martingale

**Kat-M** · November 29th 2009, 05:57 PM

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 $L^2$ ?

**Focus** · November 30th 2009, 12:17 AM

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 $\sup E[X_n^2]=\infty$ .

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