I am trying to prove the Riesz decomposition of a supermartingale as an exercise out of Brownian Motion and Stochastic Calculus from Karatzas and Shreve.
3.18 Exercise. Let $(\mathfrak{F}_t)_{0\leq t<\infty}$ satisfy the usual conditions (right continuous, $\mathfrak{F}_0$ contains all the $\mathbb{P}$ null sets). Then every right-continuous, uniformly integrable supermartingale $(X_t,\mathfrak{F}_t)_{0\leq t<\infty}$ admits the Riesz decomposition $X_t = M_t + Z_t$, $\mathbb{P}$-as, as the sum of a right-continuous, uniformly integrable martinagle $(M_t, \mathfrak{F}_t)_{0\leq t<\infty}$ and a potential $(Z_t, \mathfrak{F}_t)_{0\leq t<\infty}$.
I tried to definie $Z_t = X_t^+ - A_t$ where $A_t$ should be defined so that $Z_t$ becomes a potential. I couldn't definie $A_t$ appropriate so I looked up a solution in the internet, as there is no solution of this exercise available in the book. I found the proof of a similar theorem in Kazamaki : Remarks on the Riesz decomposition for supermartingales (can be found here). In the proof of this theorem they define $M_t=\lim_{n\to\infty}\mathbb{E}[X_n|\mathfrak{F}_t]$ and claim that this is a martingale. They use the monotone convergence theorem for conditional expectation to prove this.
- Could it be possible to instead define $Z_t$ in a similar fashion as I did and continue the proof like this?
I tried to prove myself that the monotonicity of the conditional expectations is true. I found this where somebody discusses the monotonicity of a sequence of conditional expectations of a submartingales in the discrete case. In the answer it is written that one simply uses the (sub) supermartingale property and the monotonicity of the conditional expectation. I think an inductive argument is needed to build an inequality chain, because we want for all $N$: $m\geq n\geq N$ that $\mathbb{E}[X_m|\mathfrak{F}_N]\leq \mathbb{E}[X_n|\mathfrak{F}_N]$($\leq \mathbb{E}[X_N|\mathfrak{F}_N]=X_N$).
- Do I really need an inductive argument in the time-discrete case?
Now I need to generalise to the time-continuous case. Luckily it is enough to pass from $N\in\mathbb{N}$ to $t\in[0;\infty)$. I have that $Y_n^t = \mathbb{E}[X_n|\mathfrak{F}_t]\leq X_t$ by the supermartingale property. Then since $(Y_m^t,\mathfrak{F}_t)_{m\geq n}$ is itself a supermartingale and using the discrete result I have that $m\geq n\implies Y_m^t\leq Y_n^t$($\leq X_t$).
Is the proof of the statement that the sequence $(Y_n^t)_{n\geq t} = (\mathbb{E}[X_n|\mathfrak{F}_t])_{n\geq t}$ is nonincreasing valid?
3.1 Is it possible to prove the statement above in an easier way in the sense that the result for the time-discrete case is not needed?
3.2 Is it possible to prove an analogous statment in the continuous case; i.e. can we show that for all $t_1\geq t_2\geq s\geq 0$ we have that: $Y_{t_1}^s\leq Y_{t_2}^s$($\leq Y_s^s=X_s$)?
In my proof I didn't use any requirements of the $\sigma$-algebras $(\mathfrak{F}_t)_{0\leq t<\infty}$ (e.g. I didn't use that the collection satisfies the usual conditions). Did I miss something w.r.t that?
Concerning the 'the' in the Exercise of Karatzas, Shreve, do we really have uniqueness of such a composition (up to distinguishability)? Am I supposed to prove also uniqueness in this exercise?
5.1 Is the uniqueness connected to the different requirement that $(X_t)_t$,$(M_t)_t$ are uniformly integrable?
5.2 In Kazamaki it is required the different requirement that $-\infty <\inf_t{\mathbb{E}[X_t]}$. In what sense are the requirements of the beforementioned requirement and uniform integrability connected?