Law of iterated expectations, Binomial Dist.

Question

• In each period, $X_i$ are drawn from i.i.d. Binomial ($N_i$,$\alpha$)
• The number of trials in period $i+1$ (i.e.$N_{i+1}$) depends on the number of successes in period $i$
• In particular, $N_{i+1}$ is equal to realized $X_i$
• Further let's assume that $N_1=1$

In this stochastic process, I would like to compute the following:

$E\big[\frac{1+X_1+X_2+...+X_t}{1+N_1+...+N_t}\big]$

I would like to know whether the following calculation is correct..

$E\big[\frac{1+X_1+X_2+...+X_t}{1+N_1+...+N_t}\big]=E\big[E\big[\frac{1+X_1+X_2+...+X_t}{1+N_1+...+N_t}|N_1,...,N_t\big]\big]$

$=E\big[\frac{1}{1+N_1+...+N_t}\big]+\alpha E\big[\frac{N_1+...+N_t}{1+N_1+...+N_t}\big]$

Answer 1

This answer just summarizes my last comment:

1) It seems the $\{X_i\}$ are not independent and identically distributed (i.i.d.) as described, rather each $X_i$ is conditionally independent of the past, given the value of $N_i$.

2) If $X_1=N_2$ then we get $$ E[X_1+X_2|N_1, N_2] = E[N_2+X_2|N_1,N_2] = N_2 + \alpha N_2$$ and so \begin{align} E\left[\frac{1+X_1+X_2}{1+N_1+N_2}\right] &= E\left[\frac{1}{1+N_1+N_2}\right] + E\left[\frac{E[X_1+X_2|N_1, N_2]}{1+N_1+N_2}\right] \\ &= E\left[\frac{1}{1+N_1+N_2}\right] + E\left[\frac{N_2(1+\alpha)}{1+N_1+N_2}\right] \end{align}