Right-continuity of the expectation of a supermartingale

Question

I start with a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t),P)$. I assume that the filtration is right-continuous. On this probability space I define a supermartingale $M$. Now suppose that I can find another process $N$ such that $N$ is a cadlag modification of $M$. Hence it is also a supermartingale. Does this imply that the mapping $t\mapsto E[M_t]$ is right-continuous?

I reasoned as follows. I consider an arbitrary sequence of rationals $t_n \downarrow t$ for some $t\geq 0$. I need to show $\lim_{t_n \downarrow t} E[M_{t_n}-M_t] = 0$.

$$\lim_{t_n \downarrow t} E[M_{t_n}-M_t] = \lim_{t_n \downarrow t} E[M_{t_n}-M_t+N_t-N_t] $$ It feels very tempting to say $\lim_{t_n \downarrow t} E[M_{t_n}-N_t] = 0$ because of something like Levy-Doob downward convergence theorem. Because the $$\lim_{t_n \downarrow t} E[M_{t_n}-M_t] = \lim_{t_n \downarrow t} E[-M_t+N_t] = 0 $$ by the fact that $M$ and $N$ are modifications.

Can someone tell me whether what I am saying is sensible and if possible, what I would need to do to make the argument rigorous?

Answer 1

Because $N$ is a modification of $M$, $E(M_t)=E(N_t)$ for all $t\ge 0$. What do you know about the right continuity of $t\mapsto E(N_t)$?

Answer 2

I would rather go for this decomposition $M_{t_n}-M_t=M_{t_n}-N_{t_n}+N_{t_n}-N_t+N_t-M_t$. Then as $E[M_{t_n}-N_{t_n}]$ and $E[M_{t}-N_{t}]$ are null by the modification hypothesis, we only have to prove the result for the process $N$, i.e. :

$lim_{n\to\infty} E[N_{t_n}-N_t]=0$

But $N_t$ is a super martingale so that $N_t\geq E[N_{t_n}|\mathcal{F}_t]$ and then $E[N_t]\geq E[N_{t_n}]=E[E[N_{t_n}|\mathcal{F}_t]]$, also note that $N_{t_n}\to N_t$ as $N$ is càd implies that $E[N_{t_n}|\mathcal{F}_t]\to N_t$ almost surely, so we we can apply monotone convergence theorem to $E[N_{t_n}|\mathcal{F}_t]$ to get the result.

best regards.