Optimal Weighted Wilcoxon–Mann–Whitney Test for Prioritized Outcomes

Abstract

We consider a two-group randomized clinical trial of prioritized endpoints, where mortality affects the assessment of a follow-up continuous outcome. With the continuous outcome as the principal outcome, we combine it with mortality via the worst-rank paradigm into a single composite endpoint. Then, we develop a weighted Wilcoxon–Mann–Whitney test statistic to analyze the data. We determine the optimal weights for the Wilcoxon–Mann–Whitney test statistic that maximize its power. We provide the rationale for the weights and their implications in the application of the method. In addition, we derive a formula for its power and demonstrate its accuracy in simulations. Finally, we apply the method to data from an acute ischemic stroke clinical trial of normobaric oxygen therapy.

Appendices

Appendix 1: Proof of Theorem 1.1

We consider $\widetilde {X}_{ij}=\delta _{ij}(\eta +T_{ij})+(1-\delta _{ij})X_{ij},~ i=1, 2~\mbox{and} ~j=1,\ldots , N_i$

$$\displaystyle \begin{aligned} \begin{array}{rcl} I(\widetilde{X}_{1k}<\widetilde{X}_{2l})&\displaystyle =&\displaystyle \delta_{1k}\delta_{2l}I({T}_{1k}<{T}_{2l}|T_{1k}\leq T_{max}, ~T_{2l}\leq T_{max})+\delta_{1k}(1-\delta_{2l})\\ &\displaystyle &\displaystyle + (1-\delta_{1k})(1-\delta_{2l})I({X}_{1k}<{X}_{2l}),\\ I(\widetilde{X}_{1k}=\widetilde{X}_{2l})&\displaystyle =&\displaystyle \delta_{1k}\delta_{2l}I({T}_{1k}={T}_{2l}|T_{1k}\leq T_{max}, ~T_{2l}\leq T_{max})\\ &\displaystyle &\displaystyle + (1-\delta_{1k})(1-\delta_{2l})I({X}_{1k}={X}_{2l}). \end{array} \end{aligned} $$

For q _i = 1 − p _i, we have

where

$$\displaystyle \begin{aligned} \begin{array}{rcl} \pi_{t1}&\displaystyle =&\displaystyle P({T}_{1k}<{T}_{2l}|T_{1k}\leq T_{max}, ~T_{2l}\leq T_{max})\\ &\displaystyle &\displaystyle +\frac{1}{2}P({T}_{1k}={T}_{2l}|T_{1k}\leq T_{max}, ~T_{2l}\leq T_{max})\\ \pi_{x1}&\displaystyle =&\displaystyle P({X}_{1k}<{X}_{2l})+\frac{1}{2}P({X}_{1k}={X}_{2l}). \end{array} \end{aligned} $$

Define $U_{kl}=I(\widetilde {X}_{1k}<\widetilde {X}_{2l})+\frac {1}{2}I(\widetilde {X}_{1k}=\widetilde {X}_{2l}),$ for k = 1, …, N ₁ and l = 1, …, N ₂. The binary variable $U_{kl}=I(\widetilde {X}_{1k}<\widetilde {X}_{2l})$ follows Bernoulli distribution with probability π _U1. Its mean and variance, respectively, E(U _kl) = π _U1 and $~~Var(U_{kl})=E(U_{kl})\left [1-E(U_{kl})\right ]=\pi _{U1}(1-\pi _{U1})$. Thus, we can use these results to derive the variance of U using the following formula:

$$\displaystyle \begin{aligned} Var(U)&=(N_1N_2)^{-2}\left[\sum_{k=1}^{N_1}\sum_{l=1}^{N_2}Var(U_{kl})+\sum_{k=1}^{N_1}\sum_{l=1}^{N_2}\sum_{k'=1}^{N_1}\sum_{l'=1}^{N_2} Cov(U_{kl}, U_{k'l'})\right]\\ &= (N_1N_2)^{-1}\left[Var(U_{kl})+(N_1-1)Cov(U_{kl}, U_{k'l})\right.\\ &\qquad \left.+(N_2-1)Cov(U_{kl}, U_{kl'})\right]. \end{aligned} $$

Note that when k≠k′ and l≠l′, the covariance

$$\displaystyle \begin{aligned}Cov(U_{kl}, U_{k'l'})=E(U_{kl} U_{k'l'})-E(U_{kl})E (U_{k'l'})=0.\end{aligned}$$

When k≠k′ or l≠l′, we have

$$\displaystyle \begin{aligned} Cov(U_{kl}, U_{k'l})=E(U_{kl} U_{k'l})-E(U_{kl})E (U_{k'l})= \pi_{U2}-\pi_{U1}^2;\\ Cov(U_{kl}, U_{kl'})=E(U_{kl} U_{kl'} )-E(U_{kl})E (U_{kl'}) =\pi_{U3}-\pi_{U1}^2. \end{aligned} $$

where $\pi _{U2}=E(U_{kl} U_{k'l})$ and $\pi _{U3}=E(U_{kl} U_{kl'}).$

Therefore,

$$\displaystyle \begin{aligned} Var(U) &= (N_1N_2)^{-1}\left[\pi_{U1}\left(1-\pi_{U1}\right)+(N_1-1)(\pi_{U2}-\pi_{U1}^2)\right.\\ &\qquad \qquad \qquad \left.+(N_2-1)(\pi_{U3}-\pi_{U1}^2)\right] \end{aligned} $$

• 1. No ties:

  When there are no ties, $I(\widetilde {X}_{1k}=\widetilde {X}_{2l})=0.$ In which case, $U_{kl}=I(\widetilde {X}_{1k}<\widetilde {X}_{2l})= \delta _{1k}\delta _{2l}I({T}_{1k}<{T}_{2l}|T_{1k}\leq T_{max}, ~T_{2l}\leq T_{max})+\delta _{1k}(1-\delta _{2l}) + (1-\delta _{1k})(1-\delta _{2l})I({X}_{1k}<{X}_{2l})$, for k = 1, …, N ₁ and l = 1, …, N ₂, . We have

  $$\displaystyle \begin{aligned} E(U_{kl}U_{k'l})&= P(T_{1k}<T_{2l}, T_{1k'}<T_{2l}|\delta_{1k}\delta_{1k'}\delta_{2l}=1)E(\delta_{1k}\delta_{1k'}\delta_{2l}=1)\\ &+ P(X_{1k'}<X_{2l})P(\delta_{1k}=1, ~\delta_{1k'}=\delta_{2l}=0)\\ &+P(X_{1k}<X_{2l})E(\delta_{1k}=\delta_{2l}=0)E(\delta_{1k'}=1)\\ &+ P(X_{1k}<X_{2l}, X_{1k'}<X_{2l})E(\delta_{1k}=\delta_{1k'}=\delta_{2l}=0)\\& +E(\delta_{1k}\delta_{1k'}=1)E(\delta_{2l}=0)\\ &= p_1^2p_2\pi_{t2}+2p_1q_1q_2\pi_{x1}+q_1^2q_2\pi_{x2}+p_1^2q_2\equiv \pi_{U2}\\ E(U_{kl}U_{kl'})&= P(T_{1k}<T_{2l},~ t_{1k}<t_{2l'}|\delta_{1k}\delta_{2l}\delta_{2l'}=1)E(\delta_{1k}\delta_{2l}\delta_{2l'}=1)\\ &+ P(T_{1k}<T_{2l}|\delta_{1k}\delta_{2l}=1, ~\delta_{2l'}=0)E(\delta_{1k}\delta_{2l}=1)E(\delta_{2l'}=0)\\ &+ P(t_{1k}<t_{2l'}|\delta_{1k}=1, ~\delta_{2l}=0, ~~\delta_{2l'}=1)E(\delta_{1k}\delta_{2l'}=1)E(\delta_{2l}=0)\\ &+ P(X_{1k}<X_{2l}, X_{1k}<X_{2l'})E(\delta_{1k}=\delta_{2l}=\delta_{2l'}=0)\\ &+ E(\delta_{1k}=1)E(\delta_{2l}=\delta_{2l'}=0)\\ &= p_1p_2^2\pi_{t3}+2p_1p_2q_2\pi_{t1}+q_1q_2^2\pi_{x3}+p_1q_2^2\equiv \pi_{U3}\\ \text{with} ~~~ \pi_{t2}&= P(T_{1k}<T_{2l}, T_{1k'}<T_{2l}|T_{1k}\leq T_{max},~ T_{1k'}\leq T_{max}, ~T_{2l}\leq T_{max}),\\ \pi_{x2}&= P(X_{1k}<X_{2l}, X_{1k'}<X_{2l}), \\~~ \pi_{t3} &= P(T_{1k}<T_{2l}, t_{1k}<t_{2l'}| T_{1k}\leq T_{max},~ T_{2l}\leq T_{max}, ~T_{2l'}\leq T_{max}), \\~~ \pi_{x3}&= P(X_{1k}<X_{2l}, X_{1k}<X_{2l'}) . \end{aligned} $$

  Under the null hypothesis of no difference between the two groups, with respect to survival and nonfatal outcome, we have F ₁ = F ₂ = F, G ₁ = G ₂ = G and p ₁ = p ₂ = p, q ₁ = q ₂ = q. This implies

  $$\displaystyle \begin{aligned} \begin{array}{rcl} \pi_{t1}&\displaystyle =&\displaystyle P(T_{1k}<T_{2l}|T_{1k}\leq T_{max}, T_{2l}\leq T_{max})\\ &\displaystyle =&\displaystyle \frac{1}{2p^2}\left[F(T_{max})^2-F(0)^2\right]=\frac{1}{2}\\ \pi_{t2}&\displaystyle =&\displaystyle P(T_{1k}<T_{2l}, T_{1k'}<T_{2l}|T_{1k}\leq T_{max},T_{1k'}\leq T_{max},T_{2l}\leq T_{max})\\ &\displaystyle =&\displaystyle \frac{1}{p^3}\int_0^{T_{max}}F(t)^2dF(t)\\ &\displaystyle =&\displaystyle \frac{1}{3p^3}\left[F(T_{max})^3-F(0)^3\right]=\frac{1}{3}\\ \pi_{t3} &\displaystyle =&\displaystyle P(T_{1k}<T_{2l}, T_{1k}<T_{2l'}|T_{1k}\leq T, T_{2l}\leq T, T_{2l'}\leq T))\\ &\displaystyle =&\displaystyle \frac{1}{p^3}\int_0^{T_{max}}\left[1-F(t)\right]^2dF(t)\\ &\displaystyle =&\displaystyle \frac{1}{3p^3}\left\{[1-F(T_{max})]^3-[1-F(0)]^3\right\}=\frac{1}{3}\\ \pi_{x1}&\displaystyle =&\displaystyle P(X_{1k}<X_{2l})=\int_{-\infty}^{\infty}G(x)dG(x)=\frac{1}{2}\left[G(x)^2\right]_{-\infty}^{\infty}=\frac{1}{2}\\ ~~\pi_{x2}&\displaystyle =&\displaystyle P(X_{1k}<X_{2l}, X_{1k'}<X_{2l})=\int_{-\infty}^{\infty}G(t)^2dG(t)=\frac{1}{3}\left[G(x)^3\right]_{-\infty}^{\infty}=\frac{1}{3}\\ \pi_{x3}&\displaystyle =&\displaystyle P( X_{1k}<X_{2l}, X_{1k}<X_{2l'})\int_{-\infty}^{\infty}[1-G(t)]^2dG(t)\\ &\displaystyle =&\displaystyle -\frac{1}{3}\left\{[1-G(x)]^3\right\}_{-\infty}^{\infty}=\frac{1}{3}. \end{array} \end{aligned} $$

  Therefore,

  $$\displaystyle \begin{aligned} \begin{array}{rcl} \pi_{U1}&\displaystyle =&\displaystyle p_1p_2\pi_{t1}+p_1q_2+q_1q_2\pi_{x1}\\&\displaystyle =&\displaystyle \frac{1}{2}p^2+pq+\frac{1}{2}q^2=\frac{1}{2}(p+q)^2=\frac{1}{2}\\ \pi_{U2}&\displaystyle =&\displaystyle p_1^2q_2+p_1^2p_2\pi_{t2}+2p_1q_1q_2\pi_{x1}+q_1^2q_2\pi_{x2}\\ &\displaystyle =&\displaystyle p^2q+\frac{1}{3}p^3+pq^2+\frac{1}{3}q^3=\frac{1}{3}(p+q)^3=\frac{1}{3}\\ \pi_{U3}&\displaystyle =&\displaystyle p_1q_2^2+p_1p_2^2\pi_{t3}+2p_1p_2q_2\pi_{x1}+q_1q_2^2\pi_{x3}\\ &\displaystyle =&\displaystyle pq^2+\frac{1}{3}p^3+p^2q+\frac{1}{3}q^3=\frac{1}{3}(p+q)^3=\frac{1}{3}. \end{array} \end{aligned} $$

  The mean and variance become

  $$\displaystyle \begin{aligned} \begin{array}{rcl} \mu_0&\displaystyle =&\displaystyle E_0(U) =\pi_{U1}=\frac{1}{2};\\ \sigma^2_0&\displaystyle =&\displaystyle Var_0(U)\\ &\displaystyle =&\displaystyle (N_1N_2)^{-1}\left[\pi_{U1}\left(1-\pi_{U1}\right)+(N_1-1)\left(\pi_{U2}-\pi_{U1}^2\right)\right.\\ &\displaystyle &\displaystyle \qquad \qquad \left.+(N_2-1)\left(\pi_{U3}-\pi_{U1}^2\right)\right] \\ &\displaystyle =&\displaystyle (N_1N_2)^{-1}\left[\frac{1}{2}\left(1-\frac{1}{2}\right)+(N_1-1)\left(\frac{1}{3}-\left(\frac{1}{2}\right)^2\right)\right.\\ &\displaystyle &\displaystyle \qquad \qquad \left.+(N_2-1)\left(\frac{1}{3}-\left(\frac{1}{2}\right)^2\right)\right] \\ &\displaystyle =&\displaystyle (N_1N_2)^{-1}\left[\frac{1}{4}+\frac{1}{12}(N_1-1)+\frac{1}{12}(N_2-1)\right] =\frac{N_1+N_2+1}{12N_1N_2}. \end{array} \end{aligned} $$

• 2. Ties are present: More generally, we can approximate the probabilities $\pi _{U2}=E(U_{kl} U_{k'l})$ and $\pi _{U3}=E(U_{kl} U_{kl'})$ using their unbiased estimators.

  Following Hanley and McNeil (1982), we can show that the variance V ar(U) can be estimated by:

  $$\displaystyle \begin{aligned}(N_1N_2)^{-1}\left[\widehat\pi_{U1}\left(1-\widehat\pi_{U1}\right)+(N_1-1)(\widehat\pi_{U2}-\widehat\pi_{U1}^2)+(N_2-1)(\widehat\pi_{U3}-\widehat\pi_{U1}^2)\right]\end{aligned}$$

  where $\widehat \pi _{U1}=\displaystyle (N_1N_2)^{-1}\sum _{k=1}^{N_1} \sum _{l=1}^{N_2} U_{kl}, ~\widehat \pi _{U2}=\displaystyle (N_1N_2^2)^{-1}\sum _{k=1}^{N_1} U_{k\bullet }^2,~$ and $~\widehat \pi _{U3}=\displaystyle (N_1^2N_2)^{-1}\sum _{l=1}^{N_2} U_{\bullet l}^2.$ In absence of ties, $\widehat \pi _{U2}$ and $\widehat \pi _{U3}$ are, respectively, estimates of π _U3 and π _U3.

  One can also consider other possible approximations of the variance of U using the exposition provided by Newcombe (2006).

  As we know,

  $$\displaystyle \begin{aligned}P(\widetilde{X}_{1k}<\widetilde{X}_{2l})+P(\widetilde{X}_{1k}>\widetilde{X}_{2l})+P(\widetilde{X}_{1k}=\widetilde{X}_{2l})=1.\end{aligned}$$

  Under the null hypothesis, i.e., $\widetilde {X}_{1k}$ and $\widetilde {X}_{2l}$ are identically distributed, we have $P(\widetilde {X}_{1k}<\widetilde {X}_{2l})=P(\widetilde {X}_{1k}>\widetilde {X}_{2l})$ which implies $P(\widetilde {X}_{1k}<\widetilde {X}_{2l})+\frac {1}{2}P(\widetilde {X}_{1k}=\widetilde {X}_{2l})=\frac {1}{2}.$ Therefore,

  $$\displaystyle \begin{aligned}E(U) =E(U_{kl})=P(\widetilde{X}_{1k}<\widetilde{X}_{2l})+\frac{1}{2}P(\widetilde{X}_{1k}=\widetilde{X}_{2l})=\frac{1}{2}.\end{aligned}$$

  The variance reduces to:

  $$\displaystyle \begin{aligned} \begin{array}{rcl} \sigma^2_0&\displaystyle =&\displaystyle Var_0(U)=\frac{1}{12N_1N_2}\left( N_1+N_2+1-\frac{\displaystyle\sum_{\nu=1}^{g}t_{\nu}(t_{\nu}^2-1)}{(N_1+N_2)(N_1+N_2-1)}\right) \end{array} \end{aligned} $$

  where t _ν is the number of observations with the same value in the ν-th block of tied observations sharing the same value and g is the number of such blocks (see, for instance, Rosner 2015).

Appendix 2: Mean and Variance of the Weighted U-Statistic

Consider the weights w = (w ₁, w ₂), we define the vector $\mathbf {c}'=(c_1, c_2, c_3)=\left (w_1^2, w_1w_2, w_2^2\right )$. Let $\displaystyle \widetilde {X}_{1k}=w_1\delta _{1k}(\eta +t_{1k})+w_2(1-\delta _{1k}) X_{1k},$ for k = 1, …, N ₁ and $\widetilde {X}_{2l}=w_1\delta _{2l}(\eta +t_{2l})+w_2(1-\delta _{2l})X_{2l},$ for l = 1, …, N ₂. We define the weighted WMW U-statistic by: c ′ U = (U _t, U _tx, U _x) where U ′ = (U _t, U _tx, U _x) and

$$\displaystyle \begin{aligned} \begin{array}{rcl} U_t&\displaystyle =&\displaystyle (N_1N_2)^{-1}\sum_{k=1}^{N_1}\sum_{l=1}^{N_2}\delta_{1k}\delta_{2l}\left[ I({T}_{1k}<{T}_{2l})+\frac{1}{2}I({T}_{1k}={T}_{2l})\right], ~\text{with} \\ &\displaystyle &\displaystyle T_{1k}\leq T_{max}, ~T_{2l}\leq T_{max} \\ U_{tx}&\displaystyle =&\displaystyle (N_1N_2)^{-1}\sum_{k=1}^{N_1}\sum_{l=1}^{N_2}\delta_{1k}(1-\delta_{2l}){}\\ U_x&\displaystyle =&\displaystyle (N_1N_2)^{-1}\sum_{k=1}^{N_1}\sum_{l=1}^{N_2} (1-\delta_{1k})(1-\delta_{2l})\left[ I({X}_{1k}<{X}_{2l})+I({X}_{1k}={X}_{2l})\right] \end{array} \end{aligned} $$

$$\displaystyle \begin{aligned} E(\mathbf{U})&=(E(U_t), E(U_{tx}), E(U_x))' \\ &=\Big( E(\delta_{1k}=1)E(\delta_{2l}=1)\bigg[ P({T}_{1k}<{T}_{2l}|T_{1k}\leq T_{max}, ~T_{2l}\leq T_{max})\\ &\left.+\frac{1}{2}P({T}_{1k}={T}_{2l}|T_{1k}\leq T_{max}, ~T_{2l}\leq T_{max})\right] , ~E(\delta_{1k}=1)E(\delta_{2l}=0), \\ &\left. ~E(\delta_{1k}=0)E(\delta_{2l}=0)\left[ P(X_{1k}<X_{2l})+ \frac{1}{2}P(X_{1k}=X_{2l})\right] \right)'\\ &=\left(p_1p_2\pi_{t1}, p_1q_2, q_1q_2\pi_{x1}\right)' \end{aligned} $$

In absence of ties, the variance $Var(\mathbf {U})={\varSigma } = (N_1N_2)^{-1}\left (\varSigma _{ij}\right )_{\substack {1\leq i, j\leq 3}}$ is a 3 × 3 matrix such that

$$\displaystyle \begin{aligned} \begin{array}{rcl} \varSigma_{11}&\displaystyle =&\displaystyle E[(U_t-p_1p_2\pi_{t1})(U_t-p_1p_2\pi_{t1})]\\ &\displaystyle =&\displaystyle p_1p_2\left[\pi_{t1}(1-p_1p_2\pi_{t1})+p_1(N_1-1)(\pi_{t2} -p_2\pi_{t1}^2)+p_2(N_2-1)(\pi_{t3} -p_1\pi_{t1}^2)\right],\\ \varSigma_{12}&\displaystyle =&\displaystyle \varSigma_{21}=E[(U_t-p_1p_2\pi_{t1})(U_{tx}-p_1q_2)]=\pi_{t1}p_1p_2q_2\left[(N_2-1)q_1-N_1p_1\right] ,\\ \varSigma_{13}&\displaystyle =&\displaystyle \varSigma_{31}=E[(U_t-p_1p_2\pi_{t1})(U_{x}-q_1q_2\pi_{x1})]= -\pi_{t1}\pi_{x1}(N_1+N_2-1)p_1q_1p_2q_2,\\ \varSigma_{22}&\displaystyle =&\displaystyle E[(U_{tx}-p_1q_2)(U_{tx}-p_1q_2)]= p_1q_2\left[(1-p_1q_2)+(N_1-1)p_1p_2+(N_2-1)q_1q_2\right]\\ \varSigma_{23}&\displaystyle =&\displaystyle \varSigma_{32}=E[(U_{tx}-p_1q_2)(U_{x}-q_1q_2\pi_{x1})]=\pi_{x1}p_1q_1q_2\left[(N_1-1)p_2 -N_2q_2\right],\\ \varSigma_{33}&\displaystyle =&\displaystyle E[(U_{x}-q_1q_2)(U_{x}-q_1q_2\pi_{x1})]\\ &\displaystyle =&\displaystyle q_1q_2\left[ \pi_{x1}(1-q_1q_2\pi_{x1})+q_1(N_1-1)(\pi_{x2} -q_2\pi_{x1}^2)+q_2(N_2-1)(\pi_{x3} -q_1\pi_{x1}^2)\right]. \end{array} \end{aligned} $$

Therefore,

$$\displaystyle \begin{aligned}Var(\mathbf{c}'\mathbf{U})=\mathbf{c}'{\varSigma}\mathbf{c}.\end{aligned}$$

Under the null hypothesis of no difference between the two groups, with respect to both survival and nonfatal outcome, we have p ₁ = p ₂ = p, q ₁ = q ₂ = q = 1−p, π _t1 = π _x1 = 1∕2, and π _t2 = π _x2 = π _t3 = π _x3 = 1∕3.Thus,

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} E_0(\mathbf{U})= \frac{1}{2}\left(p^2, 2pq, q^2\right)'~~~\mbox{and}~~~Var_0(\mathbf{U})={\varSigma_0}, \end{array} \end{aligned} $$

(1.17)

where ${\varSigma _0}=(N_1N_2)^{-1}\left (\varSigma _{0ij}\right )_{\substack {1\leq i, j\leq 3}}$ is a symmetric matrix with

$$\displaystyle \begin{aligned} \begin{array}{rcl} \varSigma_{011}&\displaystyle =&\displaystyle \frac{p^2}{12}A(p), ~\varSigma_{012}=\frac{p^2q}{2}\left[ (N_2{-}1)q{-}N_1p\right], ~~\varSigma_{013}={-}\frac{p^2q^2}{4}(N_2+N_1{-}1)\\ \varSigma_{022}&\displaystyle =&\displaystyle pq\left[ 1{-}pq+(N_2{-}1)q^2+(N_1{-}1)p^2\right], ~~ \varSigma_{023}= \frac{pq^2}{2}\left((N_1{-}1)p{-}N_2q\right), \\ \varSigma_{033}&\displaystyle =&\displaystyle \frac{q^2}{12}A(q),~~\text{where}~A(x)=6+4(N_2+N_1{-}2)x{-}3(N_2+N_1{-}1)x^{2}. \end{array} \end{aligned} $$

Moreover, since V ar ₀(c ′ U) = c ′Σ ₀ c ≥ 0 by definition, the matrix Σ ₀ is positive semi-definite. In practice, p is estimated by the pooled sample proportion $\hat p=(N_1\widehat p_1+N_2\widehat p_2)/(N_1+N_2)$, and both E ₀(U) and V ar ₀(U) are calculated accordingly.

Finally, when ties are present, the foregoing formulas can be modified easily as we did in the non-weighted case to account for the ties in the variance estimations.

Appendix 3: Optimal Weights

From Eq. (1.15), we have

$$\displaystyle \begin{aligned}\displaystyle \mu_{1w}-\mu_{0w}=c_1\left(\pi_{t1}p_1p_2-\frac{1}{2}p^2\right)+c_2\left(p_1q_2-pq\right)+c_3\left(\pi_{x1}q_1q_2-\frac{1}{2}q^2\right)= \mathbf{c}'\boldsymbol{\mu}\end{aligned}$$

where $\boldsymbol {\mu }'=\left (\pi _{t1}p_1p_2-\frac {1}{2}p^2, p_1q_2-pq , \pi _{x1}q_1q_2-\frac {1}{2}q^2\right ), \mathbf {c}'=(c_1, c_2, c_3)$ with c ₁ + 2c ₂ + c ₃ = 1.

We assume that det(Σ ₀) > 0, i.e., Σ ₀ is positive definite. Maximizing $\displaystyle \frac {|\mu _{1w}-\mu _{0w}|}{\sigma _{0w}},$ subject to c ₁ + 2c ₂ + c ₃ = 1, with respect to c corresponds to maximizing the Lagrange function:

$$\displaystyle \begin{aligned}O( \mathbf{c}, \lambda)= \displaystyle \left| \mathbf{c}'\mu\right|\left( \mathbf{c}'\varSigma_0 \mathbf{c}\right)^{-\frac{1}{2}}-\lambda( \mathbf{c}'\mathbf{b}-1)\end{aligned}$$

with respect to the vector c and λ, where λ is the Lagrange multiplier and b ′ = (1, 2, 1). Let $K(\mathbf {c})=sign(\mathbf {c}'\mu )[( \mathbf {c}'\varSigma _0 \mathbf {c})^{-\frac {3}{2}}]$, we have

$$\displaystyle \begin{aligned} \frac{\partial}{\partial \mathbf{c}} O( \mathbf{c}, \lambda)&= K(\mathbf{c})\left[ ( \mathbf{c}'\varSigma_0 \mathbf{c})\mu-(\varSigma_0 \mathbf{c})( \mathbf{c}'\mu)\right]-\lambda \mathbf{b}=0{} \end{aligned} $$

(1.18)

$$\displaystyle \begin{aligned} \frac{\partial}{\partial \lambda} O( \mathbf{c}, \lambda)&=\mathbf{c}'\mathbf{b}-1=0 {} \end{aligned} $$

(1.19)

From (1.18) and (1.19), we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} 0&\displaystyle =&\displaystyle \mathbf{c}'\left\{K(\mathbf{c})\left[ ( \mathbf{c}'\varSigma_0 \mathbf{c})\mu-(\varSigma_0 \mathbf{c})( \mathbf{c}'\mu)\right]-\lambda \mathbf{b}\right\}\\ &\displaystyle =&\displaystyle K(\mathbf{c})\left[ ( \mathbf{c}'\varSigma_0 \mathbf{c}) \mathbf{c}'\mu-( \mathbf{c}'\varSigma_0 \mathbf{c})( \mathbf{c}'\mu)\right]-\lambda \mathbf{c}'\mathbf{b}=\lambda, \end{array} \end{aligned} $$

because both (c ′Σ ₀ c) and (c ′μ) are scalars and c ′ b = c ₁ + 2c ₂ + c ₃ = 1.

Then, Eq. (1.18) implies (c ′Σ ₀ c)μ = (Σ ₀ c)(c ′μ), i.e., $\displaystyle \mu =(\varSigma _0 \mathbf {c})\frac {( \mathbf {c}'\mu )}{( \mathbf {c}'\varSigma _0 \mathbf {c})}=\varSigma _0\frac {( \mathbf {c}'\mu )}{( \mathbf {c}'\varSigma _0 \mathbf {c})}\mathbf {c}.$ Since we assume that the matrix $\varSigma _0^{-1}$ exists, this implies

$$\displaystyle \begin{aligned} \displaystyle\varSigma_0^{-1}\mu=\frac{( \mathbf{c}'\mu)}{( \mathbf{c}'\varSigma_0 \mathbf{c})}\mathbf{c} \end{aligned} $$

(1.20)

and thus, $\displaystyle \mathbf {b}'\varSigma _0^{-1}\mu =\frac {( \mathbf {c}'\mu )}{( \mathbf {c}'\varSigma _0 \mathbf {c})}\mathbf {b}'\mathbf {c}=\frac {( \mathbf {c}'\mu )}{( \mathbf {c}'\varSigma _0 \mathbf {c})}$.

Replacing $\displaystyle \frac {( \mathbf {c}'\mu )}{( \mathbf {c}'\varSigma _0 \mathbf {c})}$ by $\displaystyle \mathbf {b}'\varSigma _0^{-1}\mu$ in Eq. (1.20) yields $\displaystyle \varSigma _0^{-1}\mu =\displaystyle (\mathbf {b}'\varSigma _0^{-1}\mu )\mathbf {c}.$ Therefore, the optimal weight-vector is

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} {\mathbf{c}}_{opt}=\frac{\varSigma_0^{-1}\mu}{\mathbf{b}'\varSigma_0^{-1}\boldsymbol{\mu}}, \end{array} \end{aligned} $$

(1.21)

as long as $\mathbf {b}'\varSigma _0^{-1}\boldsymbol {\mu }\neq 0$. In addition,

Since Σ ₀ is positive definite, we can show that the border-preserving principal minors of order k > 2 have sign (−1)^k. Therefore, $\displaystyle {\mathbf {c}}_{opt}=\frac {\varSigma _0^{-1}\mu }{\mathbf {b}'\varSigma _0^{-1}\boldsymbol {\mu }}$ maximizes O(c).

Let us define two vectors d ₁ ′ = (1, 1, 0) and d ₂ ′ = b ′−d ₁ ′ = (0, 1, 1). To calculate w ₁ and w ₂, we just need to consider the relationships $\mathbf {c}=(w_1^2, w_1w_2, w_2^2)$ and w ₁ + w ₂ = 1. We have $\mathbf {d_1}'\mathbf {c}=w_1^2+w_1(1-w_1)=w_1.$ Therefore, using the result given in Eq. (1.21), we can deduce $\displaystyle w_1=\mathbf {d_1}'\mathbf {c}=\frac {\mathbf {d_1}'\varSigma _0^{-1}\mu }{\mathbf {b}'\varSigma _0^{-1}\boldsymbol {\mu }}$ and $\displaystyle w_2=1-\mathbf {d_1}'\mathbf {c}=\frac {(\mathbf {b}'-\mathbf {d_1}')\varSigma _0^{-1}\mu }{\mathbf {b}'\varSigma _0^{-1}\boldsymbol {\mu }}=\frac {\mathbf {d_2}'\varSigma _0^{-1}\mu }{\mathbf {b}'\varSigma _0^{-1}\boldsymbol {\mu }}.$

Appendix 4: Conditional Probabilities

1.1 Exponential Distribution

Suppose that the death times t ₁, t ₂ follow exponential distributions with hazards λ ₁, λ ₂, respectively, and denote $\displaystyle \theta =\frac {\lambda _1}{\lambda _2},~~q_1=q_2^{\theta }$, and $q_2=e^{-T\lambda _2}.$ Given that P(δ _1k = 1) = p ₁, P(δ _2l = 1) = p ₂, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} \pi_{t1}&\displaystyle =&\displaystyle P(T_{1k}<T_{2l}|\delta_{1k}=\delta_{2l}=1)= (p_1p_2)^{-1}\int_0^{T_{max}} (1-e^{-\lambda_1u})\lambda_2e^{-\lambda_2u}du\\ &\displaystyle =&\displaystyle \frac{1}{(1-q_2^{\theta})}\left[1- \frac{1-{q_2^{(1+\theta)}}}{(1+\theta)(1-q_2)}\right];\\ \pi_{t2}&\displaystyle =&\displaystyle P(T_{1k}<T_{2l}, T_{1k'}<T_{2l}|\delta_{1k}=\delta_{1k'}=\delta_{2l}=1)\\ &\displaystyle =&\displaystyle p_1^{-2}p_2^{-1}\int_0^{T_{max}} (1-e^{-\lambda_1u})^2\lambda_2e^{-\lambda_2u}du\\ &\displaystyle = &\displaystyle (1-q_2^{\theta})^{-2}\left\{1+\frac{1}{(1-q_2)}\left[\frac{1-q_2^{(1+2\theta)}}{1+2\theta}-\frac{2(1-q_2^{(1+\theta)})}{1+\theta}\right]\right\} \\ \pi_{t3}&\displaystyle =&\displaystyle P(T_{1k}<T_{2l}, t_{1k}<t_{2l'}|\delta_{1k}=\delta_{2l}=\delta_{2l'}=1)\\ &\displaystyle =&\displaystyle p_1^{-1}p_2^{-2}\int_0^{T} (e^{-\lambda_{2}T}-e^{-\lambda_2u})^2\lambda_1e^{-\lambda_1u}du\\ &\displaystyle =&\displaystyle \left(\frac{q_2}{1-q_2}\right)^2\left[1 +\frac{\theta(1-q_2^{(2+\theta)})}{(2+\theta)(1-q_2^{\theta})q_2^2}-\frac{2\theta(1-q_2^{(1+\theta)})}{(1+\theta)(1-q_2^{\theta})q_2}\right] \end{array} \end{aligned} $$

1.2 Normal Distribution

Suppose that the nonfatal outcomes X ₁, X ₂ follow normal distributions $N(\mu _{x_1}, \sigma _{x_1})$ and $N(\mu _{x_2}, \sigma _{x_2})$, respectively.

Consider $\displaystyle \varDelta _{x}{=}\frac {\mu _{x_2}-\mu _{x_1} }{\sqrt {\sigma _{x_1}^2+\sigma _{x_2}^2}}~$, $\displaystyle \rho _{x_j}{=}\frac {\sigma _{x_j}^2}{\sigma _{x_1}^2+\sigma _{x_2}^2}$, and $\displaystyle Z_{kl}= \frac {X_{1k}-X_{2l}-(\mu _{x_1}-\mu _{x_2})}{\sqrt {\sigma _{x_1}^2+\sigma _{x_2}^2}}$.

We can show that

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \pi_{x1}=P(X_{1k}<X_{2l})=\varPhi(\varDelta_{x}),\\ &\displaystyle &\displaystyle \pi_{x2}=P(X_{1k}<X_{2l}, X_{1k'}<X_{2l})=P(Z_{kl}<\varDelta_{x}, ~ Z_{k'l}<\varDelta_{x}),\\ &\displaystyle &\displaystyle \pi_{x3}=P(X_{1k}<X_{2l}, X_{1k}<X_{2l'})=P(Z_{kl}<\varDelta_{x}, ~Z_{kl'}<\varDelta_{x}), \end{array} \end{aligned} $$

$(Z_{kl},Z_{k'l})~\sim N\left ( \left (\begin {array}{l} 0\\ 0 \end {array}\right ) , \left (\begin {array}{lr} 1&\rho _{x_2}\\ \rho _{x_2}&1 \end {array}\right )\right )$and $\allowdisplaybreaks (Z_{kl},Z_{kl'})~\sim N\left ( \left (\begin {array}{l} 0\\ 0 \end {array}\right ) , \left (\begin {array}{lr} 1&\rho _{x_1}\\ \rho _{x_1}&1 \end {array}\right )\right ).$