# Supermartingale question

I'm wondering if the following statement is true:

Let $$(M_n)_{n \in \mathbb{N}}$$ and $$(N_n)_{n \in \mathbb{N}}$$ are non-negative random processes, and let $$(\mathcal{F}_n)$$ be a filtration. Let $$\lambda > 0$$

The assumptions

• For each $$n$$, $$M_n$$ and $$N_n$$ are $$\mathcal{F}_n$$-measurable.

• $$(N_n)_{n \in \mathbb{N}}$$ is supermartingale with respect to the filtration $$(\mathcal{F}_n)$$ (i.e. for each $$n\geqslant 1$$ : $$\mathbb{E}[ N_n \mid\mathcal{F_{n-1}}] \leqslant N_{n-1}$$)

• for each $$n\geqslant 1$$, $$\mathbb{E}[ M_n \mid\mathcal{F_{n-1}}] \leqslant \mathbb{E}[ N_n \mid\mathcal{F_{n-1}}] \leqslant N_{n-1}$$

Do we have, $$\mathbb{P}\left[\sup_{ 0 \leqslant k \leqslant n} M_k \geqslant \lambda \right] \leqslant \frac{1}{\lambda} \mathbb{E}[ N_0 ]?$$

Or do we have, $$\mathbb{E}[M_T] \leqslant \mathbb{E}[N_0]$$ where $$T$$ is a stopping time with respect to same filtration ?

Let $$N_k=1$$ for all $$k \ge 0$$. Let $$T$$ be uniform in $$\{1,2,\dots,n\}$$ and suppose that for all $$k \ge 0$$, the $$\sigma$$-field determined by $$T \wedge (k+1)$$ is denoted by $$\mathcal F_k$$. Finally, let $$M_k=(n+1-k)\cdot{\bf 1}_{T=k} \,.$$ Then $$T$$ is a stopping time for the filtration $$\{\mathcal F_k\}$$, the variables $$M_k,N_k$$ are $$\mathcal F_k$$- measurable, and $$E[M_k | \mathcal F_{k-1}] = (n+1-k) P(T=k | \mathcal F_{k-1}) \le (n+1-k) P(T=k | T \wedge k =k)=1\,.$$
However, $$P[\sup_{ 0 \leqslant k \leqslant n} M_k \geqslant n/2] = P[T \le 1+n/2] \ge 1/2 > \frac{2}{n}\mathbb{E}[ N_0 ] \,,$$
and $$E[M_T] =E(n+1-T) \ge n/2 \,.$$
• Thanks a lot. what if, we add the assumption $M_0 \leqslant N_0$, it still false ? Nov 15, 2022 at 8:18
• In my example, we can set $M_0=1$. Nov 15, 2022 at 8:36
• Thanks again, this is my last question, (i will not add more stupid questions). what, if we add another assumption : $\mathbb{E}[M_n | \mathcal{F}_{n-1}] \leqslant C M_{n-1}$ where $C>1$ is a constant. My question de we have $\lambda \mathbb{P}\left[\sup_{ 0 \leqslant k \leqslant n} M_k \geqslant \lambda \right] \leqslant C \mathbb{E}[ N_0 ]?$ Nov 16, 2022 at 9:17
• No. Let $N_k=1$ for all $k$ as before. Make $T$ uniform in $[1,n]$. Define $M_k=2^{k-T}$ if $T<n/2$ and $k \in [T,T+\log_4 n]$, Define $M_k=1$ for all other $k$. Nov 16, 2022 at 9:33