# martingale

• Nov 29th 2009, 05:57 PM
Kat-M
martingale
a Martingale $\displaystyle X=(X_n)_{n \geq 0}$ is bounded in $\displaystyle L^2$ if $\displaystyle sup_n E(X_n ^2) \leq \infty$.
Let $\displaystyle X$ be a martingale with $\displaystyle X_n \in L^2$ for each $\displaystyle n$. Show that $\displaystyle X$is bounded in $\displaystyle L^2$ $\displaystyle iff$ $\displaystyle \Sigma_{n \geq 0} ^{\infty} E((X_n-X_{n-1})^2) < \infty$.

here is what i want to know. If we know that $\displaystyle X$ be a martingale with $\displaystyle X_n \in L^2$ for each $\displaystyle n$, it means that $\displaystyle E(X_n ^2) < \infty$ for each $\displaystyle n$, doesnt it? so doesn't this already imply that $\displaystyle X$is bounded in $\displaystyle L^2$?
• Nov 30th 2009, 12:17 AM
Focus
Quote:

Originally Posted by Kat-M
here is what i want to know. If we know that $\displaystyle X$ be a martingale with $\displaystyle X_n \in L^2$ for each $\displaystyle n$, it means that $\displaystyle E(X_n ^2) < \infty$ for each $\displaystyle n$, doesnt it? so doesn't this already imply that $\displaystyle X$is bounded in $\displaystyle L^2$?

What if $\displaystyle \mathbb{E}[X_n^2]=n<\infty$? Then $\displaystyle \sup E[X_n^2]=\infty$.