I have a random variable $Z(t)$ (which represents the number of cells at time $t$). I know that $Z(t+\Delta t) = \sum_{k=1}^{\Delta t} Z^{k}(t)$ with $Z^{k}(t)$ all independent.
Now it's using this in a calculation involving expectations and probability generating functions that is a problem...
So I have: $$f(x, t + \Delta t) = \mathbb{E} [x^{Z(t+ \Delta t)}] \\ = \mathbb{E} [x^{\sum_{k=1}^{\Delta t} Z^{k}(t)}] \\ = \mathbb{E}[x^{Z^1(t)} \times x^{Z^2(t)} \times \cdots \times x^{Z^{\Delta t}(t)}] \\ =^{?} \mathbb{E} \prod_{k=1}^{\Delta(t)} \mathbb{E}[x^{Z^k(t)}]$$
My confusion is with the last line. I thought the expectation of a product is the product of expectations, but this is not what has happened. Someone we have managed to get an extra expectation out of the front.
Any ideas?
Edit: attached a picture which is similar where they also turn an expectation of a product in the expectation of a product of expectations. It's a bit differently laid out but the result they've used is identical to what this leads to...
Edit 2: still unsure about what proof is being used.
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$\begingroup$Summing up the comment thread. Given a deterministic product of $n$ RVs, $Y_i$, provided they are independent, one has$$\mathbb{E}\left ( \prod_1^n Y_i\right)=\prod_1^n \mathbb{E}(Y_i)$$If the number of factors itself is an RV, say $N$, and also independent of the $Y_i$ then one can use the law of total expextation to compute the mean:$$\mathbb{E}\left(\prod_1^N Y_i \right)=\mathbb{E}\mathbb{E}\left(\prod_1^n Y_i | N=n\right)$$When the $Y_i$ are IID things simplify further in the usual way before taking the second expectation over $N=n$. In the posted image it appears this is what they are doing by first conditioning on $Z(\Delta t)=K$. Comment for further clarification or any potential errors/typos!
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