Recently I was working on this probability problem:
In a factory, the number of accidents that happen every week is a random variable (say $N$), that has a mean of $\mu_{N}$ and a variance of $\sigma^2_{N}$. The number of the workers that are injured in each accident is also a random variable with mean $\mu_X$ and variance $\sigma^2_X$. If all these random variables are independent of each other, what is the expected value and the variance of "the number of workers that get injured in a week".
This is one approach to solve this problem:
Take $X_i$ as the number of workers that get injured in the $i$th accident during a week. then what we want will be $E(\sum_{i=1}^{N}X_i)$ and $Var(\sum_{i=1}^{N}X_i)$. We can compute these two using conditional expectation and variance:
$$E(\sum_{i=1}^{N}X_i)=E_N(E(\sum_{i=1}^{N} X_i|N))=E_N(N\mu_x)=\mu_N\mu_X$$
$$Var(\sum_{i=1}^{N}X_i)=Var_N(E(\sum_{i=1}^{N} X_i|N))+E_N(Var(\sum_{i=1}^{N} X_i|N))=\mu_X^2\sigma^2_N+\mu_N\sigma^2_X$$
But one might think of another approach to solve this problem. Let $X$ be the number of injuries in each accident. The number of injuries in a week can be expressed as $Y=NX$. Then, because of independence we can write:
$$E(Y)=E(NX)=E(N)E(X)=\mu_N\mu_X$$
Well, everything is good till now, but:
$$E(X^2)=Var(X)+E^2(X)=\sigma^2_X+\mu^2_X$$ $$E(N^2)=Var(N)+E^2(N)=\sigma^2_N+\mu^2_N$$ $$\Rightarrow Var(Y)=E(Y^2)-E^2(Y)=E(N^2X^2)-E^2(NX)=E(N^2)E(X^2)-E^2(N)E^2(X)$$ $$\Rightarrow Var(Y)=(\sigma^2_N+\mu^2_N)(\sigma^2_X+\mu^2_X) - \mu^2_N\mu^2_X=\sigma^2_N\sigma^2_X+\sigma^2_N\mu^2_X+\sigma^2_X\mu^2_N$$
And this result is not the same with the result of the previous approach. Which approach is the wrong one? And why?
Let $Y_1 := \sum_{i=1}^N X_i$ be the "sum" approach, and $Y_2:=NX$ be the "product" approach.
The "sum" approach is more accurate because you're interested in "the number of workers that get injured in a week", which is the sum of workers that are injured in each accident.
It is possible to use a product-based approach like $Y_3:=N\bar{X}$, but then you need to use the average, $\bar{X}=\frac{1}{N}\sum_{i=1}^N X_i$, and now you're back to using the "sum" approach.
Variance difference
It's quite natural that the "product" approach gives a higher variance than the "sum" approach.
In the "sum" approach, you treat each accident as independent of a different accident. However, in the "product" approach, you assume that each accident is equal to a different accident.
In the "sum" approach, intuitively, the variance is lower than that of the "product" approach, because a higher casualty rate in one accident might be offset by a lower casualty rate in a different accident. You see this straightforwardly in the example that, for i.i.d. $X_1, X_2$, we have $$Var(X_1+X_2) = 2Var(X_1) \le 4Var(X_1) = Var(2X_1) = Var(X_1+X_1).$$