Could anyone tell me how to prove the following:
Let $(Y_k, G_k)$ be a super-martingale and it dominates $\{E (Z|G_k), G_k\}, E(|Z|)<\infty$, if $t_1$ and $t_2$ are two stopping times such that $t_1\le t_2$ almost surely then could anyone tell me how to show $E(Y_{t_2|G_{t_1})}\le Y_{t_1}$ almost surely.
is this a standard result available in some probability book? If so could you refer me one or help me to solve it? Thanks
Yes, this is a standard equivalent formulation for supermartingale. An outline goes like this.
It suffices to do the case where the stopping times have finitely many values. (Use that dominating process to pass to a limit for infinitely many values.)
Next, it suffices to do the case where $t_1$ is constant. (Decompose the space into sets $\{t_1 = s\}$.)
The main case is: $t_1 = s$ for a fixed $s$ and $t_2$ has finitely many values. We proceed by induction on the number of values of $t_2$. If $t_2$ has only one value, we are done, this is the definition of supermartingale. Suppose that $t_2$ has more than one value. Say $n$ is the largest value of $t_2$, and $m<n$ is the next-to-largest value. The set $\{t_2 = n\}$ is actually in $G_m$, since the complement is a union of sets $\{t_2 = k\}$ with $k \le m$. Now we want to use the supermartingale property to replace $t_2$ by $$ t_3 = \begin{cases} t_2,\quad&\text{if } t_2 < n,\\ m,\quad&\text{if } t_2 = n \end{cases} $$ This is done so that $t_3$ as fewer values then $t_2$.
Here is the "replace" part.
From $\mathbb E[X_n\mid G_m] \le X_m$ we get $\mathbb E[X_n -X_m \mid G_m] \le 0$. Now using $\{t_2=n\} \in G_m$, we get $\mathbb E[(X_n-X_m)\mathbf1_{\{t_2=n\}}\mid G_m] \le 0$. Compute
$$\begin{align} \mathbb E[X_{t_2} \mid G_m] &= \mathbb E[(X_n-X_m)\mathbf1_{\{t_2=n\}}+X_{t_3} \mid G_m] \\ &= \mathbb E[(X_n-X_m)\mathbf1_{\{t_2=n\}}\mid G_m]+\mathbb E[X_{t_3} \mid G_m] \\ & \le \mathbb E[X_{t_3} \mid G_m] \end{align}$$ But $s \le m$ so $G_s \subseteq G_m$ and thus $$ \mathbb E[X_{t_2} \mid G_s] = \mathbb E\big[\mathbb E[X_{t_2} \mid G_m]\mid G_s\big] \le \mathbb E\big[\mathbb E[X_{t_3} \mid G_m]\mid G_s \big] = \mathbb E[X_{t_3} \mid G_s] $$
Since $t_3$ has fewer values than $t_2$, the induction hypothesis tells us $\mathbb E[X_{t_3} \mid G_s] \le X_s$. So we get $\mathbb E[X_{t_2} \mid G_s] \le X_s$. This completes the induction.
plug
A thorough examination of the "stopping time" point of view.
Edgar, G. A.; Sucheston, Louis, Stopping times and directed processes., Encyclopedia of Mathematics and Its Applications 47. Cambridge: Cambridge University Press (ISBN 978-0-521-13508-5/pbk). xii, 428 p. (2010). ZBL1189.60074.