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I am doing survival analysis and was wondering if there was a name for the following expression:

$$H(t)=\int_{0}^{\infty}{g(t) \cdot S(t) \cdot dt}.$$

It appears to be very similar to the equation for computing the expectation w.r.t. a pdf only with the complementary cdf instead of the pdf.

whuber
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statkun
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    Integrating by parts will re-express this as an integral against the probability element $f(t)\mathrm dt = -\mathrm dS(t).$ Presumably, your variable has non-negative support. – whuber Dec 19 '22 at 20:01

1 Answers1

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Define

$$G(t) = \int_0^t g(x)\,\mathrm dx.$$

Assuming $\lim_{t\to\infty} G(t)S(t)=0$ and writing $f(t) = -S^\prime(t)$ for the density function, integration by parts gives

$$H = \int_0^\infty g(t)S(t)\,\mathrm dt = G(0)S(0) + \int_0^\infty G(t) f(t)\,\mathrm dt.$$

(Notice $H$ is a number, not a function.)

If $X$ is a random variable with survival function $S(t) = \Pr(X \gt t),$ the right hand integral is related to the conditional expectation of $G(X)$ given that $X \ge 0$ because

$$E[G(X)\mid X\ge 0] = \frac{1}{1-S(0)} \int_0^\infty G(t)f(t)\,\mathrm dt = \frac{H - G(0)S(0)}{1-S(0)}.$$

In survival analysis, $X$ usually has non-negative support and (most often) $S(0)=0.$ In that case $H = E[G(X)]$ is the expectation of the integral of $g.$

The commonest practical application of this observation is to the case $g(t)=1,$ giving $G(X)=X.$ In that case $H = E[X]$ is a convenient expression for the expectation of a non-negative random variable.

The analysis at https://stats.stackexchange.com/a/222497/919 generalizes this result to any random variable, non-negative or not.

whuber
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