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1 Markov Chains: Transitional Modeling Qi Liu

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2 content Terminology Transitional Models without Explanatory Variables Transitional Models without Explanatory Variables Inference for Markov chains Data Analysis :Example 1 (ignoring explanatory variables)Example 1 Transitional Models with Explanatory Variables Data Anylysis: Example 2 (with explanatory variables)Example 2

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3 Terminology Transitional models Markov chain K th-order Markov chain Tansitional probabilities and Tansitional matrix Tansitional probabilities and Tansitional matrix

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4 Transitional models {y0,y1,…,yt-1} are the responses observed previously. Our focus is on the dependence of Yt on the {y0,y1,…,yt-1} as well as any explanatory variables. Models of this type are called transitional models.

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5 Markov chain A stochastic process, for all t, the conditional distribution of Yt+1,given Y0,Y1, …,Yt is identical to the conditional distribution of Yt+1 given Yt alone. i.e, given Yt, Yt+1 is conditional independent of Y0,Y1, …,Yt-1. So knowing the present state of a Markov chain,information about the past states does not help us predict the future P(Yt+1|Y0,Y1, … Yt)=P(Yt+1|Yt)

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6 K th-order Markov chain For all t, the conditional distribution of Yt+1 given Y0,Y1,…,Yt is identical to the conditional distribution of Yt+1,given (Yt,…,Yt-k+1) P(Yt+1|Y0,Y1,…Yt)=P(Yt+1|Yt-k+1,Yt-k+2,….Yt) i.e, given the states at the previous k times, the future behavior of the chain is independent of past behavior before those k times. We discuss here is first order Markov chain with k=1.

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7 Tansitional probabilities

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8 Transitional Models without Explanatory Variables At first, we ignore explanatory variables. Let f(y0,…,yT) denote the joint probability mass function of (Y0,…,YT),transitional models use the factorization: f(y0,…,yT) =f(y0)f(y1|y0)f(y2|y0,y1)…f(yT|y0,y1,…,yT-1) This model is conditional on the previous responses. For Markov chains, f(y0,…,yT) =f(y0)f(y1|y0)f(y2|y1)…f(yT|yT-1) (*) From it, a Markov chain depends only on one-step transition probabilities and the marginal distribution for the initial state. It also follows that the joint distribution satisfies loglinear model (Y0Y1, Y1Y2,…, YT-1YT) For a sample of realizations of a stochastic process, a contingency table displays counts of the possible sequences. A test of fit of this loglinear model checks whether the process plausibly satisfies the Markov property.

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9 Inference for Markov chains

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10 Inference for Markov chains(continue)

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11 Example 1 (ignoring explanatory variables) A study at Harvard of effects of air pollution on respiratory illness in children. The children were examined annually at ages 9 through 12 and classified according to the presence or absence of wheeze. Let Yt denote the binary response at age t, t=9,10,11,12. 1 wheeze;2 no wheeze y9y10y11y12coun t y9y10y11y12count 111194211119 111230211215 1121 212110 112228212244 121114221117 121212221242 122112222135 1222632222572

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12 Code of Example 1 Code of 11.7 data breath; input y9 y10 y11 y12 count; datalines; 1 1 1 1 94 1 1 1 2 30 1 1 2 1 15 1 1 2 2 28 1 2 1 1 14 1 2 1 2 9 1 2 2 1 12 1 2 2 2 63 2 1 1 1 19 2 1 1 2 15 2 1 2 1 10 2 1 2 2 44 2 2 1 1 17 2 2 1 2 42 2 2 2 1 35 2 2 2 2 572 ; proc genmod; class y9 y10 y11 y12; model count= y9 y10 y11 y12 y9*y10 y10*y11 y11*y12 /dist=poi lrci type3 residuals obstats; run; proc genmod; class y9 y10 y11 y12; model count= y9 y10 y11 y12 y9*y10 y9*y11 y10*y11 y10*y12 y11*y12 y9*y10*y11 y10*y11*y12/dist=poi lrci type3 residuals obstats; run; proc genmod; class y9 y10 y11 y12; model count= y9 y10 y11 y12 y9*y10 y9*y11 y9*y12 y10*y11 y10*y12 y11*y12 /dist=poi lrci type3 residuals obstats; run; data breath_new;set breath; a=y9*y10+y10*y11+y11*y12; b=y9*y12+Y10*y12+y9*y11; proc genmod; class y9 y10 y11 y12; model count= y9 y10 y11 y12 a b /dist=poi lrci type3 residuals obstats; run;

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13 Data analysis The loglinear model (y9y10,y10y11,y11y12) a first order Markov chain. P(Y11|Y9,Y10)=P(Y11|Y10) P(Y12|Y10,Y11)=P(Y12|Y11) G²=122.9025, df=8, with p-value<0.0001, it fits poorly. So given the state at time t, classification at time t+1 depends on the states at times previous to time t.

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14 Data analysis (cont…) Then we consider model (y9y10y11, y10y11y12),a second-order Markov chain, satisfying conditional independence at ages 9 and 12, given states at ages 10 and 11. This model fits poorly too, with G²=23.8632,df=4 and p-value<0.001.

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15 Data analysis (cont) The loglinear model (y9y10,y9y11,y9y12,y10y11,y10y12,y11y12) that permits association at each pair of ages fits well, with G²=1.4585,df=5,and p-value=0.9178086. Parameter Estimate Error Limits Square Pr > ChiSq y9*y10 1.8064 0.1943 1.4263 2.1888 86.42 <.0001 y9*y11 0.9478 0.2123 0.5282 1.3612 19.94 <.0001 y9*y12 1.0531 0.2133 0.6323 1.4696 24.37 <.0001 y10*y11 1.6458 0.2093 1.2356 2.0569 61.85 <.0001 y10*y12 1.0742 0.2205 0.6393 1.5045 23.74 <.000 y11*y12 1.8497 0.2071 1.4449 2.2574 79.81 <.0001

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16 Data analysis (cont) From above, we see that the association seems similar for pairs of ages1 year apart, and somewhat weaker for pairs of ages more than 1 year apart. So we consider the simpler model in which It also fits well, with G²=2.3, df=9, and p-value= 0.9857876.

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17 Estimated Conditonal Log Odds Ratios

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18 Transitional Models with Explanatory Variables

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20 Data Anylysis Example 2 (with explanatory variables) At ages 7 to 10, children were evaluated annually on the presence of respiratory illness. A predictor is maternal smoking at the start of the study, where s=1 for smoking regularly and s=0 otherwise.

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21 Child ’ s Respiratory Illness by Age and Maternal Smoking

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22 Data analysis (cont)

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23 Code of Example 2 data illness; input t tp ytp yt s count; datalines; 8 7 0 0 0 266 8 7 0 0 1 134 8 7 0 1 0 28 8 7 0 1 1 22 8 7 1 0 0 32 8 7 1 0 1 14 8 7 1 1 0 24 8 7 1 1 1 17 9 8 0 0 0 274 9 8 0 0 1 134 9 8 0 1 0 24 9 8 0 1 1 14 9 8 1 0 0 26 9 8 1 0 1 18 9 8 1 1 0 26 9 8 1 1 1 21 9 8 1 0 0 26 9 8 1 0 1 18 9 8 1 1 0 26 9 8 1 1 1 21 10 9 0 0 0 283 10 9 0 0 1 140 10 9 0 1 0 17 10 9 0 1 1 12 10 9 1 0 0 30 10 9 1 0 1 21 10 9 1 1 0 20 10 9 1 1 1 14 ; run; proc logistic descending; freq count; model yt = t ytp s/scale=none aggregate; run;

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24 Output from SAS Deviance and Pearson Goodness-of-Fit Statistics Criterion DF Value Value/DF Pr > ChiSq Deviance 8 3.1186 0.3898 0.9267 Pearson 8 3.1275 0.3909 0.9261 Analysis of Maximum Likelihood Estimates Standard Wald Parameter DF Estimate Error Chi-Square Pr > ChiSq Intercept 1 -0.2926 0.8460 0.1196 0.7295 t 1 -0.2428 0.0947 6.5800 0.0103 ytp 1 2.2111 0.1582 195.3589 <.0001 s 1 0.2960 0.1563 3.5837 0.0583

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25 Analysis

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26 The model fits well, with G²=3.1186, df=8, p- value=0.9267. The coefficient of is 2.2111 with SE 0.1582, Chi- Square statistic 195.3589 and p-value <.0001,which shows that the previous observation has a strong positive effect. So if a child had illness when he was t-1, he would have more probability to have illness at age t than a child who didn’t have illness at age t-1. The coefficient of s is 0.2960, the likelihood ratio test of H0 :=0 is 3.5837,df=1,with p-value 0.0583. There is slight evidence of a positive effect of maternal smoking.

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27 Interpratation of Paramters ß

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28 Thank you !

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