Firstly, I know the basic theory of conditional expectation given $\sigma$-algebra.
I'm reading a book on stochastic epidemic models. In Reed-Frost model there's a sequence of r.v's $(X_{t})$ where $X_{t+1}$ has binomial distribution with $n = X_{t}$ and $\pi = \alpha$. It's a Markov chain. The book says pretty much that it's obvious that $\mathbb{E}[X_{t+1}|X_{t}] = \alpha{X_{t}}$ and $\mathbb{E}[X_{t}|X_{0} = x_{0}] = \alpha^{t}x_{0}$. ($X_{0} = x_{0}$). Well, it's not obvious to me. Seems okay intuitively, but i would like to understand it.
I tried two approaches. First, since all $X_{t}$ take values in set $\{0,\cdots,N\}$, $\mathbb{E}[X_{t + 1}|X_{t}] = \mathbb{E}[X_{t + 1}|\sigma(X_{t})] = \mathbb{E}[X_{t + 1}|\sigma(\{X_{t} = k\}_{k})] = \sum_{k}(\frac{1}{\mathbb{P}(X_{t} = k)}\int_{\{X_{t} = k\}}X_{t + 1}d\mathbb{P})\mathbb{1}_{\{X_{t}=k\}}$ and then I guess i can use the fact that it's a M.C. to calculate that probability (right?) but I don't know how to find $\int_{\Omega}X_{t+1}\mathbb{1}(X_{t} = k)d\mathbb{P}$.
So, first question: what is that integral? how to calculate it? The set is the problem for me, not the fact that it's a Lebesgue integral.
Second approach: write $X_{t+1}$ as a random sum of iid $\text{Ber}(\alpha)$ variables.
So, second question: is there a point and how to calculate conditional expectation $\mathbb{E}[\sum_{j = 0}^{X_{t}}\chi_{i}|X_{t}]$.
Third question: how is $\mathbb{E}[X|X_{0} = x_{0}]$ related to $\mathbb{E}[X_{t}|X_{t-1}]$ and to $\mathbb{E}[X_{t}|X_{0}]$?
What oher method could/should I use?
Given that $X_{t+1}|X_t = n \sim \text{Binomial}(n, \alpha)$, and by the property of Binomial distribution, it is well known that $\mathbb{E}[X_{t+1}|X_t = n] = n\alpha$, that's why the author say that $\mathbb{E}[X_{t+1}|X_t] = \alpha X_t$ is obvious.
Once the above recursive relationship is established, the next claim follows from induction: $$ \mathbb{E}[X_t|X_0 = x_0] = \mathbb{E}[\mathbb{E}[X_t|X_{t-1}]|X_0 = x_0] = \alpha \mathbb{E}[X_{t-1}|X_0 = x_0]$$ So you repeatedly apply the double expectation to obtain the final answer.