Hints: Kaplan-Meier estimator and Greenwood's formula

To make this problem simpler, you may assume that there are no tied event-times. This makes the calculations easier, but the general structure of the solution is the same (in this case all the \(d\)s below will be equal to 1).

Note first that the number at risk at time \(T_j\) is \(Y(T_j) = n - (d_1 + \ldots + d_{j-1})\). Use this in the expression for the tie-adjusted Kaplan-Meier estimator in \((3.32)\).

Now use induction to prove that when \(t \in [T_k,T_{k+1})\), we have

\[\hat S(t) = \frac{n - (d_1 + \ldots + d_{k})}{n},\]

and show that this is equal to \(1-\hat F(t)\) regardless of the value of \(k\).

In (b) consider the steps

\[\frac{\tilde{\tau}(T_{k+1})^2}{\hat S(T_{k+1})^2} - \frac{\tilde{\tau}(T_{k})^2}{\hat S(T_{k})^2 }\]

and

\[\frac{1-\hat S(T_{k+1})}{n \hat S(T_{k+1})} - \frac{1-\hat S(T_k)}{n \hat S(T_k)}.\]

Some careful arithmetic should show that these are equal.