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.


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: $$\begin{align*} {\rm Var}[S] &= {\rm Var}[{\rm E}[S \mid N]] + {\rm E}[{\rm Var}[S \mid N]] \\ &= {\rm Var}[N \, {\rm E}[X]] + {\rm E}[N \, {\rm Var}[X]] \\ &= {\rm E}[X]^2 {\rm Var}[N] + {\rm E}[N]{\rm Var}[X]. \end{align*}$$


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