= Solution
For the population identity, write $X=\min(T,C)$, $V=\mathbf1_{\{T\leq C\}}$, and $G(t)=\mathbb P(C\geq t)$. Assume <independent censoring>, with $f_T(t)=h(t)S(t)$. Then
$$
\mathbb EV=\int_0^\infty f_T(t)G(t)dt,
$$
while the <Tonelli theorem> and the <cumulative hazard function> give
$$
\mathbb EH(X)=\mathbb E\int_0^\infty h(t)\mathbf1_{\{X\geq t\}}dt
=\int_0^\infty h(t)S(t)G(t)dt=\mathbb EV.
$$
This is <mean accumulated hazard under independent censoring>, proving $\mathbb E[V-H(X)]=0$. For the fitted <martingale residual> $Y=V-\widehat H(X)$,
$$
\boxed{\mathbb EY=-\mathbb E[\widehat H(X)-H(X)],\qquad
|\mathbb EY|\leq\mathbb E|\widehat H(X)-H(X)|.}
$$
Hence its <expectation> is approximately zero when the fitted hazard error at the observed time is small in mean. This spells out the required sense of a good estimate; pointwise consistency alone does not control a divergent tail. In particular, a <Kaplan–Meier estimator> with zero final survival gives infinite transformed times there and cannot supply finite residuals without a tail restriction. Neither exact finite-sample zero mean nor validity under <informative censoring> is being assumed.
To check an omitted explanatory variable $Z$, plot the <martingale residuals> against $Z$ or compare groupwise mean residuals, using a smooth trend or appropriate uncertainty intervals. Under a correct conditional survival model and conditionally <independent censoring>, $\mathbb E[Y\mid Z]$ should be approximately zero. A systematic positive trend indicates more observed failures than the fitted accumulated hazard predicts; a negative trend indicates fewer. Such patterns suggest adding $Z$, a nonlinear term or an interaction and reassessing the fit. The bounded-above, often long negative tail of <martingale residuals> means that the diagnostic is a mean-pattern check, not a requirement for symmetric Gaussian residuals.
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