Darth Vadar's Mean Survival Time

I find your lack of cookie disturbing

In this post we are interested in proving the Darth Vadar Rule:

\[E(X) = \int_{\mathbb{R}_+}S(x)dx \qquad x\in \mathbb{R}_+\]

where \(S(x) = 1 - F(x)\) is the survival function. We also want to illustrate its usage in showing the closed form expression of mean survival time \(E(t-t_0|t>t_0)\).

First, a Calculus proof of Darth Vadar Rule:

\[\begin{align} E(X) &= \int_{\mathbb{R}_+} x*f(x) dx \\ & = \lim_{t \rightarrow \infty} \int_0^t x*f(x) dx \qquad \mbox{(note } \frac{\partial}{\partial x}S(x) = -f(x) ) \\ &= \lim_{t \rightarrow \infty} \Big[ -t*S(t) + \int_0^t S(x) dx \Big] \quad \mbox{(integration by parts)} \\ &= \underline{- \lim_{t \rightarrow \infty} t*S(t)} + \int_{\mathbb{R}_+} S(x) dx \qquad \mbox{(claim } \lim_{t \rightarrow \infty} t*S(t) = 0) \\ & = \int_{\mathbb{R}_+} S(x) dx \end{align}\]

Now only need to show the claim \(\lim_{x \rightarrow \infty} x*S(x) = 0\). We can show this by showing \(0 \leq \lim_{x \rightarrow \infty} x*S(x) \leq 0\):

\[\begin{align} \lim_{x \rightarrow \infty} x*S(x) &\geq 0 \\ \lim_{x \rightarrow \infty} x*S(x) &= \lim_{x \rightarrow \infty} x*\int_x^\infty f(t) dt \\ &= \lim_{x \rightarrow \infty} \int_x^\infty x*f(t) dt \qquad \mbox{(note x is constant w.r.p. t)} \\ &\leq \lim_{x \rightarrow \infty} \int_x^\infty t*f(t) dt = 0 \end{align}\]

and conclude \(\lim_{x \rightarrow \infty} x*S(x) = 0\) by the squeeze theorem.

Now, Mean Survival Time:

Want to show:

\[E(T-t_0|T>t_0) = \int_{t_0}^\infty \frac{S_T(t)}{S_T(t_0)}dt \qquad T > t_0\]

Note if we define \(R = T-t_0\), above formula is just a straightforward application of the Darth Vadar rule, i.e.

\[E(T-t_0|T>t_0) = E(R|R>0) = E(R) = \int_{\mathbb{R}_+} S_R(r) dr\]

we only need to express \(S_R(r)\) in terms of \(S_T(t)\):

\[\begin{align} S_R(r) &= P(R\geq r|R \geq 0) \\ &= P(T-t_0 \geq r|T-t_0 \geq 0) = P(T \geq r+t_0|T \geq t_0) \\ & = \frac{P(T \geq r+t_0)}{P(T \geq t_0)}\\ & = \frac{S_T(t_0+r)}{S_T(t_0)} \end{align}\]

and plug this expression into \(\int_{\mathbb{R}_+} S_R(r) dr\):

\[\begin{align} E(T-t_0|T>t_0) &= \int_{\mathbb{R}_+} S_R(r) dr\\ &= \int _0^\infty \frac{S_T(t_0+r)}{S_T(t_0)} dr\\ & (\mbox{now define } t = r + t_0, \mbox{do change of variable} )\\ & = \int _{t_0}^\infty \frac{S_T(t)}{S_T(t_0)} dt\\ \end{align}\]