# martingale

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

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

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

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