Assume we have (k+1) positive non-i.i.d. random variables $(X_0, X_1,X_2,...,X_k)$ where $k$ is a constant. Moreover, we are given $E[X_i|X_{i-1}]$ is positive for any $i\geq1$. Additionally, there is another integer random variable $\tau$ such that if $\forall i\ge\tau$, $E[X_i|X_{i-1}]=c>0$ where $c$ is a constant. Furthermore, we know $1\leq\tau<k$.
Is it possible that $$ E[\sum_{s=0}^{k}X_s] \geq \sum_{s=1}^{k}E[X_s] = \sum_{s=1}^{k}E[E[X_s|X_{s-1}]]\geq \sum_{s=\tau}^{k}E[E[X_s|X_{s-1}]] = \sum_{s=\tau}^{k}E[c]=(k-\tau+1)c$$
It is a little strange for me since this lower bound is a random variable. If it is possible, how can we interpret it? I think the first inequality is due to the positive property of $X_i$, and the second equality is due to the law of total expectation. However, the other operators are not clear to me.
The second inequality is not valid. The condition $\mathbf{E}[X_s \mid X_{s-1}] = c$ for $i \geq \tau$ holds pointwise, so has to be used when such a pointwise equality makes sense.
Rectifying this issue, you would want
\begin{align*} \mathbf{E}\left[ \sum_{s=0}^{k} X_s \right] &\geq \sum_{s=1}^{k} \mathbf{E}\left[ X_s \right] \\ &= \sum_{s=1}^{k} \mathbf{E}\left[ \mathbf{E}[X_s \mid X_{s-1}] \right] \\ &= \mathbf{E}\left[ \sum_{s=1}^{k} \mathbf{E}[X_s \mid X_{s-1}] \right] \\ &\geq \mathbf{E}\left[ \sum_{s=\tau}^{k} \mathbf{E}[X_s \mid X_{s-1}] \right] \\ &= \mathbf{E}\left[ \sum_{s=\tau}^{k} c \right] = k + 1 - \mathbf{E}[\tau]. \end{align*}