<p>I am going to use a <span class="math-container">$t$</span>-continuos scenario, and then discretization follows naturally.</p>

<p>Call <span class="math-container">$f(t)$</span> the distribution function of the variable <span class="math-container">$t$</span>. The survival function is just</p>

<p><span class="math-container">$$
S(t) = \int_t^{+\infty}f(u)~{\rm d}u \tag{1} = 1 - \int_0^t f(u)~{\rm d}u 
$$</span></p>

<p>It is pretty clear from here that </p>

<p><span class="math-container">$$
\frac{{\rm d}S}{{\rm d}t} = -f(t) \tag{2}
$$</span></p>

<p>and that <span class="math-container">$\lim_{t\to\infty}S(t) = 0$</span>. Now consider the integral</p>

<p><span class="math-container">\begin{eqnarray}
\require{cancel}
\mathbb{E}[t] &amp;=&amp; \int_0^{+\infty} t f(t) ~{\rm dt} 
\stackrel{(2)}{=} -\int_0^{+\infty} t \frac{{\rm d}S}{{\rm d}t}~{\rm d}t  \\
&amp;=&amp; -\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 \\
&amp;=&amp; -\cancelto{0}{\strut{t S}\big\rvert_{0}^{+\infty}} +  \int_0^{+\infty}S(t)~{\rm d}t 
\end{eqnarray}</span></p>

<p>So in summary</p>

<p><span class="math-container">$$
\mathbb{E}[t] = \int_0^{+\infty}S(t)~{\rm d}t \tag{3}
$$</span></p>

<p>Now you can move to discrete time and represent this in the form</p>

<p><span class="math-container">$$
\mathbb{E}[t] = \sum_{t = 0}^{+\infty} S(t) \delta 
$$</span></p>

<p>for some small number <span class="math-container">$\delta$</span>. In the paper you cite, they choose <span class="math-container">$\delta = 1$</span></p>

CC BY-SA 4.0
2019-01-15T23:25:19.993Z
2019-01-15T23:25:19.993Z
caverac
<p>In his <a href="https://faculty.wharton.upenn.edu/wp-content/uploads/2012/04/Fader_hardie_jim_07.pdf" rel="nofollow noreferrer">How To Project Customer Retention</a>, Fader computes the expected tenure of a customer according to <span class="math-container">$$E = \sum_{t=0}^{\infty}S(t)$$</span> where <span class="math-container">$S(t)$</span> is the <a href="https://en.wikipedia.org/wiki/Survival_function" rel="nofollow noreferrer">Survival Function</a>. </p>

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

<p>Can anyone shed some light on this? </p>

<microeconomics><finance><marketing>
Calculating expected tenure of a customer
user20783
# Calculating expected tenure of a customer

(Asked by: username_0 on 2019-01-15)

In his [How To Project Customer Retention](https://faculty.wharton.upenn.edu/wp-content/uploads/2012/04/Fader_hardie_jim_07.pdf), Fader computes the expected tenure of a customer according to $$E = \sum\_{t=0}^{\infty}S(t)$$ where $S(t)$ is the [Survival Function](https://en.wikipedia.org/wiki/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?

**Tags:** microeconomics, finance, marketing

---

## 1 answers

### Answer by username_1 on 2019-01-15. Score: 1

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 $\delta = 1$


---

