2003 (�����)

2003 (�����) - 2003...

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Time number of airline passenger 1950 1952 1954 1956 1958 1960 100 200 300 400 500 600 2003 ASSIGNMENT Q1 (a) P(Xn=2|Xn-1=0)=1/4 Xn-1=e n-1+e n-2 (en-1=-1 and e n-2=1or en-1=1 and e n-2=-1 ) If Xn=2 so en=1 and en-1=1 as well and en-1=1 and e n-2=-1 in Xn-1 The probability of en=1 is 1/2, en-1=1/2 as well,so the answer is 1/4. (b) P(Xn=2|Xn-1=0,Xn-2=2)=0 For Xn-2=2 en-2=1 and e n-3=1 Xn-1=0, e n-2=1and en-1=-1 Xn=2 en=1 and en-1=1 but en-1 =-1 according to Xn-1 (c) The process is not Markov chain, because the probability is not only depend on one past state. It is also depend on the several past states. Q2 (a) There is clearly an increasing trend for the data, but there are some seasonal fluctuates in the data. (b) Since the data is clearly an increasing trend, it did not suggest any transformation, one possible transformation may be log transformation. This lead to smaller variance.
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with log transformation day number of airline passenger 1950 1952 1954 1956 1958 1960 5.0 without log transformation day 1950 1952 1954 1956 1958 1960 100 V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 200 300 400 500 600 boxplot for yearly running median (c) median [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [1,] 223 214.5 251.5 252 252 289.5 333 320 285.5 251.5 220 253.5 There is an increasing trend for every year. The median keep increasing. (d)
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Time WW 1950 1952 1954 1956 1958 1960 150 200 250 300 350 400 450 Time A - WW 1950 1952 1954 1956 1958 1960 -50 0 50 100 (e) St=Xt-Tt (f) d=Xt-St BB is the seasonal component estimate using moving average method. So A-BB is the de-seasonal data.
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Time A - BB 1950 1952 1954 1956 1958 1960 100200300 400 500 600 Time UU 1950 1952 1954 1956 1958 1960 800 1000 1400 UU is the data that after using the moving average filter on A-BB. It is the re-estimate trend. It
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This note was uploaded on 06/12/2011 for the course ASB 2003 taught by Professor Kim during the Three '11 term at University of New South Wales.

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2003 (�����) - 2003...

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