I am looking at branching processes in Dobrow 2016 (p. 160), where the author states that the "mean of the offspring distribution" is $\mu =\sum_{k=0}^{\infty} k a_k$. I want to know why the following equation holds:
$$\sum_{k=0}^{\infty} \mathbb{E} \Big( \sum_{i=1}^k X_i \Big) = \sum_{k=0}^{\infty} k \Big( \sum_{k=0}^{\infty} k a_k \Big).$$
How is this result obtained?
The paper specifies that $X_1,X_2,X_3,... \sim \text{IID Offspring}$ are values from the offspring distribution. It also gives the mean of the offspring distribution as:
$$\mathbb{E}(X_i) = \mu = \sum_{k=0}^\infty k a_k.$$
Hence, using the linearity of the expectation operator, you have:
$$\begin{equation} \begin{aligned} \sum_{k=0}^\infty \mathbb{E} \Big( \sum_{i=1}^k X_i \Big) &= \sum_{k=0}^\infty \sum_{i=1}^k \mathbb{E}(X_i) \\[6pt] &= \sum_{k=0}^\infty \sum_{i=1}^k \mu \\[6pt] &= \sum_{k=0}^\infty k \mu \\[6pt] &= \sum_{k=0}^\infty k \Big( \sum_{k=0}^\infty k a_k \Big). \\[6pt] \end{aligned} \end{equation}$$
