'&$%Outline• Application of discrete-time likelihood model to hy pothesis test:detect human-to-human transmission of infectious disease• Modeling missing infection outcome: a Bayesian perspective• Modeling asymptomatic infection: a Bayesian perspective• Summary'&$%A Resampling-Based Test to Dete ct Person-To-PersonTransmission of Infectious Dise ase'&$%Introduction• Problem under investigation.– to identify a valid statistical test for person-to-persontransmission of infectious diseases. For example, highpathogenic H5N1 avian influenza outbreaks in Indonesia.• Why do we need it?– To alert health practioners in a timely fashion. Time is crucialto contain or mitigate pandemics.– When there is no person-to-person transmission, statisticalestimates and 95% CIs of the transmission probability may notbe meaningful and could be misleadin g.– Existing methods are not satisfactory.'&$%Statistical Model• Consider a community composed of hou s eholds, an d three channelsthrough which infect i on may occur.– Close contact within households: the probability that asusceptible is infected by an infective in the same household inone day is p1.– Casual contact within community: t h e p robabi l i ty that asusceptible is infected by an infective in the same communitybut different household in one day is p2.– Common source of infection (e.g., zoonotic source or visitinginfectives from outside of the community): th e probability thata susceptible is infected by the c ommon source in one day is b.'&$%• Hypothesis to be tested.H0: p1= p2= 0 vs.H1: p1> 0 or p2> 0,• Test statisticλ = −2 logsupbL0(b|˜ti, i = 1, . . . , N)supb,p1,p2L(b, p1, p2|˜ti, i = 1, . . . , N),where˜tiis the symptom on s et time for person i, and N is thepopulation size.• Challenge 1: boundary values under testing.'&$%• More assumptions:– Random mixing in households and in the community.– The latent period coinc i des with the incubation period.– Distributions of the latent period (δ) and the infectious period(η) are known:∗ δ ∼ g(l) = Pr(δ = l), l = δmin, δmin+ 1, . . . , δmax.∗ η ∼ f(l) = Pr(η > l), l = ηmin, ηmin+ 1, . . . , ηmax.– Observation starts from day 1 to day T , and exposure to thecommon source starts from day 1 to day S ≤ T .• Challenge 2: When S < T − δmin, parameters under testing affec tdata domain. Asymptotic method does not work.'&$%• The probability that an infective j infe cts an susceptible i on day tispji(t) = pI(Hj=Hi)1pI(Hj6=Hi)2f(t −˜tj), (1)where Hiis the set of hou s ehold members of pe rson i.• The probability that subject i escapes infec ti on from all infectivesources on day t isei(t) = (1 − b)I(t≤S)NYj=1(1 − pji(t)). (2)'&$%• A likelihood for b, p1and p2contributed by person i isLi(b, p1, p2|˜tj, j = 1, . . . , N)=QTt=1ei(t), not infected,Ptg(˜ti− t)1 − ei(t)Qtτ =1ei(τ), otherwise,(3)• When p1= p2= 0, (2) reduces toei(t) = (1 − b)I(t≤S).'&$%• Statistical tests– Asymptotic null distribution∗ Not available for testing p1= p2= 0.∗ If we assume p2= 0 and let S ≥ T , λ ∼12χ20+12χ21underH0: p1= 0.– Simple permutation test∗ L0(b|˜ti, i = 1, . . . , N) is invariant to permutation of infectionstatus and associated symptom onset ti mes across the wholepopulation.∗ Let λ[k]be the te s t statistic calcul ated from he kthpermutedsample (˜t[k]1,˜t[k]2, . . . ,˜t[k]N), k = 1, . . . , M. P-value is gi ven by1MPkI(λ ≥ λ[k]).'&$%– Refined permutation test∗ Identify a larger class of data sets that have the samelikelihood under H0.∗ Consider the situation without latent period, where˜tiisobserved infection time. The likeli hood under H0isL0(b|˜ti, i = 1, . . . , N) =Yi∈D(1 − b)S×Yi∈D{(1 − b)˜ti−1b}= b˜N(1 − b)(N−˜N)S−˜N+Pi∈D˜ti,(4)where D is the set of˜N infe cted cases.∗ L0(b|˜ti, i = 1, . . . , N) is invariant to changes in D and {˜ti}, aslong asPi∈D˜tiand˜N remain the same.'&$%∗ An equivalent problem: sampling with equal probability fromall distinct arrangements of n balls (sum of infection times)into m boxes (infected cases), each box with a fixed volume ofv (S).∗ Let W (n, m, v) be the number of all possible di sti nctarrangements for such condition. Solve the recursive systemW (n, m, v) =min(n,v)Xk=0W (n − k, m − 1, v) (5)with the stopping rules W (n, 0, v) = 0, W (0, m, v) = 1 andW (n, 1, v) = I(n ≤ v).'&$%∗ An arrangement can be sample d with equal probabilitythrough the following procedure:1. Start with the box labelled i = 1, and there are N1= nballs to be d i strib uted.2. In step i, let Nibe the number of b all s not distribut ed yet.Randomly choose an integer nifrom (0, 1, . . . , r) accordin gto the weights W (Ni− k, m − i, v), k = 0, 1, . . . , r, wherer = min(Ni, v), and assign niballs to box i. LetNi+1= Ni− ni, and go to box i + 1.3. In the last step, distribute all the remaining Nmballs tobox m.∗ Nmwill not exceed v for sure.'&$%Simulation Study1 2 3 4 5 6 7 8 9 11 13 15 17 19 21 23 25 27 29 31 33DayNumber of Cases0.0 0.5 1.0 1.5 2.0 2.5 3.0CPI = 0.03R0= 0.32SAR1= 0.055SAR2= 2e−04Figure 1: A sample epidemic curve for b = 0.001, p1= 0.014 and p2=0.00005. Cases from the same household have the same color.Table 1: Comparison of Type I Error and Power between the Permuta-tion Test and the Asymptotic Test for Mod els with Only b and p1. Thecommunity is composed of 4 households of size 5. Results are based on5000 simulations. 2000 permuted samples were drawn for each test.Simple Refinedb CPI p§1SAR1N†idxN‡totAsymptotic Permutation Permutation.01 .26 .0 .0 3 5 .029 .039 .050.02 .078 3 6 .21 .22 .26.05 .18 3 8 .60 .57 .63.08 .28 3 10 .85 .81 .85.02 .45 .0 .0 4 9 .034 .046 .049.02 .078 4 10 .21 .21 .24.05 .18 4 12 .60 .54 .63.08 .28 4 14 .87 .79 .87.03 .6 .0 .0 4 11 .048 .049 .048.02 .078 4 13 .18 .19 .22.05 .18 4 15 .55 .48 .58.08 .28 4 16 .80 .67 .81§ Type I errors are reported when p1= 0.† Nidxis the avera ge number of index ca ses.‡ Ntotis average total number of ca ses.Table 2: Comparison of Type I Error and Power between the Permuta-tion Test and the Asymptotic Test for Mod els with Only b and p1. Thecommunity is c omposed of 100 hou seholds of size 5. Results are basedon 2000 simulations. 2000 permuted samples were drawn for each test.Simple Refinedb CPI p§1SAR1N†idxN‡totAsymptotic Permutation Permutation.0005 .015 .0 .0 7 7 .037 .042 .046.010 .039 7 8 .51 .52 .53.020 .078 7 9