Solution: Dropping variables from the additive model

We have the additive model

\[\alpha(t | X = x) = \beta_0(t) + \beta_1(t) x_1 + \ldots + \beta_p(t) x_p\],

and thus survival function

\[S(t) = P(T > t | X = x) = \exp\left[-\int_0^t\alpha(s | X = x)ds\right] = \exp\left[-(B_0(t) + B_1(t) x_1 + \ldots + B_p(t) x_p)\right],\]

where \(B_i(t) = \int_0^t\beta_i(s)ds\).

The model where a covariate, say \(X_p\), is dropped, has survival function

\[S^*(t) = P(T > t | X_{-p} = x_{-p}),\]

where the index \(-p\) means the vector contains all its original elements except from the one with index \(p\). Using the law of total expectation to marginalize out \(X_p\) (assuming it is real-valued), we get

\begin{align*} P(T > t | X_{-p} = x_{-p}) &= \int_{-\infty}^{\infty} f(T > t, x_p | X_{-p} = x_{-p}) dx_p \\ &= \int_{-\infty}^{\infty} P(T > t | X = x) f_{X_p | X_{-p} = x_{-p}}(x_p) dx_p \\ &= \exp\left[-(B_0(t) + B_1(t) x_1 + \ldots + B_{p-1}(t) x_{p-1})\right]\int_{-\infty}^{\infty} \exp(-B_p(t)x_p) f_{X_p}(x_p) dx_p \\ &= \exp(-B_0(t)) \exp\left[-(B_1(t) x_1 + \ldots + B_{p-1}(t) x_{p-1})\right]E_{X_p}\left[\exp(-B_p(t)X_p)\right].\end{align*}

From line 2 to 3 we used that \(f_{X_p | X_{-p} = x_{-p}} = f_{X_p}\), since the \(X_i\) are independent. Note that \(\exp(-B_0(t)) \) and \(E_{X_p}\left[\exp(-B_p(t)X_p)\right]\) are both positive functions of only \(t\), so that we can write \(\exp(-B_0(t))E_{X_p}\left[\exp(-B_p(t)X_p)\right] = \exp(-\widetilde{B_0}(t))\), and thus we see that

\[\underline{\underline{S^*(t) = \exp\left[-(\widetilde{B_0}(t) + B_1(t) x_1 + \ldots + B_{p-1}(t) x_{p-1})\right]}},\]

corresponding to an additive model where only \(\beta_0\) is changed.

Now we consider the case where the \(X_i\) are multivariate normal, with means \(\mu_i\) and some covariance matrix \(\Sigma\). For the model where \(X_p\) is dropped, following the argument above, we get survival function

\begin{align*} S^*(t) &= \exp\left[-(B_0(t) + B_1(t) x_1 + \ldots + B_{p-1}(t) x_{p-1})\right]\int_{-\infty}^{\infty} \exp(-B_p(t)x_p) f_{X_p | X_{-p} = x_{-p}}(x_p) dx_p .\end{align*}

Since the \(X_i\) are multivariate normal, we have that \(X_p | X_{-p} = x_{-p}\) is normally distributed with mean

\[\mu_p^*=\mu_p + \Sigma_{p,-p} \Sigma_{-p,-p}^{-1}(x_{-p}-\mu_{-p})\]

and variance

\[\sigma^2_p = \Sigma_{p,p} - \Sigma_{p,-p} \Sigma_{-p,-p}^{-1}\Sigma_{-p,p},\]

where \(\Sigma_{p,p} = Var(X_p)\), \(\Sigma_{p,-p} = (Cov(X_1,X_p),\ldots,Cov(X_{p-1},X_p))\), \(\Sigma_{-p,p}={\Sigma_{p,-p}}^T\) and \(\Sigma_{-p,-p}\) is the covariance matrix of \(X_{-p}\).

Now notice that

\[\int_{-\infty}^{\infty} \exp(-B_p(t)x_p) f_{X_p | X_{-p} = x_{-p}}(x_p) dx_p = E_{X_p | X_{-p} = x_{-p}}[\exp(-B_p(t)X_p)]\]

is the moment generating function of \(X_p | X_{-p} = x_{-p}\), and thus equal to

\[\exp(-B_p(t)\mu_p^*+ \sigma^2_p B_p(t)^2/2).\]

From above we know that \(\mu_p^*\) is linear in \(x_{-p}\), and hence the expression for \(S^*(t)\) still has the form

\[S^*(t) = \exp\left[-(\widetilde{B_0}(t) + \widetilde{B_1}(t) x_1 + \ldots + \widetilde{B_{p-1}}(t) x_{p-1})\right],\]

and we still have an additive model, where potentially all the \(\beta_i(t)\), \(i = 0, 1, \ldots p-1\), are changed.