Hardy Weinberg Model- 6 Genotypes

Transcription

1 Hardy Weinberg Model- 6 Genotypes Silvelyn Zwanzig Hardy -Weinberg with six genotypes. In a large population of plants (Mimulus guttatus there are possible alleles S, I, F at one locus resulting in six genotypes labeled SS, II, FF, SI, SF. Let θ, θ 2, θ 3 denote the probabilities of S,I,F with θ + θ 2 + θ 3. The Hardy Weinberg model specifies that the six genotypes have probabilities: Genotype Genotype SS II F F SI SF IF ( Probability θ 2 θ2 2 θ3 2 2θ θ 2 2θ θ 3 2θ 2 θ 3 Consider an i.i.d sample (X,.., X n according to the distribution (.. Belongs the distribution ( to a strictly(! k-parametric exponential family? Theorem 3.2 implies, that it is enough to consider one observation of X from (. It holds with P (X x θ 2I (x θ 2I 2(x 2 θ 2I 3(x 3 (2θ θ 2 I 4(x (2θ θ 3 I 5(x (2θ 2 θ 3 I 6(x θ 2I (x+i 4 (x+i 5 (x θ 2I 2(x+I 4 (x+i 6 (x 2 θ 2I 3(x+I 5 (x+i 6 (x 3 2 I 4(x+I 5 (x+i 6 (x I k (x { for x k 0 for x k for k,..., 6 (2

2 Introduce I S (x 2I (x + I 4 (x + I 5 (x I I (x 2I 2 (x + I 4 (x + I 6 (x 2 for x for x 4, 5 0 else 2 for x 2 for x 4, 6 0 else 2 for x 3 I F (x 2I 3 (x + I 5 (x + I 6 (x for x 5, 6 0 else and reformulate the probability function, P (X x θ I S(x θ I I(x 2 θ I F (x 3 2 I 4(x+I 5 (x+i 6 (x. (3 Note thus 6 I k (x for all x, k I S (x + I I (x + I F (x 2I (x + I 4 (x + I 5 (x + 2I 2 (x + I 4 (x + I 6 (x + 2I 3 (x + I 5 (x + I 6 (x 6 2 I k (x 2. k Furthermore θ 2 + θ2 2 + θ (θ θ 2 + θ 2 θ 3 + θ 3 θ 2. It holds θ 2 + θ2 2 + θ (θ θ 2 + θ θ 2 + θ θ 2 (θ + θ 2 + θ 3 2. We obtain θ + θ 2 + θ 3 2

3 Rewrite the probability function P (X x θ I S(x θ I I(x 2 ( θ θ 2 2 I S(x I I (x 2 I 4(x+I 5 (x+i 6 (x as 2-parametric exponential family P (X x h(xa(θ exp (ς (θt (x + ς 2 (θt 2 (x, θ (θ, θ 2 (4 with h(x 2 I 4(x+I 5 (x+i 6 (x, A(θ exp (2 ln( θ θ 2 (5 and ( T (x I S (x, ς (θ ln ( T 2 (x I I (x, ς 2 (θ ln θ θ θ 2 θ 2 θ θ 2 (6. (7 Properties, using (2 T (x 2I (x + I 4 (x + I 5 (x T 2 (x 2I 2 (x + I 5 (x + I 6 (x We know with p k P (X k that { for k k I k (XI k2 (X 2 k 0 else, { pk for k EI k (X p k, EI k (XI k2 (X k 2 k 0 else (8 Hence ET 2θ 2 + 2θ θ 2 + 2θ θ 3 2θ (θ + θ 2 + θ 3 2θ (9 ET 2 2θ θ θ 2 + 2θ 2 θ 3 2θ 2 (θ 2 + θ + θ 3 2θ 2 3

4 and ET 2 E (2I (x + I 4 (x + I 5 (x 2 (0 4EI (x + EI 4 (x + EI 5 (x 2EI (x + ET 2θ 2 + 2θ 2θ ( + θ Cov(T, T 2 E (T 2θ (T 2 2θ 2 E (T T 2 4θ θ 2 E (T T 2 E (2I (x + I 4 (x + I 5 (x (2I 2 (x + I 4 (x + I 6 (x ( EI 4 (x 2θ θ 2 Thus Cov(T, T 2 2θ θ 2 Summarizing, the distribution of (T, T 2, T 3 is multinomial with. multinomial(2, θ, θ 2, θ 3 where T 3 (x 2I 3 (x + I 5 (x + I 6 (x Solution: The i.i.d sample (X,.., X n according to the distribution ( belongs to a exponential family. 2. Derive k and the minimal sufficient statistics. Compare: (4 - (7 Apply Theorem 3.2 and Theorem 4.0. Solution: k 2 and T (X (T (X, T 2 (X, n T (X T (X i 2n + n 4 + n 5 T 2 (X i n T 2 (X i 2n 2 + n 4 + n 6 i 4

5 where n k # {x i k, i,..., n} 3. Are the minimal sufficient statistics complete? Solution: Yes. Because: Apply Theorem 5.5 and the natural parametrization with (6 and (7. The natural parameter space is defined by { } Z (ς, ς 2 : exp (ς T (x + ς 2 T 2 (x h(xdv < Here exp (ς T (x + ς 2 T 2 (x h(xdv thus Z R 2. x,2,3,4,5,6 exp (ς T (x + ς 2 T 2 (x h(x < 4. Derive the MLE for all unknown parameters: θ,mle, θ 2,MLE, θ 3,MLE. The log likelihood function is l(θ T (x ln(θ + T 2 (x ln(θ 2 + T 3 (x ln( θ θ 2 + const with T 3 (x 2n 3 + n 5 + n 6. l(θ T (x T 3 (x! 0, (2 θ θ θ θ 2 l(θ T 2 (x T 3 (x! 0 θ 2 θ 2 θ θ 2 Using θ + θ 2 + θ 3 θ 3 T (x θ T 2 (x 0 (3 θ 3 T 2 (x θ 2 T 3 (x 0 and adding both equations gives θ 3 (T + T 2 ( θ 3 T 3 0. Because T (x + T 2 (x + T 3 (x 2n we obtain θ 3,MLE T 3 2n 2n 3 + n 5 + n 6. 2n Analogously θ,mle T 2n 2+n 4 +n 5, θ2,mle T 2 2n 2+n 4 +n 6 2n 2n 2n 2n 5

6 5. Give the MLE for the probability of SI. We have p SI 2θ θ 2 Thus (Def. 5.2 p SI MLE 2 θ,mle θ2,mle T T 2 2n ( 2 6. Are the MLE θ,mle, θ 2,MLE, θ 3,MLE optimal? In which sense? (Give the definition! The MLE are functions of sufficient and complete statistics. We have to check the unbiasedness. Using (9 E θ,mle E T 2n ET 2n n2θ 2n θ. By symmetry θ 2,MLE, θ 3,MLE are also unbiased. Thus by Theorem of Lehman Scheffe the maximum likelihood estimators are best unbiased estimators. (Def These estimators have minimal variance in the class of all unbiased estimators. 7. Derive an alternative moment estimator for probability of SI, basing on the number of plants belonging to the genotype SI only. Number of plants belonging to the genotype SI : n i I 4 (x i n 4 Note the probability of SI equals p 4. It holds by (8 En 4 n EI 4 (x i np 4 i Thus p SI n 4 n is an reasonable unbiased estimator. 8. Compare the estimators in p SI MLE and p SI. Which is better? Why? First we have to study the unbiasedness of p SI MLE. E p SI MLE E T (x T 2 (x 2n 2 6

7 Because of (8 and ( it holds ( n E (T (x T 2 (x E T (x i i n T 2 (x j j ( ( E T (x i T 2 (x i + E T (x i T 2 (x j ne (T T 2 + (n 2 net ET 2 n2θ θ 2 + (n 2 n2θ 2θ 2 2θ θ 2 (2n 2 n. Thus E p SI MLE E T (x T 2 (x 2θ 2n 2 θ 2 ( 2n. The maximum likelihood estimator is not unbiased! Propose an bias corrected estimator: p SI T (x T 2 (x 2n 2 n This estimator p SI is BUE, because of the Theorem of Lehman Scheffe. 9. Calculate the Fisher information matrix for (. We have an i.i.d. sample. Consider only one observation. The score function is given by (compare (2 V (θ ( θ l(θ θ 2 l(θ ij ( θ T θ 3 T 3 θ 2 T 2 θ 3 T 3 ( Check: EV (θ 0. We have E θ T θ 3 T Fisher information matrix: I X (θ Cov (V (θ Using ( we calculate i j 7

8 and ( E T 2 T 3 θ θ 3 E (T θ ET T 3 + ET 2 θ θ 3 θ θ θ 2 ( + θ 2 2θ θ θ θ θ 3 θ3 2 3 ( + θ 3 ( 2 + θ θ 3 ( E T ( T 3 T 2 T 3 θ θ 3 θ 2 θ 3 ET T 2 ET T 3 ET 2 T 3 + ET 2 θ θ 2 θ θ 3 θ 2 θ 3 θ θ θ 2 2θ θ 3 2θ 2 θ 3 + 2θ θ θ 2 θ θ 3 θ 2 θ 3 θ3 2 3 ( + θ 3 2 θ 3 Because of Corollary 4. I X (θ ni X (θ I X (θ 2n ( θ + θ 3 θ 3 θ 3 θ 2 + θ 3 0. Derive the Cramer Rao Bound for unbiased estimators of p SI γ. Apply formulary ( for all unbiased estimators V ar ( γ D θ g T I X (θ D θ g. Here g(θ 2θ θ 2, thus ( 2θ2 D θ g 2θ and 8

9 ( θ + θ 3 θ 3 Thus θ 3 θ 2 + θ 3 ( ( (θ3 + θ 2 θ θ θ 2 ( θ θ θ θ 2 θ θ 2 (θ 3 + θ θ 2 θ θ 2 ( θ 2 θ 2 I X (θ ( ( θ θ θ θ 2 2n θ θ 2 ( θ 2 θ 2. (4 Remind : θ 3 + θ 2 + θ Then the CRB is ( T ( θ2 ( θ θ θ θ 2 2 n θ θ θ 2 ( θ 2 θ 2 ( θ2 θ 2θ θ 2 θ 2 4θ θ 2 +θ n. Calculate the efficiencies. Consider first the moment estimator p SI n 4 n. V ar ( p SI n 2 V ar ( n i I 4 (x i n V ar (I 4 (x n 2θ θ 2 ( 2θ θ 2 efficiency: e( p SI, p SI θ 2 4θ θ 2 +θ n CRB V ar ( p SI 2θ θ 2 2θ n θ 2 ( 2θ θ 2 θ 2 4θ θ 2 + θ 2θ θ 2 Consider now the bias corrected maximum likelihood estimator: p SI T (x T 2 (x 2n 2 n. It holds V ar ( p SI (2n 2 n 2 V ar (T (x T 2 (x Observe (T (x, T 2 (x, T 2 (x multinomial(2n, θ, θ 2, θ 3 9

10 Presentation as sums of independent Bernoulli variables T j (x i Z i,j, Z i,j Ber(θ j i.i.d., j, 2, 3, Z i, +Z i,2 +Z i,3 Need V ar(t (x T 2 (x E (T (x T 2 (x 2 (ET (x T 2 (x 2 ( 2 T 2 (x Z i, Thus T 2 (x T2 2 (x Z i, ET (x T 2 (x (2n 2 n2θ θ 2 Z i, Z i2, i i i 2 i ( i Z i, Z i, + Z i, + ( Z i2,2 + Z i,z i2, i i 2 i 3 Z i2,2 i i 3 + Z i, Z i3,2z i4,2 i i 3 i 4 + Z i,z i2, Z i2,2 i i 2 i 3 + Z i2,2 i i 3 i because: Z i,z i,2 0 Z i, i Z i,z i2, Z i3,2z i4,2 i i 2 i 3 i 4 Z i,z i,2 + i 3 i 4 Z i3,2z i4,2 i i 3 Z i,z i2,2 Z i,z i3,2z i4,2 i i 3 i 4 Z i,z i2, i i 2 Z i3,2z i4,2 i 3 i 4 Z i,z i2,2 i i 3 0

11 Z i,z i2, i i 2 i 3 i 4 Z i3,2z i4,2 Z i,z i2,z i3,2z i4,2 i i 2 i 3 i 4 T 2 (x T 2 2 (x Z i,z i2,2 i i Z i,z i3,2z i4,2 i i 3 i 4 Z i2,z i3,z i4,2 i 2 i 3 i 4 i i 2 i 3 i 4 Z i,z i2,z i3,2z i4,2 ET 2 (x T2 2 (x 2n (2n θ θ 2 (5 +2n (2n (2n 2θ θ2 2 (6 +2n (2n (2n 2θ 2 θ 2 +2n (2n (2n 2(2n 3θθ V ar ( T 2 (x T2 2 (x ET 2 (x T2 2 (x (n (2n 2θ θ 2 2 2n (2n (2n 2(2n 3 (2n (2n 2 V ar ( p SI (2n 2 n V ar (T 2 (x T 2 (x 2n (2n θ θ 2 + 2n (2n (2n 2θ θ2 2 +2n (2n (2n 2θ (2n 2 n 2 2 θ 2 + n (2n (2n 2(2n 32θθ (n (2n 2θ θ 2 2 2θ θ 2 (2n 2 n ( + (2n 2(θ 2 + θ + (6 8nθ θ 2 Efficiency e( p SI, p SI 2n2 n θ 2θ θ 2 4θ θ 2 +θ 2 n 2θ θ 2 ( + (2n 2(θ 2 + θ + (6 8nθ θ 2

12 (2n θ 2 4θ θ 2 +θ (+(2n 2(θ 2 +θ +( 8n+6θ θ 2?????? Conclusion, also the corrected maximumlikelihood estimator is not efficient. 2. Calculate the efficiences of θ,mle, θ 2,MLE. We have θ,mle T, θ2,mle T 2. Furthermore T 2n 2n, T 2 are multinomial distributed. Thus and (T (x, T 2 (x, T 2 (x multinomial(2n, θ, θ 2, θ 3 ( θ ( θ Cov (T, T 2 2n θ θ 2 θ θ 2 θ 2 ( θ 2 Cov ( θ,mle, θ 2,MLE Cov ( θ,mle, θ 2,MLE 2n 4n Cov (T, T 2 2 ( θ ( θ θ θ 2 θ θ 2 θ 2 ( θ 2 comparing with CRBI X (θ see (4. We get that θ,mle, θ 2,MLE are efficient. 2