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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?

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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 $\delta = 1$

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