Question

(Continuation of Example 3.1.4) (a) Show that the independent censoring condition of \(\mathrm{ABGK}\) (3.5) reduces to (3.8) in the case of right-censored survival data as described in the example. (b) Assume that \(T^{*}\) and \(U\) are conditionally independent given an explanatory variable \(X\), and that the distribution of \(T^{*}\) and \(U\) depends on \(X\). Show that the right-censoring induced by \(U\) is independent. (c) Assume that \(T^{*}\) and \(U\) are conditionally independent given \(X\), but that we never observe \(X\). So \(N^{*}(t)\) has intensity \(\lambda^{*}(t)\) with respect to \(\mathcal{F}_{t}^{*}=\mathcal{F}_{t}^{N^{*}} .\) Is the filtering of \(N^{*}(t)\) generated by \(U\) independent?

Short answer

Answer: No, the filtering of N*(t) generated by U cannot be guaranteed to be independent without observing X. Additional information is required to establish independence.

Step-by-step solution

Part (a)

First, we recall that the independent censoring condition of \(\mathrm{ABGK} (3.5)\) is given by: \(E \left[ \frac{d N^{*}(t)}{Y(t)} | \mathcal{F}_{t-} \right] = E \left[ \frac{d N(t)}{Y(t)} | \mathcal{F}_{t-} \right].\) Now, for right-censored survival data, we have \(N^{*}(t) = N(t)1_{\{T \le U\}}\) and \(Y(t) = 1_{\{T^* > t\}}1_{\{U > t\}}\). Replacing these expressions in the independent censoring condition, we obtain: $$E \left[ \frac{d N(t)1_{\{T \le U\}}}{1_{\{T^* > t\}}1_{\{U > t\}}} | \mathcal{F}_{t-} \right] = E \left[ \frac{d N(t)}{1_{\{T^* > t\}}1_{\{U > t\}}} | \mathcal{F}_{t-} \right].$$ Comparing this with the formula \((3.8)\), we see that the independent censoring condition of \(\mathrm{ABGK}\) reduces to this condition in the case of right-censored survival data.

Part (b)

If \(T^{*}\) and \(U\) are conditionally independent given the explanatory variable \(X\), and the distribution of \(T^{*}\) and \(U\) depends on \(X\), then \(T^{*} \perp U | X\), and we can write their joint density as $$f_{T^*, U | X}(t^*, u | x) = f_{T^* | X}(t^* | x) \cdot f_{U | X}(u | x).$$ Now, suppose that \(T^{*}\) and \(U\) are both right-censored by time \(t\). Then, the right-censoring induced by \(U\) is independent if \(\frac{d N^{*}(t)}{1_{\{T^* > t\}}1_{\{U > t\}}}\) is independent of \(\frac{d N(t)}{1_{\{T^* > t\}}1_{\{U > t\}}}\). From the joint density formula, we can establish that \(\frac{d N^{*}(t)}{1_{\{T^* > t\}}1_{\{U > t\}}}\) and \(\frac{d N(t)}{1_{\{T^* > t\}}1_{\{U > t\}}}\) are indeed independent as they depend only on \(X\).

Part (c)

If \(T^{*}\) and \(U\) are conditionally independent given \(X\), then the intensity of \(N^{*}(t)\) has a form \(\lambda^{*}(t) = \lambda(t | X)\). If we do not observe \(X\), the filtering of \(N^{*}(t)\), denoted by \(\mathcal{F}_{t}^{*}\) with respect to \(\mathcal{F}_{t}^{N^{*}}\), cannot be determined precisely. However, if \(\lambda^{*}(t)\) can still be expressed solely as a function of \(Y(t)\), then the filtering of \(N^{*}(t)\) would still be generated by \(U\). Unfortunately, since we do not observe \(X\), this is not the case, and we conclude that the filtering of \(N^{*}(t)\) generated by \(U\) cannot be guaranteed to be independent without additional information.