Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 207 / 6

Condition on the event time $T_A=u$. Independence makes the chance that $T_B$ is still later equal to its survivor function $F_B(u)$. HenceThe continuous distributions give zero probability of a tie. The complementary probability is ${\pi }_B={\int }_0^{\infty }{F}_A(u)f_B(u)du=1-{\pi }_A$.

For the survivor functions, $F_A^{\prime }(u)=-f_A(u)$ and $F_B^{\prime }(u)=-f_B(u)$. The product rule givesIntegrating from 0 to $t$, using $F_A(0)=F_B(0)=1$, yieldsThe three terms partition the possible histories: A occurs first by time $t$, B occurs first by time $t$, or neither has occurred by $t$. The first two are the cumulative incidence functions for the two competing risks, while the third is the survival probability of their minimum. The initial value assumes nonnegative event times with no atom at zero.

Let ${\pi }_A=\mathbb{P}(T_A<T_B)$, and assume $0<{\pi }_A<1$ so that both conditional distributions are defined. By independence, the joint density of $T_A=u$ and $T_A<T_B$ is $f_A(u)F_B(u)$, whereas the joint density of $T_A=u$ and $T_B<T_A$ is $f_A(u)[1-F_B(u)]$. Normalizing givesBoth are densities of the latent event time $T_A$. The second is not the density of the observed first event $T_B$ in cases where B wins. Their integrals are 1 by the definitions of ${\pi }_A$ and its complement.

The minimum exceeds $t$ exactly when both times exceed $t$. By independence,Differentiating its survivor function givesThe tail integral formula for moments then givesEquivalently, integration by parts in ${\int }_0^{R}t{f}_X(t)dt$ gives $-R{F}_A(R)F_B(R)+{\int }_0^R{F}_A(t)F_B(t)dt$. The assumed $t{F}_B(t)\to 0$ makes the boundary term vanish, since $0\le F_A\le 1$.

The stated tail condition does not itself guarantee a finite mean. For example, the proper survivor functionssatisfy $t{F}_B(t)\to 0$, but their product behaves as $1/[t\log t\log \log t]$, whose integral diverges. Thus the displayed expectation is valid as an extended nonnegative value; finiteness needs integrability of the product. The exponential distribution in the next part guarantees that finiteness.

For the exponential distribution with mean $\mu$, $f_A(t)=F_A(t)/\mu$ and $F_A(t)=e^{-t/\mu }$. ThereforeSince ties have probability zero, $\mathbb{P}(X=T_B)=1-\mathbb{E}X/\mu$. Taking expected values in the definition of $U$ givesThere is also an imputation interpretation. If A is observed first, $T_A=X$; if B censors A, memorylessness of the exponential distribution and independence imply that the remaining mean lifetime is $\mu$. Thus $U$ is the conditional expectation of $T_A$ given the observed minimum and censoring indicator. The tower property of conditional expectation gives the same equality. This does not say that $U$ and $T_A$ have the same distribution.