 Solution

Put $U=H(T)$ and $V=H(C)$, so $Y=\min (U,V)$ and $0\le \mathbb{E}Y\le \mathbb{E}U=1$. Its expectation is not generally one and depends on censoring. Under independent censoring, condition on $V=v$: the exponential distribution givesWriting $\Delta =1_{T\le C}$ for the observed event indicator, it follows that $\mathbb{E}Y=\mathbb{E}\Delta$. The exponential mean imputation under independent censoring therefore gives a modified Cox–Snell residual with known mean:On censoring, the added one is the expected remaining transformed lifetime, by memorylessness of the exponential distribution. This gives a mean-one variable, not necessarily another exponential variable. Equivalently, $M=\Delta -Y=1-Y^{\ast }$ is a martingale residual with mean zero. Independent censoring, or its appropriate conditional version, is essential for these identities; arbitrary informative censoring does not justify the correction.

Fit a model using the relevant covariates, calculate ${\hat{M}}_i=v_i-{\hat{H}}_i(x_i)$, and plot these martingale residuals against a candidate explanatory variable or against an included variable's value. A smooth systematic trend away from zero can indicate an omitted effect or an unsuitable functional form, such as a nonlinear age effect. After fitting an appropriate effect, the residual trend should diminish. The equivalent plot of ${\hat{Y}}_i^{\ast }$ has mean-one reference rather than zero.