Solution: Finding mean survival time from the survival function

We need to find \(E(T) = \int_0^{\infty}t\cdot f(t)dt\), where \(f(t)\) is the density of \(T\). Recall the definition of the survival function \(S(t) = P(T > t)\). We can express the latter probability as an integral of the density \(f(t)\), namely \(S(t) = \int_t^{\infty}f(x)dx\). Using the hint provided in the problem text we can write \begin{align*}E(T) &= \int_0^{\infty}t\cdot f(t)dt \\ &= \int_0^{\infty}\left(\int_0^{\infty}I(t>u)du\right)\cdot f(t)dt \\ &= \int_0^{\infty}\left(\int_0^{\infty}I(t>u)\cdot f(t)du\right)dt.\end{align*} Changing the order of integration, and using the definition of \(I(t > u)\) we get \begin{align*}E(T) &= \int_0^{\infty}\left(\int_0^{\infty}I(t>u)\cdot f(t)dt\right)du \\ &= \int_0^{\infty}\left(\int_0^{u}0\cdot f(t)dt+\int_u^{\infty}1\cdot f(t)dt\right)du \\ &= \int_0^{\infty}\left(\int_u^{\infty}f(t)dt\right)du,\end{align*} and finally using the expression above for \(S(t)\), or rather \(S(u),\) we obtain \[\underline{\underline{ E(T) = \int_0^{\infty}S(u)du.}}\]

Using the relation we proved above, we find the mean survival time for the exponential distribution and the Weibull distribution.

Exponential distribution

With survival function \(S(t) = e^{-\lambda t}\) we get \begin{align*} E(T) &= \int_0^{\infty}S(u)du \\ &= \int_0^{\infty}e^{-\lambda u}du\\ &= -\frac{1}{\lambda}e^{-\lambda u}\Big|_{u=0}^{\infty} \\ &= 0 - \left(-\frac{1}{\lambda}\right),\end{align*} and so the mean survival time is \[\underline{\underline{ E(T) = \frac{1}{\lambda}.}}\]

Weibull distribution

With survival function \(S(t) = \exp\left\{-\frac{bt^k}{k}\right\}\) we get \begin{align*} E(T) &= \int_0^{\infty}S(u)du \\ &= \int_0^{\infty}\exp\left\{-\frac{bu^k}{k}\right\}du.\end{align*} Now we use the substitution \(t(u)=\frac{bu^k}{k}\), so that \(u(t) = \left(\frac{k}{b}\right)^{\frac{1}{k}} t^{\frac{1}{k}}\), and so \(\frac{du}{dt} = \left(\frac{k}{b}\right)^{\frac{1}{k}} \frac{1}{k} t^{\frac{1}{k}-1}\) (note that the limits remain the same). Inserting this in the above integral yields \begin{align*} E(T) &= \int_0^{\infty}\exp\left\{-\frac{bu^k}{k}\right\}du \\ &= \int_0^{\infty}e^{-t}\left(\frac{k}{b}\right)^{\frac{1}{k}} \frac{1}{k} t^{\frac{1}{k}-1}dt \\ &= \left(\frac{k}{b}\right)^{\frac{1}{k}} \frac{1}{k} \int_0^{\infty}e^{-t} t^{\frac{1}{k}-1}dt,\end{align*} where we can recognize the last integral as a gamma function, since \(\Gamma(\alpha) = \int_0^{\infty}e^{-t} t^{\alpha-1}dt.\) Thus the mean survival time is \[\underline{\underline{ E(T) = \left(\frac{k}{b}\right)^{\frac{1}{k}} \frac{1}{k} \Gamma\left(\frac{1}{k}\right).}}\] If for example \(b=\frac{1}{4}\) and \(k=\frac{1}{2}\), we would get \begin{align*} E(T) &= \left(\frac{1/2}{1/4}\right)^{\frac{1}{1/2}} \frac{1}{1/2} \Gamma\left(\frac{1}{1/2}\right) \\ &= \left(2\right)^{2}\cdot 2\cdot \Gamma(2)\\ &= 8 \cdot 1! \\ &= 8.\end{align*}