I am trying to go through a past exam paper but I don't know how to deal with stopping times since we only did 2 exercises in class... I got stuck, so I would really appreciate if someone could help me. And if the final result is correct, why would that be less than or equal to 0? Thanks a lot in advance!
QUESTION: Let $(\Omega,\mathcal{F},\{\mathcal{F}\}_{n\ge 0},\mathbb{P})$ be a filtered probability space ($n\in\mathbb{N}$). Let $X:=\{X_{n}\}_{n\ge 0}$ be a non-negative submartingale with respect to $\{\mathcal{F}\}_{n\ge 0}$ such that $X_0=0$. Moreover, let $c:=\{c_n\}_{n \ge 0}$ be a non-increasing sequence of positive numbers, $\tau :=\min\{n\ge 0:c_nX_n\ge x\}$ be a stopping time and $x>0$.
Show that the following inequality holds: $$\mathbb{E}\left[c_{\tau\wedge n}X_{\tau\wedge n}-\sum_{i=1}^{\tau\wedge n}c_i\mathbb{E}(X_i-X_{i-1} \mid \mathcal{F}_{i-1})\right]\le 0$$ ATTEMPT:
Since $\tau\wedge n$ is $\mathcal{F}_n$-measurable, so is $X_{\tau\wedge n}$, which is also a submartingale with respect to $\mathcal{F}_n$ and $\mathcal{F}_{\tau\wedge n}$.
Now, $$\mathbb{E}\left[c_{\tau\wedge n}X_{\tau\wedge n}-\sum_{i=1}^{\tau\wedge n}c_i\mathbb{E}(X_i-X_{i-1}\mid\mathcal{F}_{i-1})\right]=c_{\tau\wedge n}X_{\tau\wedge n}-\mathbb{E}\left[\sum_{i=1}^{\tau\wedge n}c_i\mathbb{E}(X_i-X_{i-1}\mid\mathcal{F}_{i-1})\right]=c_{\tau\wedge n}X_{\tau\wedge n}+\sum_{i=1}^{\tau\wedge n}\mathbb{E}\left[c_iX_{i-1}\right]-\mathbb{E}\left[\sum_{i=1}^{\tau\wedge n}c_i\mathbb{E}(X_i\mid\mathcal{F}_{i-1})\right]\le c_{\tau\wedge n}X_{\tau\wedge n}+\sum_{i=1}^{\tau\wedge n}c_iX_{i-1}-\sum_{i=1}^{\tau\wedge n}c_iX_{i-1}=c_{\tau\wedge n}X_{\tau\wedge n}$$
Note, for this all you need is that $\tau$ is a stopping time, $c_n$ are non increasing, $X_n \geq 0$ and in $L^1$. $$ c_nX_n = \sum^n_{i=1} [c_i X_i - c_{i-1}X_{i-1}] \leq \sum^n_{i=1} c_i (X_i-X_{i-1}) = \sum^\infty_{i=1} 1_{i \leq n}\ c_i(X_i - X_{i-1}) . $$ Now substitute $\tau \wedge n$ for $n$. $$ c_{\tau \wedge n}X_{\tau \wedge n} \leq \sum_{i=1}^\infty 1_{i \leq \tau\wedge n} c_i(X_i - X_{i-1}) $$ and take expecations $$ Ec_{\tau \wedge n}X_{\tau \wedge n} \leq \sum_{i=1}^\infty E 1_{i \leq \tau\wedge n} c_i(X_i - X_{i-1}). $$ For each $i$, use repeated conditioning on $\mathcal{F_{i-1}}$ and the fact that $\{ i \leq \tau \wedge n \} = \{\tau \wedge n \leq i -1\}^c$. $$ Ec_{\tau \wedge n}X_{\tau \wedge n} \leq \sum_{i=1}^\infty E 1_{i \leq \tau\wedge n} c_iE[X_i - X_{i-1}\mid \mathcal{F}_{i-1}] =E\sum_{i=1}^\infty 1_{i \leq \tau\wedge n} c_iE[X_i - X_{i-1}\mid \mathcal{F}_{i-1}] $$ $$ =\sum_{i=1}^{\tau \wedge n} c_iE[X_i - X_{i-1}\mid \mathcal{F}_{i-1}] $$