Given $B>A>0$ and $C>0$. Let $\{X\_t\}\_{t=0}^{\infty}$ be a submartingale with $X\_0=A$ and
\begin{equation}
\mathbb{E}[X\_{t+1} | \mathcal{F}\_t] \geq X\_t + C.
\end{equation} 


Let $ \tau := \min\{t:X\_t>B\}$. Under what condition can we upper bound $\mathbb{E}[\tau]$ roughly $\frac{B-A}{C}$?


For instance, define $$\eta\_t:=\frac{X\_t}{C}-t.$$
We see that $\eta\_t$ is a submartingale as well and $\eta\_{t\wedge \tau} \leq \frac{B}{C}$ almost surely for all $t\geq0$. Therefore, by the optional stopping theorem,
\begin{equation}
\begin{aligned}
\frac{A}{C}=\eta\_0 \leq \mathbb{E}[ \eta\_{\tau}] &= \mathbb{E}[\eta\_{\tau-1}] + \mathbb{E}[\eta\_{\tau}-\eta\_{\tau-1}]\\
&\leq \frac{B}{C}-\mathbb{E}[\tau]+1+\mathbb{E}[\eta\_{\tau}-\eta\_{\tau-1}].
\end{aligned}
\end{equation}
Hence,
$$
\mathbb{E}[\tau] \leq \frac{B-A}{C} +2+\frac{\mathbb{E}[X\_t-X\_{t-1}]}{C}
$$


So if we have conditions like $X\_t-X\_{t-1} \leq K $ or $\mathbb{E}[X\_t-X\_{t-1} \mid \mathcal{F}\_{t-1} ] \leq K, $


we have 
$$
\mathbb{E}[\tau] \leq \frac{B-A}{C} +2+\frac{K}{C}.
$$


However, the condition appears weird to me. Intuitively the larger $\mathbb{E}[X\_t-X\_{t-1}]$ is, the faster $X\_t$ shall reach or cross $B$. But the proof above requires an upper-bound on $\mathbb{E}[X\_t-X\_{t-1}]$, limiting its speed of increment. 


Is there other proof technique for getting $\mathbb{E}[\tau]$ upper bound roughly $\frac{B-A}{C}$, or this "weird" condition $\mathbb{E}[X\_t-X\_{t-1}] \leq K$ is actually necessary?



Expected time for a submartingale increasing from A to B
I don't think your argument works as stated. You don't have $\eta\_{t \wedge \tau} \le B/C$ because $X\_\tau$ could be a lot larger than $B$. That is, maybe when $X\_t$ first exceeds $B$, it jumps a long way past it. An almost sure upper bound on the increment $|X\_t - X\_{t-1}|$ can fix it, but a bound on the expectation of the increment cannot.


To see what can go wrong, let $0 < p\_1, p\_2, \dots < 1$ be a sequence to be chosen later, and let $\{\xi\_i\}$ be independent where $P(\xi\_i = 0) = 1 - p\_i$, $P(\xi\_i = 1/p\_i) = p\_i$, so that $E[\xi\_i] = 1$ for all $i$. Setting $X\_n = \xi\_1 + \dots + \xi\_n$, we have the desired conditions with $A=0$, $C=1$. Also note that $E[|X\_{n+1} - X\_n|] = 1$ for all $n$. 


Now taking $B=1$ we see that $P(\tau < \infty) = P(\bigcup\_i \{\xi\_i > 0\}) \le \sum\_i p\_i$ by union bound. So choosing $p\_i$ with $\sum\_i p\_i < 1$ we have $\tau = \infty$ with positive probability and in particular $E[\tau] = \infty$.


