Solution: Covariate measured with error

We consider the covariate \(X\), measured with some error, so that \(X = U + e\), where \(U \sim N(\mu,\sigma_u^2)\), \(e \sim N(0,\sigma_e^2)\). Thus \(X \sim N(\mu,\sigma_u^2+\sigma_e^2)\), and \(Cov(X,U) = Cov(U+e,U) = Cov(U,U) + Cov(e,U) = \sigma_u^2+0\).

We assume the additive model holds with respect to the true covariate \(U\), so that

\[\alpha(t|U=u) = \beta_0(t) + \beta_1(t)u.\]

We thus get survival function

\[S(t) = P(T > t | U = u) = \exp\left[-\int_0^t\alpha(s|U=u)ds\right] = \exp[-(B_0(t)+B_1(t)u)],\]

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

Note that when conditioned on \(U=u\), \(T\) and \(X\) are independent, which means that \(P(T > t | U = u, X= x) = P(T > t | U = u) \).

We now want to express the distribution of \(T\) in terms of only \(X\). We thus get survival function

\begin{align*} S^*(t) &= P(T > t | X = x) \\ &= \int_{-\infty}^{\infty} f(T > t, u | X = x) du \\ &= \int_{-\infty}^{\infty} P(T > t | X = x, U = u) f_{U | X = x}(u) du \\ &= \int_{-\infty}^{\infty} P(T > t | U = u) f_{U | X = x}(u) du \\ &= \exp(-B_0(t))\int_{-\infty}^{\infty} \exp(-B_1(t)u) f_{U|X=x}(u) du .\end{align*}

We have that \(U | X = x\) is normally distributed with mean

\[\mu^*=\mu + Cov(X,U)Var(X)^{-1}(x-\mu) = \frac{\sigma^2_u}{\sigma^2_u+\sigma^2_e}x + \frac{\sigma^2_e}{\sigma^2_u+\sigma^2_e}\mu\]

and variance

\[\sigma^{2*} = Var(U)- Cov(X,U)Var(X)^{-1}Cov(U,X) = \frac{\sigma^2_u\sigma^2_e}{\sigma^2_u+\sigma^2_e}.\]

Now notice that

\[\int_{-\infty}^{\infty} \exp(-B_1(t)u) f_{U|X=x}(u) du = E_{U | X = x}[\exp(-B_1(t)U)]\]

is the moment generating function of \(U | X= x\), and thus equal to

\[\exp(-B_1(t)\mu^*+ \sigma^{2*} B_1(t)^2/2).\]

Inserting the expressions for \(\mu^*\) and \(\sigma^{2*}\), and inserting the resulting expression above yields

\begin{align*} S^*(t) &= \exp(-B_0(t)) \exp\left[-B_1(t)\left(\frac{\sigma^2_u}{\sigma^2_u+\sigma^2_e}x + \frac{\sigma^2_e}{\sigma^2_u+\sigma^2_e}\mu\right) + \frac{\sigma^2_u\sigma^2_e}{2(\sigma^2_u+\sigma^2_e)} B_1(t)^2\right] \\ &= \exp\left[-\left(B_0(t) + B_1(t)\frac{\sigma^2_e}{\sigma^2_u+\sigma^2_e}\mu - \frac{\sigma^2_u\sigma^2_e}{2(\sigma^2_u+\sigma^2_e)} B_1(t)^2\right)- \left(B_1(t)\frac{\sigma^2_u}{\sigma^2_u+\sigma^2_e}x\right) \right] ,\end{align*}

which means the model with respect to \(X\) is still additive. To find the coefficients we use that \(\beta_i(t) = \frac{d}{dt}B_i(t)\), so that

\begin{align*}\tilde{\beta_0} &= \frac{d}{dt}\left( B_0(t) + B_1(t)\frac{\sigma^2_e}{\sigma^2_u+\sigma^2_e}\mu - \frac{\sigma^2_u\sigma^2_e}{2(\sigma^2_u+\sigma^2_e)} B_1(t)^2\right) \\ &= \beta_0(t) + \beta_1(t)\frac{\sigma^2_e}{\sigma^2_u+\sigma^2_e}\mu - \frac{\sigma^2_u\sigma^2_e}{\sigma^2_u+\sigma^2_e} B_1(t)\beta_1(t)\end{align*}

and

\begin{align*}\tilde{\beta_1} &= \frac{d}{dt}B_1(t)\frac{\sigma^2_u}{\sigma^2_u+\sigma^2_e} \\ &= \beta_1(t)\frac{\sigma^2_u}{\sigma^2_u+\sigma^2_e},\end{align*}

for the model

\[\alpha^*(t|X=x) = \tilde{\beta_0}(t) + \tilde{\beta_1}(t)x.\]