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I am trying to find the expected value of $X$, where $X$ is the number of orders a customer will make in a lifetime.

Assuming that there is a $p=.1$ chance of the customer placing an initial order, and then (given that the customer places that initial order) a $p=.9$ chance that the customer places an order after that (and each additional time the customer places an order after that as well. NB it is dependent — i.e. if the customer doesn't place an order, then he won't place any additional orders).

So I have it set up like this:


$X = \#$ of orders a customer places
$O = \#$ of orders a customer places after the first one

$E[X] = 0.1(E[O] + 1) + 0.9(0) = 0.1 + 0.1 \cdot E[O]$

$P(O = 0)$ = $0.1$
$P(O = n)$ = $0.1 \cdot 0.9^n$ for $n > 0$

So, $$ E[O] = \sum_{k=0}^\infty k \cdot P(O=k) = 0.1 \sum_{k=0}^\infty k \cdot 0.9^k = 9 \>, $$ leaving us with $E[X] = 0.1 \cdot (9 + 1) = 1$.


EDITED: using a shifted geometric distribution with $p=0.1$ for $O$

Is that correct? If not, where am I going wrong?

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    $\begingroup$ How do you know that someone is a customer if they don't place an initial order? $\endgroup$
    – Placidia
    Dec 31 '12 at 21:34
  • $\begingroup$ Just assume that out of a population of X's, 10% of them will place an initial order. Does that answer your question? (in the actual scenario, we're assuming a customer has already made an order, order 0, and now there's a .1 chance he places a second order, and a .9 chance that if he places a second, he will place a third, and a .9 chance that if he places a third, he will place a fourth…etc) $\endgroup$ Dec 31 '12 at 21:35
  • $\begingroup$ OK. I think my answer below works for that scenario. The expected number does not include order 0 of course. $\endgroup$
    – Placidia
    Dec 31 '12 at 21:44
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This looks like a shifted geometric distribution with an initial coin toss. i.e. with probability 0.9, there are no purchases, and with probability 0.1, there is a purchase, with the possibility of a geometric number of further purchases. You are looking at the number of failures before the first success (i.e. refusal to purchase) - which is anything from 0 up.

$E(X)=0.9 \cdot 0 + 0.1 \cdot (1+9)=1$

The expectation of a shifted geometric $p$ is $(1-p)/p$, where $p=0.1$ in this case. That's the probability that a person ceases to be a customer. Once they're in the system, you expect 9 purchases (after the initial one.)

Thanks to Cardinal for a helpful comment. So yes, you got it right!

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  • $\begingroup$ Why do you expect 10 purchases? I expect 90 purchases (E[O], as written above). We agree on the expression for E[X], just not E[O]. Is mine incorrect above? $\endgroup$ Dec 31 '12 at 22:24
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    $\begingroup$ The geometric probability $p$ is the probability of a "success" and you are looking for the number of purchases until the first decision to purchase no more - so $p$ is the probability of not buying - which is 0.1, if I understood correctly. $\endgroup$
    – Placidia
    Dec 31 '12 at 23:21
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    $\begingroup$ Regarding your answer: Conditional on the first success, do you not want the number of failures (purchases) before the first success (no purchase)? That is the geometric on the set $\{0,1,\ldots\}$ not $\{1,2,\ldots\}$ and so the expected value of this r.v. would be $\mu = 0.9/0.1 = 9$, yielding $\mathbb E X = 1$ as @Isaac has calculated in his question. $\endgroup$
    – cardinal
    Jan 1 '13 at 0:00
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    $\begingroup$ Yes. That's correct. It is the shifted geometric. I'll fix my answer. $\endgroup$
    – Placidia
    Jan 1 '13 at 0:31
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    $\begingroup$ (+1) This method of proof, in particular, is one of my favorites and accords much more closely with one's intuition than other approaches. Even for finding the mean of the standard geometric distribution, this sort of "one-step" analysis allows for a quick proof that can easily be made rigorous and avoids the need of manipulating infinite series or generating functions. $\endgroup$
    – cardinal
    Jan 1 '13 at 16:04

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