Show that a martingale $\{X_n\}$ is bounded in $L^2$ if and only if $EX_n^2<\infty$ for each $n$ and $\sum_{n\ge1}E(X_{n+1}-X_n)^2<\infty$

Question

A martingale $\{X_n\}$ is bounded in $L^2$ by definition if $\sup\limits_nEX_n^2<\infty$. Show that a martingale $\{X_n\}$ is bounded in $L^2$ if and only if $EX_n^2<\infty$ for each $n$ and \begin{align*} \sum_{n\ge1}E(X_{n+1}-X_n)^2<\infty. \end{align*}

This question was previously asked here but I have not been able to work out the details for how either answer leads to the solution. If anyone is willing to work out those details or provide a different solution, that would be much appreciated.

Answer 1

Claim: Suppose that $(X_n)_{n\in\mathbb N}$ is a square-integrable martingale adapted to the $\sigma$-algebras $(\mathscr F_n)_{n\in\mathbb N}$. Then, for any $n\in\mathbb N$, \begin{align*} \mathbb E[(X_{n+1}-X_n)^2]=\mathbb E[X_{n+1}^2]-\mathbb E[X_n^2]. \end{align*}

Proof: It will be sufficient to show that $\mathbb E[X_nX_{n+1}]=\mathbb E[X_n^2]$; the rest is basic algebra. By the law of iterated expectations, the martingale property, and the fact that $X_n$ is $\mathscr F_n$-measurable, \begin{align*} \mathbb E[X_nX_{n+1}]&=\mathbb E\big[\mathbb E[X_nX_{n+1}|\mathscr F_n]\big]\\ &=\mathbb E\big[X_n\mathbb E[X_{n+1}|\mathscr F_n]\big]\\ &=\mathbb E[X_n\times X_n]\\ &=\mathbb E[X_n^2], \end{align*} as sought. $\enspace\blacksquare$

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It follows that for any $N\in\mathbb N$, \begin{align*} \sum_{n=1}^N\mathbb E[(X_{n+1}-X_n)^2]=\mathbb E[X_{N+1}^2]-\mathbb E[X_1^2]\tag{$\star$} \end{align*} by telescoping.

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Now suppose that $K\equiv\sup_{n\in\mathbb N}\mathbb E[X_n^2]<\infty$. Clearly, for each given $n\in\mathbb N$, we have $\mathbb E[X_n^2]\leq K<\infty$ and ($\star$) implies that \begin{align*} \sum_{n=1}^{\infty}\mathbb E[(X_{n+1}-X_n)^2]&=\limsup_{N\to\infty}\left\{\sum_{n=1}^N\mathbb E[(X_{n+1}-X_n)^2]\right\}\\ &=\limsup_{N\to\infty}\left\{\mathbb E[X_{N+1}^2]-\mathbb E[X_1^2]\right\}\\ &\leq K-\mathbb E[X_1^2]<\infty. \end{align*}

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Conversely, suppose that $\mathbb E[X_n^2]<\infty$ for each $n\in\mathbb N$ and that \begin{align*} L\equiv\sum_{n=1}^{\infty}\mathbb E[(X_{n+1}-X_n)^2]<\infty. \end{align*} Again, ($\star$) implies for every $N\in\mathbb N$ that \begin{align*} \mathbb E[X_{N+1}^2]=\sum_{n=1}^{N}\mathbb E[(X_{n+1}-X_n)^2]+\mathbb E[X_1^2]\leq L+\mathbb E[X_1^2], \end{align*} so \begin{align*} \sup_{n\in\mathbb N}\mathbb E[X_n^2]\leq L+\mathbb E[X_1^2]<\infty. \end{align*}

Answer 2

Hint: It is easy to verify from the definition of a martingale that $(X_{n+1}-X_n)$ is an orthogonal sequence. The result now follows from general Hilbert space Theory: $\sum\limits_{n=1}^{N}\|X_{n+1}-X_n\|^{2}=\|x_N\|^{2}-\|x_1\|^{2}$ which is bounded iff $(\|x_n\|)$ is bounded.

[For example, to show that $E(X_{n+2}-X_{n+1})(X_{n+1}-X_n)=0$ condition on $X_{n+1}$, pull out $X_{n+1}-X_n$ and observe that $(EX_{n+2}-X_{n+1})=0$].