I have the following review problem I've been working through and would appreciate any help towards solving it.

Customers enter a store according to a Poisson process of rate $\lambda$ = 5 per hour. Independently, each customer buys something with probability p = 0.8 and leaves without making a purchase with probability q = 0.2. Each customer buying something will spend an amount of money uniformly distributed between \$1 and \$101 (independently of the purchases of the other customers). What are the mean and the standard deviation of the total amount of money spent by customers within any given 10-hour day?

So far I think the mean spend over the 10 hour block is given by $\lambda t\times E[spend]$, where E[spend] = E[spend|buy]*P(buy) = 40.8 and $\lambda_{p} t = 10\times4 = 40$ ($\lambda_{p} = 0.8 \times 5 = 4$).

$\therefore$ mean spend is 40.8 $\times$ 40 = \$1631

I'm not 100% certain this is correct, and I'm also unsure on how to approach the standard deviation part of the question.


1 Answer 1


Here are formulas for the expectation and variance for a Compound Poisson Process.

Define random variables:

\begin{eqnarray*} X &=& \text{Total amount spent in one 10-hour day} \\ N(t) &=& \text{Number of customers up to time $t$} \\ Y_i &=& \text{Amount spent by $i^{th}$ customer (even if he buys nothing).} \\ \end{eqnarray*}

Now applying those formulas, we get:

\begin{eqnarray*} E(X) &=& E(N(10)) E(Y_1) \\ &=& (5\times 10) \times \left(0.8\times \dfrac{101+1}{2}\right) \\ &=& 2040. \end{eqnarray*}

\begin{eqnarray*} \sqrt{Var(X)} &=& \sqrt{E(N(10)) E(Y_1^2)} \\ &=& \sqrt{(5\times 10) \times 0.8\times \left(\dfrac{101+1}{2}\right)^2} \\ &=& 102\sqrt{10} \\ &\approx& 322.55. \end{eqnarray*}

I think what you got wrong was applying the $p=0.8$ factor to $\lambda_p$ instead of just to $E(\text{spend})$.

  • $\begingroup$ Thank you for your help :) $\endgroup$
    – Arbit
    Commented May 15, 2015 at 3:10

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