# If $Y_n=\min\{M_n,7\}$ and $\{M_n\}$ is a martingale wrt ${X_n}$, show that ${Y_n}$ is a supermartingale wrt ${X_n}$

If $Y_n=\min\{M_n,7\}$ and $\{M_n\}$ is a martingale wrt ${X_n}$, show that ${Y_n}$ is a supermartingale wrt ${X_n}$

I tried doing cases for $M_n<7$ and for $M_n>7$, but I couldn't get that $E[M_{n+1}-M_n|X_{\le n}]\le 0$

## 1 Answer

The function $f(x)=\min\{x,7\}$ is concave, hence by Jensen's inequality for conditional expectation $$\mathbb{E}[Y_{n+1}|\mathcal{F}_n]=\mathbb{E}[f(X_{n+1})|\mathcal{F}_n]\leq f(\mathbb{E}[X_{n+1}|\mathcal{F}_n])=f(X_n)=Y_n$$

• I don't know what Fn stands for. – ak87 Apr 12 '16 at 2:05
• $\mathcal{F}_n$ is whatever filtration that $X_n$ is a martingale with respect to. So it looks like $\mathcal{F}_n=\sigma(X_1,\dots,X_n)$ in your case. – carmichael561 Apr 12 '16 at 2:07
• ok. I think I understand. So, "filtration" is the sequence ${X_i}$ of fair bets, and by using Jensen's inequality, $\mathbb{E}[Y_{n+1}-Y_n|\mathcal{F}_n]\le 0$ since I can put $Y_n$ inside the expectation function. Thanks - and correct me if I'm wrong please. – ak87 Apr 12 '16 at 2:37
• The filtration is the information you know at time $n$, which you could just replace by $X_{\leq n}$ for your problem. And my argument shows that $\mathbb{E}[Y_{n+1}-Y_n|\mathcal{F_n}]\leq 0$. – carmichael561 Apr 12 '16 at 2:40