0
$\begingroup$

The number of paying customers per day has mean 400 and variance 100. The average a customer spend is 5, with variance 4. Assume that amount each customer spent is independent to number of customers a day, and also assume that the amount customers spent are independent of each other.

I know that the mean of revenue would be E[XY]=E[X]E[Y], but how do I find the variance of the revenue?

Thank you.

$\endgroup$
1
$\begingroup$

The reason why you are having problems is because you have not understood how to set up the model properly. The correct way is as follows:

Let $N$ be the random variable that counts the number of paying customers. Let $X_i$ be the amount that the $i^{\rm th}$ such customer spends, for each $i = 1, 2, \ldots, N$. Then the total amount that is spent in one day at the store is $$S = X_1 + X_2 + \cdots + X_N.$$ We wish to compute the expected value and variance of the random variable $S$. Note you cannot simply take the random variable describing a single customer's spending and multiply that by the random number of customers to describe the total expenditures; this will give you a correct mean due to the linearity of expectation, but it will give you an incorrect variance.

If $X_1, X_2, \ldots, X_N$ are IID, and $N$ is independent of all the $X_i$, call $X$ the common distribution of each of the $X_i$s.

The expected value of $S$ is given by $${\rm E}[S] = {\rm E}[{\rm E}[S \mid N]] = {\rm E}[N \, {\rm E}[X]] = {\rm E}[N]{\rm E}[X],$$ using the law of total expectation (also called iterated expectation, or double expectation). For the variance, we need the law of total variance: $$\begin{align*} {\rm Var}[S] &= {\rm Var}[{\rm E}[S \mid N]] + {\rm E}[{\rm Var}[S \mid N]] \\ &= {\rm Var}[N \, {\rm E}[X]] + {\rm E}[N \, {\rm Var}[X]] \\ &= {\rm E}[X]^2 {\rm Var}[N] + {\rm E}[N]{\rm Var}[X]. \end{align*}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy