In this framework let \(A=a\) denote the treatment arm, \(Y\) the outcome variable, and \(t\) the time variable. The SensIAT package allows users to specify different loss functions and link functions when fitting models to their data. This flexibility enables users to tailor the modeling approach to their specific research questions and data characteristics. Let \(\mu_a(t) = E\{Y(t)\mid A=a\}\) denote the population mean outcome at time \(t\) were all participants assigned to treatment arm \(A=a\). The mean function is modeled as $_a(t) = g{-1}((t)_a) = s((t)^_a) $ where \(g\) is a link function, \(\mathbf{B}\) is a vector values basis function and \(\gamma_a\) is a vector of coefficients, distinct for each treatment arm. The function \(s = g^{-1}\) is the inverse link function. We consider three loss functions to use in order to fit the coefficient \(\gamma_a\):
The estimate for the treatment group marginal mean function, \(\widehat\mu_{\mathcal{L}, g}(t) = g^{-1}\big\{\boldsymbol{B}(t)^\prime \widehat{\boldsymbol{\beta}}\big\}\) is found by solving \(\frac{1}{n}\sum_{i=1}^n \widehat{\boldsymbol{\Psi}}_{\mathcal{L},g}(\boldsymbol{O}_i;\boldsymbol{\beta}) = 0\), for \(\beta\), where
\[ \widehat{\boldsymbol{\Psi}}_{\mathcal{L},g}(\boldsymbol{O}_i;\boldsymbol{\beta}) = \sum_{k=1}^{K_i} \Bigg\{ W_{\mathcal{L}, g}(T_{ik};\boldsymbol{\beta})\frac{\big[Y_i(T_{ik}) - \widehat{\mathbb{E}} \big\{Y(T_{ik}) \mid \overline{\boldsymbol{O}}_i(T_{ik}) \big\} \big] }{\widehat{\rho} \big \{T_{ik} \mid \overline{\boldsymbol{O}}_i(T_{ik}), Y_i(T_{ik}) \big\}} \Bigg\} + \int_{t=t_1}^{t_2} W_{\mathcal{L}, g}(t\mid\boldsymbol{\beta}) \left[ \widehat{\mathbb{E}} \left\{ Y(t)\mid \overline{\boldsymbol{O}}_i(t) \right\} - s\left\{ \boldsymbol{B}(t)^\prime \boldsymbol{\beta} \right\} \right]dt \] The \(W_{\mathcal{L}, g}(t;\boldsymbol{\beta})\) term depends on the choice of loss function and link function. Then we solve \(\frac{1}{n}\sum_{i=1}^n \widehat{\boldsymbol{\Psi}}_{\mathcal{L},g}(\boldsymbol{O}_i;\boldsymbol{\beta}) = 0\) for \(\beta\) using a vectorized root-finding algorithm.
For squared error loss in the transformed space, \(\mathcal{L}_1\), in general we have \[ W_{\mathcal{L},g}(t;\boldsymbol{\beta}) = W_1(t;\boldsymbol{\beta}) = \boldsymbol{V}_1^{-1}\boldsymbol{B}(t)\left.\frac{\partial g(z)}{\partial z} \right|_{z = s\left\{\boldsymbol{B}(t)^\prime \boldsymbol{\beta}\right\}} , \quad \boldsymbol{V}_1 = \int_{t=t_1}^{t_2} \boldsymbol{B}(t) \boldsymbol{B}(t)^\prime dt \]
For the identity link, \(g(\mu) = \mu\), \(\frac{\partial g(z)}{\partial z}\equiv 1\), so \[ W_1(t;\boldsymbol{\beta}) = \boldsymbol{V}_1^{-1}\boldsymbol{B}(t). \]
For the log link, \[ g(\mu) = \log(\mu), \]
\(\frac{\partial g(z)}{\partial z} = \frac{1}{z}\),
\[ W_1(t;\boldsymbol{\beta}) = \frac{\boldsymbol{V}_1^{-1}\boldsymbol{B}(t)} {\boldsymbol{B}(t)^\prime \boldsymbol{\beta}}. \]
For the logit link, \[ \begin{align} g(\mu) &= \log\left(\frac{\mu}{1-\mu}\right), \\ \frac{\partial g(z)}{\partial z} &= \frac{1}{z(1-z)}, \\ W_1(t;\boldsymbol{\beta}) &= \boldsymbol{V}_1^{-1}\boldsymbol{B}(t) \left.\frac{1}{z(1-z)}\right|_{z=s\{\boldsymbol{B}(t)^\prime\boldsymbol{\beta}\}} = \boldsymbol{V}_1^{-1}\boldsymbol{B}(t) \frac{\left[1+\exp\{\boldsymbol{B}(t)^\prime\boldsymbol{\beta}\}\right]^2}{\exp\{\boldsymbol{B}(t)^\prime\boldsymbol{\beta}\}}. \end{align} \]
For quasi-likelihood loss, \(\mathcal{L}_3\), in general we have \[ W_{\mathcal{L}_3,g}(t;\boldsymbol{\beta}) = W_3(t;\boldsymbol{\beta}) = \boldsymbol{V}_3(\boldsymbol{\beta})^{-1} \boldsymbol{B}(t) , \quad \boldsymbol{V}_3(\boldsymbol{\beta}) = \int_{t=t_1}^{t_2} \boldsymbol{B}(t) \boldsymbol{B}(t) ^ \prime \left. \frac{\partial s(z)}{\partial z} \right|_{z=\boldsymbol{B}(t)^\prime \boldsymbol{\beta}} dt \]
For the identity link, \(s(z) = z\), so \[ \frac{\partial s(z)}{\partial z} \equiv 1, \] and \[ W_3(t;\boldsymbol{\beta}) = \boldsymbol{V}_3^{-1}\boldsymbol{B}(t), \] where \[ \boldsymbol{V}_3 = \int_{t=t_1}^{t_2} \boldsymbol{B}(t) \boldsymbol{B}(t)^\prime dt. \]
For the log link, \(s(z) = \exp(z)\), so \[ \frac{\partial s(z)}{\partial z} = \exp(z), \] and \[ W_3(t;\boldsymbol{\beta}) = \boldsymbol{V}_3(\boldsymbol{\beta})^{-1} \boldsymbol{B}(t), \] where \[ \boldsymbol{V}_3(\boldsymbol{\beta}) = \int_{t=t_1}^{t_2} \boldsymbol{B}(t) \boldsymbol{B}(t)^\prime \exp\left(\boldsymbol{B}(t)^\prime \boldsymbol{\beta}\right) dt. \]
For the logit link, \(s(z) = \frac{\exp(z)}{1+\exp(z)}\), so \[ \frac{\partial s(z)}{\partial z} = \frac{\exp(z)}{\left\{1+\exp(z)\right\}^2}, \] and \[ W_3(t;\boldsymbol{\beta}) = \boldsymbol{V}_3(\boldsymbol{\beta})^{-1} \boldsymbol{B}(t), \] where \[ \boldsymbol{V}_3(\boldsymbol{\beta}) = \int_{t=t_1}^{t_2} \boldsymbol{B}(t) \boldsymbol{B}(t)^\prime \frac{\exp\left(\boldsymbol{B}(t)^\prime \boldsymbol{\beta}\right)}{\left\{1+\exp\left(\boldsymbol{B}(t)^\prime \boldsymbol{\beta}\right)\right\}^2} dt. \]