--- title: "Fitting with Different Loss and Link Functions" output: rmarkdown::html_vignette vignette: > %\VignetteIndexEntry{Fitting with Different Loss and Link Functions} %\VignetteEngine{knitr::rmarkdown} %\VignetteEncoding{UTF-8} --- ```{r, include = FALSE} knitr::opts_chunk$set( collapse = TRUE, comment = "#>" ) ``` ```{r setup} library(SensIAT) ``` # Introduction 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 $\mu_a(t) = g^{-1}\left(\mathbf{B}(t)^\prime\gamma_a\right) = s\left(\mathbf{B}(t)^\prime\gamma_a\right) $ 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 $\gamma_a$: * $\mathcal{L}_1(\boldsymbol{\gamma}) = \int_{t=t_1}^{t_2} \left[ g\left\{\mu(t)\right\} - \boldsymbol{B}(t)^\prime \boldsymbol{\gamma}\right]^2 dt$, squared error loss in the transformed space; * $\mathcal{L}_2(\boldsymbol{\gamma}) = \int_{t=t_1}^{t_2} \left[ \mu(t) - s\left\{\boldsymbol{B}(t)^\prime \boldsymbol{\gamma}\right\}\right]^2 dt$, squared error loss in the original space; * $\mathcal{L}_3(\boldsymbol{\gamma}) = \int_{t=t_1}^{t_2} \left[ b\{\boldsymbol{B}(t)^\prime \boldsymbol{\gamma}\} - \mu(t)\boldsymbol{B}(t)^\prime \boldsymbol{\gamma}\right] dt$, quasi-likelihood loss, where $s(z) = \frac{\partial b(z)}{\partial z }$. In the SensIAT package, we have implemented the following link functions: * Identity link: $g(\mu) = \mu$, $s(z) = z$, * Log link: $g(\mu) = \log(\mu)$, $s(z) = \exp(z)$, * Logit link: $g(\mu) = \log\left(\frac{\mu}{1-\mu}\right)$, $s(z) = \frac{\exp(z)}{1+\exp(z)}$. # Details 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. ## Squared Error Loss in the Transformed Space For squared error loss in the transformed space, $\mathcal{L}_1$, in general we have $$ W_{\mathcal{L},g}(t;\boldsymbol{\beta}) = W_1(t;\boldsymbol{\beta}) = \boldsymbol{V}_1^{-1}\boldsymbol{B}(t)\left.\frac{\partial g(z)}{\partial z} \right|_{z = s\left\{\boldsymbol{B}(t)^\prime \boldsymbol{\beta}\right\}} , \quad \boldsymbol{V}_1 = \int_{t=t_1}^{t_2} \boldsymbol{B}(t) \boldsymbol{B}(t)^\prime dt $$ ### Identity Link For the identity link, $g(\mu) = \mu$, $\frac{\partial g(z)}{\partial z}\equiv 1$, so $$ W_1(t;\boldsymbol{\beta}) = \boldsymbol{V}_1^{-1}\boldsymbol{B}(t). $$ ### Log Link For the log link, $$ g(\mu) = \log(\mu), $$ $\frac{\partial g(z)}{\partial z} = \frac{1}{z}$, $$ W_1(t;\boldsymbol{\beta}) = \frac{\boldsymbol{V}_1^{-1}\boldsymbol{B}(t)} {\boldsymbol{B}(t)^\prime \boldsymbol{\beta}}. $$ ### Logit Link For the logit link, $$ \begin{align} g(\mu) &= \log\left(\frac{\mu}{1-\mu}\right), \\ \frac{\partial g(z)}{\partial z} &= \frac{1}{z(1-z)}, \\ W_1(t;\boldsymbol{\beta}) &= \boldsymbol{V}_1^{-1}\boldsymbol{B}(t) \left.\frac{1}{z(1-z)}\right|_{z=s\{\boldsymbol{B}(t)^\prime\boldsymbol{\beta}\}} = \boldsymbol{V}_1^{-1}\boldsymbol{B}(t) \frac{\left[1+\exp\{\boldsymbol{B}(t)^\prime\boldsymbol{\beta}\}\right]^2}{\exp\{\boldsymbol{B}(t)^\prime\boldsymbol{\beta}\}}. \end{align} $$ ## Quasi-likelihood Loss For quasi-likelihood loss, $\mathcal{L}_3$, in general we have $$ W_{\mathcal{L}_3,g}(t;\boldsymbol{\beta}) = W_3(t;\boldsymbol{\beta}) = \boldsymbol{V}_3(\boldsymbol{\beta})^{-1} \boldsymbol{B}(t) , \quad \boldsymbol{V}_3(\boldsymbol{\beta}) = \int_{t=t_1}^{t_2} \boldsymbol{B}(t) \boldsymbol{B}(t) ^ \prime \left. \frac{\partial s(z)}{\partial z} \right|_{z=\boldsymbol{B}(t)^\prime \boldsymbol{\beta}} dt $$ ### Identity Link 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 $$ \boldsymbol{V}_3 = \int_{t=t_1}^{t_2} \boldsymbol{B}(t) \boldsymbol{B}(t)^\prime dt. $$ ### Log Link 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 $$ \boldsymbol{V}_3(\boldsymbol{\beta}) = \int_{t=t_1}^{t_2} \boldsymbol{B}(t) \boldsymbol{B}(t)^\prime \exp\left(\boldsymbol{B}(t)^\prime \boldsymbol{\beta}\right) dt. $$ ### Logit Link 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 $$ \boldsymbol{V}_3(\boldsymbol{\beta}) = \int_{t=t_1}^{t_2} \boldsymbol{B}(t) \boldsymbol{B}(t)^\prime \frac{\exp\left(\boldsymbol{B}(t)^\prime \boldsymbol{\beta}\right)}{\left\{1+\exp\left(\boldsymbol{B}(t)^\prime \boldsymbol{\beta}\right)\right\}^2} dt. $$