31
P32=0
P33=0
P34=0
P35=0
P36=0
P37=0
P40=0
P41=0
P42=0
P43=0
P44=0
P45
P46=0
P47
P50=0
P51=0
P52=0
P53
P54=0
P55=0
P56
P57=0
P60=0
P61=0
P62
P63=0
P64
P65=0
P66=0
P67=0
P70=0
P71=0
P72=0
P73=0
P74=0
P75
P76=0
P77
In Problem 3 we are given that P00=.8, P75=.2. By examining Table 2, we can see that we are also given that P30=.6, P23=.6, P53=.6, P14=.6, P64=.6, P47=.6, P77=.6. We can put these values into P, and use the fact that each row adds up to 1 to find the rest of the probabilities.
.8
.2
0
0
0
0
0
0
0
0
.4
0
.6
0
0
0
0
0
0
.6
0
0
.4
0
.6
.4
0
0
0
0
0
0
0
0
0
0
0
.4
0
.6
0
0
0
.6
0
0
.4
0
0
0
.4
0
.6
0
0
0
0
0
0
0
0
.2
0
.8
Rain (Y/N)
State Day i Day i -1 Day i-20
Y
Y
Y
1
N
Y
Y
2
Y
N
Y
3
Y
Y
N
4
N
N
Y
5
Y
N
N
6
N
Y
N
7
N
N
N
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