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.

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.

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.