Fitting with Different Loss and Link Functions

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 $_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 $\gamma_a$:

• $\mathcal{L}_1(\boldsymbol{\gamma}) = \int_{t=t_1}^{t_2} \left[ g\left\{\mu(t)\right\} - \boldsymbol{B}(t)^\prime \boldsymbol{\gamma}\right]^2 dt$, 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

$$\widehat{\boldsymbol{\Psi}}_j(\boldsymbol{O}_i;\boldsymbol{\beta}) = \sum_{k=1}^{K_i} \Bigg\{ W_{\mathcal{L}, g}(T_{ik};\boldsymbol{\beta})\frac{\big[Y_i(T_{ik}) - \widehat{\mathbb{E}} \big\{Y(T_{ik}) \mid \overline{\boldsymbol{O}}_i(T_{ik}) \big\} \big] }{\widehat{\rho} \big \{T_{ik} \mid \overline{\boldsymbol{O}}_i(T_{ik}), Y_i(T_{ik}) \big\}} \Bigg\} + \int_{t=t_1}^{t_2} W_{\mathcal{L}, g}(t\mid\boldsymbol{\beta}) \left[ \widehat{\mathbb{E}} \left\{ Y(t)\mid \overline{\boldsymbol{O}}_i(t) \right\} - s\left\{ \boldsymbol{B}(t)^\prime \boldsymbol{\beta} \right\} \right]dt$$ The $W_{\mathcal{L}, g}(t;\boldsymbol{\beta})$ term depends on the choice of loss function and link function.

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, $$g(\mu) = \log(\mu),$$

$$\frac{\partial g(z)}{\partial z} = \frac{1}{z},$$$$W_1(t;\boldsymbol{\beta}) = \boldsymbol{V}_1^{-1}\boldsymbol{B}(t) \left\{\boldsymbol{B}(t)^\prime \boldsymbol{\beta}\right\}^{-1}.$$

Logit Link

For the logit link, $$g(\mu) = \log\left(\frac{\mu}{1-\mu}\right),$$$$\frac{\partial g(z)}{\partial z} = \frac{1}{z(1-z)},$$$$W_1(t;\boldsymbol{\beta}) = \boldsymbol{V}_1^{-1}\boldsymbol{B}(t) \left\{\boldsymbol{B}(t)^\prime \boldsymbol{\beta} \left(1-\boldsymbol{B}(t)^\prime \boldsymbol{\beta}\right)\right\}^{-1}.$$

Squared error loss in the original space

For squared error loss in the original space, $\mathcal{L}_2$, in general we have $$W_{\mathcal{L}_2,g}(t;\boldsymbol{\beta}) = W_2(t;\boldsymbol{\beta}) = \boldsymbol{V}_2(\boldsymbol{\beta})^{-1} \frac{\partial s(\boldsymbol{B}(t)^\prime \boldsymbol{\beta})}{\partial \boldsymbol{\beta}} , \quad \boldsymbol{V}_2(\boldsymbol{\beta}) = \int_{t=t_1}^{t_2} \frac{\partial s(\boldsymbol{B}(t)^\prime \boldsymbol{\beta})}{\partial \boldsymbol{\beta}} \frac{\partial s(\boldsymbol{B}(t)^\prime \boldsymbol{\beta})}{\partial \boldsymbol{\beta}^\prime} dt$$

Identity Link

For the identity link, $s(z) = z$, so $$s(\boldsymbol{B}(t)^\prime \boldsymbol{\beta}) = \boldsymbol{B}(t)^\prime \boldsymbol{\beta},$$$$\frac{\partial s(\boldsymbol{B}(t)^\prime \boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = \boldsymbol{B}(t),$$ and $$W_2(t;\boldsymbol{\beta}) = \boldsymbol{V}_2^{-1}\boldsymbol{B}(t),$$ where $$\boldsymbol{V}_2 = \int_{t=t_1}^{t_2} \boldsymbol{B}(t) \boldsymbol{B}(t)^\prime dt.$$

Log Link

For the log link, $s(z) = \exp(z)$, so $$s(\boldsymbol{B}(t)^\prime \boldsymbol{\beta}) = \exp\left(\boldsymbol{B}(t)^\prime \boldsymbol{\beta}\right),$$$$\frac{\partial s(\boldsymbol{B}(t)^\prime \boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = \boldsymbol{B}(t) \exp\left(\boldsymbol{B}(t)^\prime \boldsymbol{\beta}\right),$$ and $$W_2(t;\boldsymbol{\beta}) = \boldsymbol{V}_2(\boldsymbol{\beta})^{-1} \boldsymbol{B}(t) \exp\left(\boldsymbol{B}(t)^\prime \boldsymbol{\beta}\right),$$ where $$\boldsymbol{V}_2(\boldsymbol{\beta}) = \int_{t=t_1}^{t_2} \boldsymbol{B}(t) \boldsymbol{B}(t)^\prime \exp\left(2\boldsymbol{B}(t)^\prime \boldsymbol{\beta}\right) dt.$$

Logit Link

For the logit link, $s(z) = \frac{\exp(z)}{1+\exp(z)}$, so $$s(\boldsymbol{B}(t)^\prime \boldsymbol{\beta}) = \frac{\exp\left(\boldsymbol{B}(t)^\prime \boldsymbol{\beta}\right)}{1+\exp\left(\boldsymbol{B}(t)^\prime \boldsymbol{\beta}\right)},$$$$ \frac{\partial s(\boldsymbol{B}(t)^\prime \boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = \boldsymbol{B}(t) \frac{\exp\left(\boldsymbol{B}(t)^\prime \bold{\beta}\right)}{\left\{1+\exp\left(\boldsymbol{B}(t)^\prime \boldsymbol{\beta}\right)\right\}^2}, $$ and $$ W_2(t;\boldsymbol{\beta}) = \boldsymbol{V}_2(\boldsymbol{\beta})^{-1} \boldsymbol{B}(t) \frac{\exp\left(\boldsymbol{B}(t)^\prime \bold{\beta}\right)}{\left\{1+\exp\left(\boldsymbol{B}(t)^\prime \bold{\beta}\right)\right\}^2}, $$ where $$\boldsymbol{V}_2(\boldsymbol{\beta}) = \int_{t=t_1}^{t_2} \boldsymbol{B}(t) \boldsymbol{B}(t)^\prime \frac{\exp\left(2\boldsymbol{B}(t)^\prime \boldsymbol{\beta}\right)}{\left\{1+\exp\left(\boldsymbol{B}(t)^\prime \boldsymbol{\beta}\right)\right\}^4} dt.$$

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 $$\boldsymbol{V}_3(\boldsymbol{\beta}) = \int_{t=t_1}^{t_2} \boldsymbol{B}(t) \boldsymbol{B}(t)^\prime \frac{\exp\left(\boldsymbol{B}(t)^\prime \boldsymbol{\beta}\right)}{\left\{1+\exp\left(\boldsymbol{B}(t)^\prime \boldsymbol{\beta}\right)\right\}^2} dt.$$