| 1. Binary |
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\begin{document}$$P(y_{ij}|\eta_i)= {\left\{ \begin{array}{ll} p(t_j) & \text{ if } y_{ij}=1 \\ 1-p(t_j) & \text{ if } y_{ij}=0 \end{array}\right. } $$\end{document}P(yij|ηi)=p(tj)ifyij=11-p(tj)ifyij=0
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\documentclass[12pt]{minimal}
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\begin{document}$$ p(t)={\text{expit}}{(b + a t)}$$\end{document}p(t)=expit(b+at)
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\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} b &= \theta_{1}+\eta_{i1} \\ a & = (\theta_{2}+\eta_{i2})(1-\theta_\text{H} z_i) \end{aligned}$$\end{document}b=θ1+ηi1a=(θ2+ηi2)(1-θHzi)
| 0.3 |
\documentclass[12pt]{minimal}
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\begin{document}$$\left( \begin{array}{cc}-1\\ 4\end{array}\right) $$\end{document}-14
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\documentclass[12pt]{minimal}
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\begin{document}$$\left( \begin{array}{cc} 0.4 & 0 \\ 0 & 4 \end{array} \right) $$\end{document}0.4004
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| 2. Time-to-event (TTE) |
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\begin{document}$$\displaystyle P(y_{i})= {\left\{ \begin{array}{ll} h(y_{i})S(y_i) & \text{ if } y_{i}<T \\ S(T) & \text{ if } y_{i}=T \end{array}\right. } $$\end{document}P(yi)=h(yi)S(yi)ifyi<TS(T)ifyi=T
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\begin{document}$$\begin{aligned}{lcl} h(t) &= \lambda \gamma (\lambda \cdot t)^{\gamma -1}\\ S(t) &= \exp (-\int_0^{t} h(x) dx) \end{aligned}$$\end{document}lclh(t)=λγ(λ·t)γ-1S(t)=exp(-∫0th(x)dx)
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\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \lambda & = \theta_{1}\exp (\theta_\text{H} z_i) \\ \gamma & = \theta_{2} \end{aligned}$$\end{document}λ=θ1exp(θHzi)γ=θ2
| 0.4 |
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\begin{document}$$\left( \begin{array}{l}0.2 \\ 2\end{array}\right) $$\end{document}0.22
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| 3. Count |
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\begin{document}$$\displaystyle P(y_{ij}|\eta_i)= \frac{\lambda (t_{j})^{y_{ij}}}{y_{ij}!}e^{-\lambda (t_{j})}$$\end{document}P(yij|ηi)=λ(tj)yijyij!e-λ(tj)
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\begin{document}$$\lambda (t) = b + A (1 - \exp (-k\cdot t))$$\end{document}λ(t)=b+A(1-exp(-k·t))
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\begin{document}$$\begin{aligned} b_i &= \theta_{1}\exp (\eta_{i1}) \\ A_i &= \theta_{2}\exp (\eta_{i2}) \\ k_i &= \theta_{3}\exp (\eta_{i3}+\theta_\text{H} \tilde{z}_i) \\ \end{aligned}$$\end{document}bi=θ1exp(ηi1)Ai=θ2exp(ηi2)ki=θ3exp(ηi3+θHz~i)
| 0.3 |
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\begin{document}$$\left( \begin{array}{ll}1\\ 4\\ 2\end{array}\right) $$\end{document}142
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\documentclass[12pt]{minimal}
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\begin{document}$$\left( \begin{array}{lll} 0.09 & 0 & 0 \\ 0 & 0.09 & 0 \\ 0 & 0 & 0.09 \end{array}\right) $$\end{document}0.090000.090000.09
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| 4. PK [12] |
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\begin{document}$$\displaystyle y_{ij}=\frac{A_1(t_j)}{V} ( 1 + \varepsilon_{ij}) $$\end{document}yij=A1(tj)V(1+εij)
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\documentclass[12pt]{minimal}
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\begin{document}$$\begin{array}{lcl} \frac{dA_1}{dt}=\frac{Q}{V_p} A_2 - \left( \frac{Cl}{V} + \frac{Q}{V} \right) {A_1} \\ \frac{d{A_2}}{dt}= \frac{Q}{V} {A_1} - \frac{Q}{V_p} {A_2} \end{array}$$\end{document}dA1dt=QVpA2-ClV+QVA1dA2dt=QVA1-QVpA2
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\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} Cl &= \theta_{1}\exp (\eta_{i1})(1+\theta_\text{H} \tilde{z}_i) \\ V & = \theta_{2}\exp (\eta_{i2}) \\ Q &= \theta_{3}\exp (\eta_{i3}) \\ V_p & = \theta_{4} \end{aligned}$$\end{document}Cl=θ1exp(ηi1)(1+θHz~i)V=θ2exp(ηi2)Q=θ3exp(ηi3)Vp=θ4
| 0.2 |
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\begin{document}$$\left( \begin{array}{ll}0.04\\ 0.14\\ 3.62\\ 2.9\end{array}\right) $$\end{document}0.040.143.622.9
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\begin{document}$$\left( \begin{array}{lll} 0.09 & 0 & 0.16\\ 0 & 0.04 & 0\\ 0.16 & 0 & 1.23 \end{array}\right) $$\end{document}0.0900.1600.0400.1601.23
| 0.06 |
| 5. PKPD [8] |
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\begin{document}$$ \begin{aligned} y_{ij}^{\prime} & =\frac{A_2(t_j)}{V} ( 1 + \varepsilon_{1,ij}) \\ y_{ij}^{\prime \prime} & =R(t_j) + \varepsilon_{2,ij} \end{aligned}$$\end{document}yij′=A2(tj)V(1+ε1,ij)yij″=R(tj)+ε2,ij
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\begin{document}$$\begin{array}{l} \frac{d{A_1}}{dt}=-k_aA_1 \\ \frac{d{A_2}}{dt}= k_aA_1 - \frac{Cl}{V} A_2 \\ \frac{d{R}}{dt}=k_o\frac{A_2}{V}/(C_{50}+\frac{A_2}{V}) - k_o R \end{array}$$\end{document}dA1dt=-kaA1dA2dt=kaA1-ClVA2dRdt=koA2V/(C50+A2V)-koR
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\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} Cl & = \theta_{1}\exp (\eta_{i1}) \\ V & = \theta_{2}\exp (\eta_{i2}) \\ k_a & = \theta_3 \\ k_o & = \theta_{4}\exp (\eta_{i3}) \\ C_{50} & = \theta_{5}\exp (\eta_{i4})(1+\theta_\text{H} z_i) \end{aligned}$$\end{document}Cl=θ1exp(ηi1)V=θ2exp(ηi2)ka=θ3ko=θ4exp(ηi3)C50=θ5exp(ηi4)(1+θHzi)
| 0.3 |
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\begin{document}$$\left( \begin{array}{l} 10\\ 100\\ 2\\ 0.2 \\ 0.3 \end{array}\right) $$\end{document}1010020.20.3
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\begin{document}$$\left( \begin{array}{llll} 0.49 & 0 & 0 & 0\\ 0 & 0.49 & 0 & 0\\ 0 & 0 & 0.49 & 0 \\ 0 & 0 & 0 & 0.49 \end{array}\right) $$\end{document}0.4900000.4900000.4900000.49
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\documentclass[12pt]{minimal}
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\begin{document}$$\left( \begin{array}{l}0.04\\ 0.04\end{array}\right) $$\end{document}0.040.04
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