Question

Let \(N(t)=\left(N_{1}(t), \ldots, N_{k}(t)\right), t \in[0, \tau]\), be a multivariate counting process with respect to \(\mathcal{F}_{t}\). It holds that the intensity $$ \lambda(t)=\left(\lambda_{1}(t), \ldots, \lambda_{k}(t)\right) $$ of \(N(t)\) is given (heuristically) as $$ \lambda_{h}(t)=P\left(d N_{h}(t)=1 \mid \mathcal{F}_{t-}\right), $$ where \(d N_{h}(t)=N_{h}((t+d t)-)-N_{h}(t-)\) is the change in \(N_{h}\) over the small time interval \([t, t+d t)\). (a) Let \(T^{*}\) be a lifetime with hazard \(\alpha(t)\) and define \(N(t)=I\left(T^{*} \leq t\right)\). Use the above \((2.29)\) to show that the intensity of \(N(t)\) with respect to the history \(\sigma\\{N(s): s \leq t\\}\) is $$ \lambda(t)=I\left(t \leq T^{*}\right) \alpha(t) . $$16 2. Probabilistic background (b) Let \(T^{*}\) be a lifetime with hazard \(\alpha(t)\) that may be right- censored at time \(C\). We assume that \(T^{*}\) and \(C\) are independent. Let \(T=T^{*} \wedge C\), \(\Delta=I\left(T^{*} \leq C\right)\) and \(N(t)=I(T \leq t, \Delta=1)\). Use the above (2.29) to show that the intensity of \(N(t)\) with respect to the history $$ \sigma\\{I(T \leq s, \Delta=0), I(T \leq s, \Delta=1): s \leq t\\} $$ is $$ \lambda(t)=I(t \leq T) \alpha(t) $$

Short answer

Question: Show that the intensity of a counting process can be represented in terms of lifetimes with hazard and its history. Answer: By considering a lifetime with hazard and analyzing the intensity of the counting process in two different cases, we can show that the intensity can be represented as: $$ \lambda(t) = I(t \leq T) \cdot \alpha(t) $$ where \(I(t \leq T)\) is an indicator function, and \(\alpha(t)\) is the hazard function. This representation holds for both non-censored and right-censored lifetime hazards.

Step-by-step solution

Part (a) Intensity of N(t) with lifetime hazard

We need to show that the intensity of N(t) is given by the formula: $$ \lambda(t)=I\left(t \leq T^{*}\right) \alpha(t) $$ where \(N(t) = I(T^* \leq t)\) and \(T^*\) is a lifetime with hazard \(\alpha(t)\). Using the given heuristic definition of intensity, we get: $$ \lambda(t) = P(dN(t) = 1 \mid \mathcal{F}_{t-}) $$ where \(dN(t) = N((t+dt)-) - N(t-)\). We know that \(N(t) = I(T^* \leq t)\). So, $$ dN(t) = I(T^* \leq (t+dt)-) - I(T^* \leq t-) $$ Now, consider the two cases when \(t \leq T^*\) and \(t > T^*\): 1) If \(t \leq T^*\), then: $$ dN(t) = I(T^* \leq (t + dt)-) - I(T^* \leq t-) $$ In this case, we have \(I(T^* \leq t-) = 1\) and \(I(T^* \leq (t+ dt)-) = 1\). Therefore, \(dN(t) = 0\). 2) If \(t > T^*\), then: $$ dN(t) = I(T^* \leq (t+ dt)-) - I(T^* \leq t-) $$ In this case, we have \(I(T^* \leq t-) = 0\) and \(I(T^* \leq (t+ dt)-) = 1\). Therefore, \(dN(t) = 1\). Thus, we can write the intensity as: $$ \lambda(t) = P(dN(t) = 1 \mid \mathcal{F}_{t-}) = I(t \leq T^*) \cdot P(T^* \leq (t+dt) \mid T^* \leq t ) = I(t \leq T^*) \cdot \alpha(t) $$ This completes part (a) of the exercise.

Part (b) Intensity of N(t) with right-censored lifetime hazard

In this part, we need to show that the intensity of the counting process is given by the formula: $$ \lambda(t)=I(t \leq T) \alpha(t) $$ where \(T = T^* \wedge C\), \(\Delta = I\left(T^{*} \leq C\right)\), \(N(t) = I(T \leq t, \Delta=1)\), and the history is \(\sigma\\{I(T \leq s, \Delta=0), I(T \leq s, \Delta=1): s \leq t\\}\). Using the given heuristic definition of intensity, just like before, we get: $$ \lambda(t) = P(dN(t) = 1 \mid \mathcal{F}_{t-}) $$ where \(dN(t) = N((t+dt)-) - N(t-)\). We can write the change in the process as \(dN(t) = I(T \leq (t+dt), \Delta=1) - I(T \leq t, \Delta=1)\). Now, as we did in part (a), we consider the two cases when \(t \leq T\) and \(t > T\): 1) If \(t \leq T\), then: $$ dN(t) = I(T \leq (t+ dt)-, \Delta=1) - I(T \leq t-, \Delta=1) $$ In this case, we have \(I(T \leq t -, \Delta=1) = 1\) and \(I(T \leq (t+dt)-, \Delta=1) = 1\). Therefore, \(dN(t) = 0\). 2) If \(t > T\), then: $$ dN(t) = I(T \leq (t+ dt)-, \Delta=1) - I(T \leq t-, \Delta=1) $$ In this case, we have \(I(T \leq t -, \Delta=1) = 0\) and \(I(T \leq (t+dt)-, \Delta=1) = 1\). Therefore, \(dN(t) = 1\). Thus, we can write the intensity as: $$ \lambda(t) = P(dN(t) = 1 \mid \mathcal{F}_{t-}) = I(t \leq T) \cdot P(T \leq (t+dt) \mid T \leq t ) = I(t \leq T) \cdot \alpha(t) $$ This completes part (b) of the exercise.