In his How To Project Customer Retention, Fader computes the expected tenure of a customer according to $$E = \sum_{t=0}^{\infty}S(t)$$ where $S(t)$ is the Survival Function.

From a purely mathematical standpoint, expectation is calculated via $$E[X] = \sum_{x \in X} xP(X=x).$$ Hence, I would expect expected tenure formula to be $$E = \sum_{t=0}^{\infty}tS(t).$$

Can anyone shed some light on this?


I am going to use a $t$-continuos scenario, and then discretization follows naturally.

Call $f(t)$ the distribution function of the variable $t$. The survival function is just

$$ S(t) = \int_t^{+\infty}f(u)~{\rm d}u \tag{1} = 1 - \int_0^t f(u)~{\rm d}u $$

It is pretty clear from here that

$$ \frac{{\rm d}S}{{\rm d}t} = -f(t) \tag{2} $$

and that $\lim_{t\to\infty}S(t) = 0$. Now consider the integral

\begin{eqnarray} \require{cancel} \mathbb{E}[t] &=& \int_0^{+\infty} t f(t) ~{\rm dt} \stackrel{(2)}{=} -\int_0^{+\infty} t \frac{{\rm d}S}{{\rm d}t}~{\rm d}t \\ &=& -\int_0^{+\infty} \left[\frac{{\rm d}(t S)}{{\rm d}t} - \cancelto{1}{\frac{{\rm d}t}{{\rm d}t}} S\right] {\rm d}t \\ &=& -\cancelto{0}{\strut{t S}\big\rvert_{0}^{+\infty}} + \int_0^{+\infty}S(t)~{\rm d}t \end{eqnarray}

So in summary

$$ \mathbb{E}[t] = \int_0^{+\infty}S(t)~{\rm d}t \tag{3} $$

Now you can move to discrete time and represent this in the form

$$ \mathbb{E}[t] = \sum_{t = 0}^{+\infty} S(t) \delta $$

for some small number $\delta$. In the paper you cite, they choose $\delta = 1$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.