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# 8.55 - Type Count Cell probabilities Starchy green(51 C72...

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Unformatted text preview: Type Count Cell probabilities Starchy green (51 C72) X1 : l997" 0 25 (2 +0) : 01 Starchy green(51W') X2=906 025(1—0)=02 0 < 0 <1 Starchy green (52 (32) X3 = 904 0 25(1- 9) = 93 Starchy green (52W) X4 = 32 0 25 0 = 04 £— 21 = 2X] = 3339 5-1 (a) We want to find NILE of 0 If X1, X2,X3,X4 denotes counts in four cells With cell probabilities 01,82,83,64 respectively Then likelihood function of ( 81,82, 83,84 ) is, ‘ x amperage.) : J—Hipf “Xvi 1—1 i-l Hence log likelihood is, 4 4 j :5 3(6) : 103m—ZXi I + 2X}, 103?: [the middle term in the loglikelihood should be summation logm‘ 1-1 1—1 ,1 :103221—in+Xliog[0.25[2+e)]+Xgiog[0 250—9)] i-l +X,iogf0 25r1+eﬂ+21< to 2501 025X1 0.252172 0.25X3 0.251s+ _ _ + : 025(2+e) 0.25[1+e) 0.25[i+e) 0.250 : X1_£_£+£=U 2+0 1—0 1—9 a => X1[e+eg]+X2[20+e’]+X3[29+09]+X.[2+20+e+e‘] = 0 => 03[+X1+Xj+X3+213]+e[Xl+2Xl+2X3+X,]+2X4 = 0 :> [X1+Xg+X3+X,]62—[X1—2X2—2X3—X‘]8—2X :0 Now, X1+X3+X3+X4 = 3339 And, Xl—2X2—2X3—X4 : —l655, X2 : 32 :2 383902+16558764 = 0 Solving this Quadratic in B We get, 0 = 0.035? 50 MLE of e is 0: 0.0357 Now, We Want to ﬁnd asymptotic variance of this MLE, a‘iogL —X1 X2 X; X, a : 2+ 2+ 2__2 ae (2+9) [1-9) [1-9) 6 _E[a‘iogr] 7 MAG) shag) E(X3)+E[X4) 862 ’(2+e)2 (1-0)” (1-0)” 62 : 1(9) : n01 _ 2202 _ n05 +ﬂ (2+0)2 (1+0)2 (1+0)2 62 :n|: 012_02+032+0_4 : 1(0) (2+8) (1—8) 9 0.25(2+e)_0.25x2x(1+e) 0250 7 2 W W+ 6* _ Q L21; ' 100 2+6 1—6 6 1 2 i « — 3 —— +— = 1(6) 4 2+0.036 1—0.036 0.036 = §[04912+2.075+27.73] = 29124765 ( n = 3339 ) 3 Assyrnptotic Variance of MLE : 1 1(0) : 3434x10'5 (b) An apprOXimate 100 (17d)% confidence interval for 8 based on part(a) is given by, Zeal} 211(0) Where 0 is MB of 0, 2 0.036: 196 J3839>< 291241.765 = 0 036:1.8535x104 =[ 0.03581,0.03619 ] 0i (c) In order to Find standard error (SE) of 0 using bootstrap technique we need to apply Simulation We are not given actual value of 8 So, we start estimated value MLE} of 0 that is 0 which is principal of bootstrap technique Hence usmg B =003‘57" we generate B(say) random samples ”*. For each of the these generated random sample we have a MLE 0 So in this case we have 01,02”...03 For B generated random samples Then by bootstrap technique standard dev1ation of 0 is given by, 2 1 s A. 1 B A. 7 Z 9: — 7292' B 5-1 31-1 Here we can take B = 1000. B should be 2 250. (d) In process of bootstrap technique as we did in part(c) We get 01,02 ...... 01:00 For this data we find upper quartile that is 0.975 Quartileﬂi) and lower quantile that is 0025 quantile (Q) Thus our approXimate 95% confidence interval is (B—E,B—5) ...
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