(4)* Each pair has two members

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Paper: Biostatistics

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Principal investigator: Dr.Bhaswati Ganguli, Department of Statistics, University of Calcutta

Paper co-ordinator: Dr.Sugata SenRoy,Department of Statistics, University of Calcutta

Content writer: Dr.Atanu Bhattacharjee, Division of Clinical Research and Biostatistics, Malabar Cancer Centre Content reviewer: Dr.Indranil Mukhopadhyay,Indian Statistical

Institute, Kolkata

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* Each pair has two members.

* Each sampled are independently provide the value of the matching covariate Z.

* The members within each pair are conditionally independent.

* The sample of matched-pairs be considered to consist ofN independent samples(strata).

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* A2×2 table is obtained from the measures of each strata.

* SupposeK represents the total number of strata and j= 1, ...., K.

* The strata may be formed through categories of a single covariate ( may be sex)

* The strata wise observation can be formed through table

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Title1:- Jth stratum Frequency Table Response Group1 Grop2 Response(+) a_j b_j m_1j

Response(-) c_j d_j m_2j n_1j n_2j N_j Title2:- Jth stratum Probability Table

Response Group1 Grop2 Response(+) π1j π2j

Response(-) 1-π1j 1-π2j

Total 1 1

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* Within thej^th stratum any of the measures of group-response association described previously may be computed:

* The risk difference RDˆ _j =p_1j−p_2j,

* Relative Risk RRˆ j = ^p_p^1j

2j = ^a_b^jⁿ^2j

jn1j

* The odds ratioORˆj = ₁₋^p^1j_p

1j/[₁₋^p^2j_p

2j] = ^a_b^j^d^j

jcj.

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2j = ^a_b^jⁿ^2j

jn1j

1j/[₁₋^p^2j_p

2j] = ^a_b^j^d^j

jcj.

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2j = ^a_b^jⁿ^2j

jn1j

1j/[₁₋^p^2j_p

2j] = ^a_b^j^d^j

jcj.

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2j = ^a_b^jⁿ^2j

jn1j

1j/[₁₋^p^2j_p

2j] = ^a_b^j^d^j

jcj.

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* Test of the null hypothesisH₀ :π_1j =π_2j(ORj= 1) for all j.

* The alternative hypothesis that the probabilities within strata differ such that there is a common odds ratio i.e. H1: (ORj =OR̸= 1)∀j= 1,2, ....K.

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* Test of the null hypothesisH₀ :π_1j =π_2j(ORj= 1) for all j.

* The alternative hypothesis that the probabilities within strata differ such that there is a common odds ratio i.e. H1: (ORj =OR̸= 1)∀j= 1,2, ....K.

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* The expected frequency for the index cell,E(a_j)

* E_j =E(a_j) = ⁿ^1j_N^m^1j

j

* Variance ofaj underH0 is

* V_cj = ^m^1j_N^m2 ^2jⁿ^1jⁿ^2j

j(Nj−1) For each statum , under H₀ asymptotically for largeNj and for fixed K

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* E_j =E(a_j) = ⁿ^1j_N^m^1j

j

* V_cj = ^m^1j_N^m2 ^2jⁿ^1jⁿ^2j

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* E_j =E(a_j) = ⁿ^1j_N^m^1j

j

* V_cj = ^m^1j_N^m2 ^2jⁿ^1jⁿ^2j

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* E_j =E(a_j) = ⁿ^1j_N^m^1j

j

* V_cj = ^m^1j_N^m2 ^2jⁿ^1jⁿ^2j

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* The terma_j−E_j ≈N(0, V_cj).

* ∑

j(aj−Ej)≈N[0,∑

jVcj]

* The stratified-adjusted Mantel-Haenszel is X_{C(M H)}² = ^[

∑

j(aj−Ej)]²

∑

jVcj = ^[a⁺_V⁻^E⁺^]²

c+

a₊ =∑_K

j=ia_j, E+=∑

jEj, and Vc+=∑

jVcj.

* It is noted that(a+−E+) is the sum of asymptotically normally distributed then X_{C(M H)}² ≈χ² on 1 d.f.

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* ∑

jVcj]

∑

j(aj−Ej)]²

∑

jVcj = ^[a⁺_V⁻^E⁺^]²

c+

a₊ =∑_K

j=ia_j, E+=∑

jEj, and Vc+=∑

jVcj.

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* ∑

jVcj]

∑

j(aj−Ej)]²

∑

jVcj = ^[a⁺_V⁻^E⁺^]²

c+

a₊ =∑_K

j=ia_j, E+=∑

jEj, and Vc+=∑

jVcj.

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* ∑

jVcj]

∑

j(aj−Ej)]²

∑

jVcj = ^[a⁺_V⁻^E⁺^]²

c+

a₊ =∑_K

j=ia_j, E+=∑

jEj, and Vc+=∑

jVcj.

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* ∑

jVcj]

∑

j(aj−Ej)]²

∑

jVcj = ^[a⁺_V⁻^E⁺^]²

c+

a₊ =∑_K

j=ia_j, E+=∑

jEj, and Vc+=∑

jVcj.

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* ∑

jVcj]

∑

j(aj−Ej)]²

∑

jVcj = ^[a⁺_V⁻^E⁺^]²

c+

a₊ =∑_K

j=ia_j, E+=∑

jEj, and Vc+=∑

jVcj.

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1:M matching

Note there are2(M+ 1) possible outcomes 2 ofn1 < mare concordant and hence excluded. Divide the remaining2M into M sets of 2 as Title1:- 2x2 contigency table

Exposed Not-exposed

Cases Exposed 1 0

Control m-1 M-m+1

m M-m+1

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1:M matching

Pn= (^M_m₋₁)p1p^m₀⁻¹(1−p0)^M⁻^m+1 (1) Title2:- 2x2 contigency table

Control Exposed Control Not-exposed

Cases Exposed 0 1

Cases Not-exposed m M-m

m M-m+1

form= 1,2, ....M

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1:M matching

P_n= (^M_m)p^m₀ (1−p₀)^M^−m(1−p₁) (2) Conditional probability of case exposed /M exposed

π_m=P r[Case exposed|mExposed] (3) πm= (^M_m₋₁)p1p^m₀ ⁻¹(1−p0)^M⁻^m+1

(^M_m₋₁)p₁p^m₀⁻¹(1−p₀)^M⁻^m+1+ (^M_m)p^m₀ (1−p₀)^M⁻^m(1−p₁) (4)

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π_m = (1−p₀)p₁

p₁(1−p₀) +^M⁻_m^m+1p₀(1−p₁) (5) πm = mψ

mψ+M −m+ 1 (6) Let,n_i,jm= i cases and jm controls exposed, i=0,1.Then,

n1,m−1 ∼Bin(n1,m−1+n0,m, πm), m= 1(1)M LE =

∏m

( mψ

mψ+M−m+ 1)ⁿ^1,m−1( M−m+ 1

mψ+M−m+ 1)ⁿ⁰^,m

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1:M matching Also,

E(n_1,m₋₁|n_1,m₋₁+n_0,m) = (n_1,m₋₁+n_0,m)mψ

mψ+M−m+ 1 (8) V ar(n_1,m₋₁|n_1,m₋₁+n_0,m) = (n1,m−1+n0,m)mψ(M−m+ 1)

(mψ+M −m+ 1)² (9)

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1:M matching NowL=log L=∑m

m=1[n1,m−1{log(mψ)−log(mψ+M −m+ 1)}+ n0,m{log(M −m+ 1)−log(mψ+M−m+ 1)}]

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1:M matching

δL δψ =

∑M m=1

1 ψ =

∑m m=1

m

mψ+M−m+ 1{n0,m+n1,m−1} (10)

∑M m=1

n1,m−1 =

∑M m=1

(n0,m+n1,m−1)mψˆ

mψˆ+M−m+ 1 (11)

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1:M matching Thus,

ψˆ_{M H} =

∑_M

m=1(M−m+ 1)n1,m−1

∑_M

m=1mn0,m

(12) To testH₀ :ψ= 1, we can see

χ²= [1∑_M

m=1(n1,m−1−⁽ⁿ^1,m⁻¹_M+1⁺ⁿ^0,m⁻¹^)m)1− ¹₂]²

1 (M+1)²

∑_M

m=1(n_1,m₋₁+n_0,m₋₁)m(M −m+ 1) (13)