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 ?


1 Answer 1


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 $$E[M_T] =E(n+1-T) \ge n/2 \,.$$

  • $\begingroup$ Thanks a lot. what if, we add the assumption $M_0 \leqslant N_0$, it still false ? $\endgroup$ Nov 15, 2022 at 8:18
  • $\begingroup$ In my example, we can set $M_0=1$. $\endgroup$ Nov 15, 2022 at 8:36
  • $\begingroup$ 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 ]?$ $\endgroup$ Nov 16, 2022 at 9:17
  • $\begingroup$ 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 $k$. $\endgroup$ Nov 16, 2022 at 9:33

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