Show that $\sup_n(|X_n|) < \infty$

martingalesprobability theory

Suppose that $X_1,X_2,….$ is a martingale satisfying $E[X_1]=0$ and $E[X_n^2] <\infty$. Assume that $\sum_n E[(X_n-X_{n-1})^2] < \infty$. Prove that $X_n$ converges with probability $1$.

I am using the martingale convergence theorem to solve this problem and it is stated as:

Let $X_1,X_2,…$ be a sub martingale. If $K=\sup_n E(|X_n|) <
\infty$
, then $X_n \rightarrow X$ with probability $1$.

I am trying to show that $\sup_n E(|X_n|) < \infty$. In this case,
\begin{eqnarray*}
X_n &=& X_n – X_{n-1}+X_{n-1}+……+X_2-X_1+X_1\\
|X_n| & \leq& \sum_{I=1}^n |X_i-X_{i-1}|+|X_1|
\end{eqnarray*}

I am confused here, how to use $\sum_n E[(X_n-X_{n-1})^2] < \infty$ to show that $\sup_n E(|X_n|) < \infty$.

Anyone can suggest some direction to prove this condition?

Best Answer

A stronger result is true: $\sup_n EX_n^{2}<\infty$. To see this note that for $n \geq m$ we have $E(X_{n+1}-X_n)(X_m-X_{m-1})=E[E(X_{n+1}-X_n)(X_m-X_{m-1})|\mathcal F_n)=0$ by martingale property. Let $Y_n=X_n-X_{n-1}$. Then $(Y_n)$ is an orthogonal sequence in $L^{2}$ and $\sum \|Y_n\|^{2} <\infty$. [The norm here is the $L^{2}$ norm]. This implies that the series $\sum Y_n$ converges in $L^{2}$. In particular, $\|Y_1+Y_1+\cdots+Y_n\|$ is bounded and this shows that $\sup_n E|X_n|^{2} <\infty$.

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