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Background

  • Let's say you own a convenience store

  • Each day $N$ customers will visit you, where $N \in (1,7 \, billion \, humans)$

  • $X_i$ is the amount that each customer will spend

  • All the $X_i$'s are iid and also independent of $N$

Let's say you want to know the expected value of total revenue made in a day. I know the following is true (Wald's equation),

$$E \bigg[ \sum_{i=1}^{N} X_i \bigg] = E[N]E[X]$$

My question

Can you explain why it is illegal to do,

$$E \bigg[ \sum_{i=1}^{N} X_i \bigg] = \sum_{i=1}^{N} E[X_i] = NE[X]$$

I know this is illegal because $NE[X] \neq E[N]E[X]$. I know the LOE doesn't always hold for infinite sums but in this situation $N$ is capped by the human population. If you could explain in lay terms I would greatly appreciate it (versus going into sigma algebras and IUT). Thanks.

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    $\begingroup$ $N$ is a random variable. $\endgroup$ – Lord Shark the Unknown Aug 31 '18 at 21:59
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    $\begingroup$ The expected value of a random variable is a real number. Your attempt at using linearity gives a random variable because $N$ is random. So you aren’t even getting the right type of object when you apply linearity this way. It’s like saying 117 * 123 = Fish; I don’t have to do any calculation to tell you that it is wrong. $\endgroup$ – guy Aug 31 '18 at 22:03
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    $\begingroup$ The problem is already in the example with the $3$s. If $N$ is a random variable, then the expected value of an $N$-fold sum of $3$s is not $3N$ but $3\mathsf E[N]$. $\endgroup$ – joriki Aug 31 '18 at 22:08
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    $\begingroup$ $N$ is a random variable: $E[X]N$ is a random variable. Expectations are numbers, not random variables. $\endgroup$ – Lord Shark the Unknown Aug 31 '18 at 22:10
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    $\begingroup$ @HJ_beginner $E[X \mid Y = y]$ is a number, yes. There are a couple of ways to be rigorous about this; one is to start by defining $E[X \mid Y = y]$ and then define $E[X \mid Y] $ to be what you get when you "plug in" $Y$ for $y$. Alternatively, you can define $E[X \mid Y]$ first as a random variable, and then let $E[X \mid Y = y]$ be the value $E[X \mid Y]$ takes on the level set $[Y = y]$. $\endgroup$ – guy Sep 1 '18 at 1:47

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