# Calculating expected tenure of a customer

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 $$\delta = 1$$