PhD course in Advanced survival analysis. One-sample tests. Properties. Idea: (ABGK, sect. V.1.1) Counting process N(t)

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1 PhD course in Advanced survival analysis. (ABGK, sect. V.1.1) One-sample tests. Counting process N(t) Non-parametric hypothesis tests. Parametric models. Intensity process λ(t) = α(t)y (t) satisfying Aalen s multiplicative intensity model. Usually, N(t) is obtained by aggregating individual counting processes, α(t) is the hazard function, and Y (t) is the number at risk. 1 October 212 Per Kragh Andersen We will test the null hypothesis H : α(t) = α (t), with α (t) a known hazard function, e.g. a population mortality. 1 2 Compare Nelson-Aalen increments dn(s) dâ(s) = Y (s) Idea: with α (s)ds. Formally: introduce predictable, non-negative weights, K(s), and consider the process: Z(t) = ( ) K(s) dâ(s) α (s)ds. Then Z(t) tends to be large (small) if α(s) is larger (smaller) than α (s) for all s [, t]. Note that we always assume that K(s) = whenever Y (s) =. Properties. If H is true then dn(s) = α (s)y (s)ds + dm(s) and Z(t) equals the stochastic integral Z(t) = K(s) Y (s) dm(s). In particular, Z(t) is a mean zero martingale with (under H ) ( ) 2 K(s) t K 2 (s) Z (t) = d M (s) = Y (s) Y (s) α (s)ds. Conditions can be found under which, by the martingale CLT, the test statistic Z(t)/ Z (t) is asymptotically standard normal under H. 3 4

2 Choice of weights. Common choice is K(t) = Y (t), cf. Ex. V.1.1. Then with Z(t) = ( ) dn(s) Y (s) Y (s) α (s)ds = N(t) E(t), Z (t) = E(t) = Y 2 (s) Y (s) α (s)ds = E(t) Y (s)α (s)ds. The one-sample logrank test statistic is then Example: Fyn diabetics. For female diabetics (Ex. V.1.3) we have, for t = 1 years, N(t) = 237 and using published Danish life-tables we get E(t) = 8.1. The one-sample logrank statistic takes the (highly significant) value: = (N(t) E(t))/ E(t) and E(t) is the expected number of events in [, t] under H (since then N(t) E(t) is a martingale). 5 6 (ABGK sec. V.2.1) k-sample tests. k-variate counting process: (N 1 (t),...,n k (t)) where, typically, each component, N h (t), is obtained by aggregating individual processes. Intensity process (λ 1 (t),...,λ k (t)) of the form: We will test the null hypothesis: λ h (t) = α h (t)y h (t), h = 1,...,k. H : α 1 (t) =... = α k (t) for all t, e.g., if mortality is the same in k groups. 7 Properties of test statistic. If H is true then N (t) = k h=1 N h(t) is a univariate counting process with intensity process: λ (t) = k λ h (t) = α(t)y (t) h=1 where α(t) is the common value of the α h (t) s under H. The idea is now to compare the Nelson-Aalen increments a priori: with those under H : dâh(t) = dn h(t) Y h (t) dâ(t) = dn (t) Y (t). 8

3 Properties of test statistic. Formally, follow the approach from the one-sample situation: introduce non-negative weights, K(s), and consider the processes (for h = 1,...,k): Z h (t) = = Note that k h=1 Z h(t) =. ( ) K(s)Y h (s) dâ h (s) dâ(s) K(s)dN h (s) (observed (weighted)) K(s) Y h(s) Y (s) dn (s) (expected (weighted)). We assume that K(s) = whenever Y (s) =. Properties of test statistic. Under H, the Z h (t) s are mean zero martingales. From their predictable (co-)variance processes, we obtain the (co-)variance estimators: σ hj (t) = K 2 (s) Y ( h(s) δ hj Y ) j(s) dn (s). Y (s) Y (s) Introduce Z(t) = (Z 1 (t),...,z k 1 (t)) T and Σ(t) = (σ hj (t)) k 1 h,j=1 and use the quadratic form as test statistic: Z(t) T Σ(t) 1 Z(t) which is χ 2 k 1 under H under suitable regularity conditions (Th. V.2.1). 9 1 General choices: Choices of weights. Nelson-Aalen estimates by tumor thickness. Log-rank (Savage): K(s) = I(Y (s) > ). Gehan-Breslow (Wilcoxon, Kruskal-Wallis): K(s) = Y (s). Tarone-Ware: K(s) = Y (s) ρ. Survival data: Peto-Prentice (Wilcoxon, Kruskal-Wallis): K(s) = ( ) u<s 1 N (u) Y (s) Y (u)+1 Y (s)+1 Harrington-Fleming: K(s) = I(Y (s) > )Ŝ(s ) ρ. The log-rank test is optimal for proportional hazards alternatives. Categories: -2, 2-5, 5+ mm 11 12

4 Melanoma data Compare mortality for tumor thickness groups: below 2mm, 2-5mm, above 5mm. Log-rank test (at t =14 years): X 2 (t) = 31.6, d.f.=2 Harrington-Fleming test (at t =14 years, ρ = 1): X 2 (t) = 33.9, d.f.=2 The two-sample case. When k = 2 the test statistic takes the form X 2 (t) = (Z 1(t)) 2 σ 11 (t) and an alternative, standard normally distributed (under H ), statistic is: U(t) = Note that U(t) has a direction. Z 1(t) σ11 (t). Example: melanoma data, compare males and females. Log-rank: U(t) = 2.54, P =.11, Gehan-Breslow: U(t) = 2.72, P =.65, Peto-Prentice: U(t) = 2.66, P = Nelson-Aalen estimates by sex. Exercise 1. Show that the 2-sample logrank test (i.e., K(s) = I(Y. (s) > )) can be written in the form Z 1 (t) = Y 1 (s)y 2 (s) Y 1 (s) + Y 2 (s) (dâ2(s) dâ1(s)). 2. Show that, under H : α 1 (t) = α 2 (t), Z 1 is a martingale. 3. Find the predictable variation process Z 1 and the variance estimate σ 11 (t). Males: dashed, females: solid 15 16

5 Parametric models. Set-up: N i (t), i = 1,...,n are counting processes with intensity processes λ i (t), i = 1,...,n w.r.t. filtration (F t ). Recall that λ i (t)dt P(dN i (t) = 1 F t ). Examples (mainly survival data) of parametric models (where Y i (t) is at-risk indicator): λ i (t) = αy i (t), constant hazard λ i (t) = αρ(αt) ρ 1 Y i (t), Weibull λ i (t) = θµ i (t)y i (t), relative mortality λ i (t) = (γ + µ i (t))y i (t), excess mortality λ i (t) = α j Y i (t), t j < t t j+1 piecewise constant hazard Likelihood: Jacod s formula. ABGK; p.58. Consider a general parametric model λ i (t) = α i (t; θ)y i (t), where θ = (θ 1,...,θ p ) is the parameter vector. To derive the likelihood for θ, let N = i N i, Y = i Y i and note that P(dN (t) = F t ) 1 λ (t; θ)dt. With τ being the terminal time of observation we have P(data) = <t τ π P(dN 1 (t),...,dn n (t) F t )P(other events at t dn(t), F t ) <t τ π P(dN 1 (t),...,dn n (t) F t ) L(θ) = P(data) = ( Likelihood. π <t τ λ 1 (t, θ) dn 1(t) λ n (t, θ) dn n(t) (1 λ (t, θ)dt) 1 dn (t) n <t τ λ i (t, θ) dni(t) ) exp( λ (u, θ)du). Here, dn i (t) = 1 if N i jumps at t and otherwise. One may show that the scores U j (θ) = θ j log L(θ) are stochastic integrals w.r.t. the counting process martingales when evaluated at the true parameter: Score: log L(θ) = θ j log L(θ) = log λ i (t, θ)dn i (t) λ (t)dt. λ ij (t, θ) τ λ i (t, θ) dn i(t) λ j(t)dt. Use the Doob-Meyer decomposition dn i (t) = λ i (t, θ )dt dm i (t) to show that the score evaluated at θ equals: λ ij (t, θ) log L(θ) = θ j λ i (t, θ) dm i(t). Thereby, using the martingale CLT, asymptotic normality of the scores may be established leading to a proof that the MLE for θ = (θ 1,...,θ q ) enjoys the usual large sample properties. 19 2

6 Sufficiency. In most of the examples above, λ i (t; θ) = α(t; θ)y i (t) in which case the likelihood is (proportional to): L(θ) = ( α(t; θ) dn (t) ) exp( <t τ It follows that (N, Y ) is sufficient for θ. α(u; θ)y (u)du). Constant hazard. (N, Y ) sufficient when the hazard is assumed constant, i.e. λ i (t; α) = αy i (t). If we let R(t) = Y (u)du be the total time at risk in [, t] then L(α) = α N (τ) exp( αr(τ)) and the MLE is the occurrence/exposure rate α = N (τ) R(τ) NOTE: The Nelson-Aalen estimator (and thereby, by properties of the product-integral, also the Kaplan-Meier and Aalen-Johansen estimators, to be defined later) may be given a MLE interpretation. and ŜD( α) = ( 2 α 2 log L( α)) 1 2 = α R(τ) Nelson-Aalen estimates for male and female melanoma patients. Example: melanoma data. Here we have, for females: α = =.356 per year with ŜD =.67 while, for males, we have α = =.689 per year with ŜD =.128. We can compare the two using the approximately χ 2 1 LR statistic which takes the value 6.15 (P =.1). The constant hazard hypothesis may be evaluated using confidence bands for the Nelson-Aalen estimates or by embedding the model in the larger Weibull model. For the Weibull model the estimated shape parameter for females is ρ = 1.197(.22) while for males it is ρ = 1.17(.166) and for both sexes the Wald statistic ( ρ 1)/ŜD is quite insignificant. Males: dashed, females: solid 23 24

7 Nelson-Aalen estimate for male melanoma patients with bands. Piecewise constant hazards. The constant hazard model is often quite restrictive, however, a simple device offers a lot of flexibility, relax the assumption to piecewise constant hazards: λ i (t) = α j Y i (t), t j < t t j+1, where = t < t 1 <... < t k 1 < t k = + are pre-specified interval cut-points. Define occurrence, D j = N (t j+1 ) N (t j ) and exposure, R j = j+1 t j Y (u)du. Then L(α) = ( j α D j j ) exp( j α j R j ), and the MLE simply becomes α j = D j R j, ŜD( α j ) = α j R j, as. indep Example: suicides among Danish non-manual workers. Ex. I.3.3, p.17. Four categories of non-manual workers: 1. academics 2. advanced non-academic training 3. extensive practical training 4. other non-manual workers Suicides and person-years tabulated by age group and work category (excerpt): Suicides Age Group 1 Group 2 Group 3 Group Males Females Person-years Age Group 1 Group 2 Group 3 Group Males Females

8 Example: suicides among Danish non-manual workers. Age-specific suicide rates (e.g., male academics, 2-24, 2/13439=.15; 25-29, 1/52787=.19): Suicide rates (per 1 years) (SD) Age Males Females (1.1) () (.6) 3.6 (1.6) (1.) 4.9 (1.9) (.9) 7.6 (2.9) (1.) 6.4 (2.6) (.9) 4.9 (2.2) (1.1) 4.6 (2.3) (1.3) 9. (3.7) (1.1) 4.1 (2.9) The standardized mortality ratio. Recall the model for relative mortality λ i (t) = θµ i (t)y i (t) where µ i (t) is a known mortality rate, e.g. an age-, sex-, and calendar time specific population mortality for invividual i at time t. If θ = 1 we have N (τ) = µ i (u)y i (u)du + M (τ), and thereby E(N (τ)) = E( n µ i(u)y i (u)du). Therefore, E(τ) = µ i (u)y i (u)du can be thought of as the (under H : θ = 1) expected number of deaths The MLE is θ = N (τ) E(τ), the standardized mortailty ratio with θ ŜD( θ) = E(τ). Example: suicides (continued), let µ i (t) = µsex(i)(t) at age t be defined as occ/exp rates from combined data (treated as known). Then we get for the academics Males: θ = 1.2(.1) Females: θ = 1.57(.25) both marginally significantly (judged from a Wald test) larger than 1. 31

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