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The following is from Klein and Moeschberger, p. 76.

Let $(T,\delta)$ be a tuple with $T = \min(X,C_r)$ and $\delta = 0$ if the lifetime X is censored and $\delta = 1$ if it is not; $C_r$ denotes the censoring time. I would like to compute the likelihood function for data with random censoring. Denote the pdf and cdf of $X$ by $f$ and $F$, respectively, and denote the pdf and cdf of $C_r$ by $g$ and $G$, respectively.

We calculate

\begin{equation*} \begin{split} P(T_i = t, \delta = 1) = P(X_i = t, X_i < C_{r,i} ) &= \frac{\mathrm{d}}{\mathrm{d}t} \int_{0}^{t} \int_{x}^{\infty} f(x) g(v) \, \mathrm{d}v \, \mathrm{d}x \\ &= \frac{\mathrm{d}}{\mathrm{d}t} \int_{0}^{t} f(x) - f(x)G(x) \, \mathrm{d}x = f(t) - f(t)G(t) \end{split} \end{equation*}

However, Klein and Moeschberger gives

\begin{equation*} P(X_i = t, X_i < C_{r,i}) = f(t)G(t) \end{equation*}

I feel like this is a very basic integration mistake, but I have looked at it and can't find anything wrong.

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  • $\begingroup$ 1. What are $f, g,$ and $G$? 2. Is the censoring random or is it actually deterministic, in the sense of "if and only if $x \geq c$, we observe $c$?" $\endgroup$
    – jbowman
    Commented Apr 28 at 20:34
  • $\begingroup$ I have edited the question. The functions $f$,$g$ and $G$ are the pdf of $X$ and the pdf and cdf of $C_r$, respectively. Censoring is assumed to be random. $\endgroup$
    – Montresor
    Commented Apr 28 at 20:37
  • $\begingroup$ Another approach that avoids explicit integration is to note that $P(X = t, C > t) = P(C>t)P(X=t)$, as $C$ and $X$ are independent, which in turn evidently equals $(1-G(t))f(t)$, as you found. I suspect this is just a mistake, whoever wrote that particular line saw the "$X_i < C_{r,i}$" and thought "cumulative density at...", but I've been unable to find an online errata. $\endgroup$
    – jbowman
    Commented Apr 28 at 20:57
  • $\begingroup$ After much scrutinizing, I also believe this to be a typographical mistake. $\endgroup$
    – Montresor
    Commented Apr 28 at 21:21
  • $\begingroup$ Which edition are you quoting from? This doesn't seem to agree with what's on page 76 of my copy of the second edition. $\endgroup$
    – EdM
    Commented Apr 28 at 21:22

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I think the confusion lies with $G(t)$. $G(C_r)$ is the survival function, i.e., 1-cdf of $C_r$, so

$$ P(X_i=t, X<C_{r,i})=P(X=t)P(C>t)=f(t)\underset{\text{survival function}}{G(t)} $$

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