Number of trials depends on the number of successes in the previous period 
*

*In each period $i$, $X_i$ are drawn from $\mathrm{Binomial}(N_i\, ,\,\alpha)$

*The number of trials in period $i+1$ (i.e.$N_{i+1}$) depends on the number of successes in period $i$

*In particular, $N_{i+1}=X_i$

*Further, $N_1=n$
In this stochastic process of $X_i$, I would like to prove the following two statements
\begin{align}
\mathrm{Eq}.1& & E\left[\frac{1+X_1+X_2+...+X_t}{1+N_1+...+N_t}\right] &> \alpha \\
\mathrm{Eq}.2& & E\left[\frac{1+X_1+X_2+...+X_t}{1+N_1+...+N_t}\right] &> E\left[\frac{1+X_1+X_2+...+X_{t+1}}{1+N_1+...+N_{t+1}}\right]
\end{align}
Intuitively, the above two statements should hold. But I have a difficult time proving this mathematically.
 A: As for (Eq.1), consider the case $n = 1$ and $t = 3$. Then we can manually compute the expectation as
\begin{align*}
\mathbf{E}\left[\frac{1+X_1+X_2+X_3}{1+1+X_1+X_2} \right]
&= \frac{3\alpha^3 + \alpha^2 + 2\alpha + 6}{12},
\end{align*}
which is in fact smaller than $\alpha$ if $\alpha > \frac{\sqrt{22}-2}{3} \approx 0.896805$. And this is in fact not restricted to small $n$. Indeed, we can check that $S_{\infty} := \sum_{t=1}^{\infty} N_t$ converges and has the same distribution as the sum of $n$ independent RVs having geometric distribution with parameter $1-\alpha$. Then
$$ \lim_{t\to\infty} \mathbf{E}\left[ \frac{1+X_1+\cdots+X_t}{1+N_1+\cdots+N_t} \right] = \mathbf{E}\left[1 - \frac{n}{1+S_{\infty}}\right]. \tag{*}$$
This can be computed in terms of the hypergeometric function, and its the  difference between $\text{(*)}$ and $\alpha$ can be plotted as follows:

So, if $\alpha$ is close to $1$ and $t$ is large, then the left-hand side of (Eq.1) will be actually smaller than $\alpha$.

As for (Eq.2), numerical simulation seems to support the validity of it, although I have no nice idea to attack this problem.
