Bounded Square-Integrable Martingale I am stumped by the following question. Suppose $(X_n, \mathbb{F}_n)_{n\ge 0}$ is a square-integrable martingale and $\sup_k |X_k| \le C$ for non-random $C$. Let $D_k = X_k - X_{k-1}$ for $k\ge 1$. Show
$$
E\left[\left(\sum^k_{j=1} D_j^2 \right)^2 \right] \le 6 C^4
$$
for any $k$.
I've tried expanding, using the fact that $D_j$ are mean-zero, uncorrelated, but I can't find a bound that doesn't grow with $k$.
 A: Since $(X_n)_{n \in \mathbb{N}}$ is a square-integrable martingale, we know from Doob's decompositon that $X_n^2- \langle X \rangle_n$ is a martingale, where $$\langle X \rangle_n := \sum_{j=1}^n \mathbb{E}(X_j^2-X_{j-1}^2 \mid \mathcal{F}_{j-1})$$ is the compensator. Because of the martingale property, we also have
$$\langle X \rangle_n = \sum_{j=1}^n \mathbb{E}((X_j-X_{j-1})^2 \mid \mathcal{F}_{j-1}) = \sum_{j=1}^n \mathbb{E}(D_j^2 \mid \mathcal{F}_{j-1}).\tag{1}$$
Since $(X_n)_n$ is bounded by $C$, it follows that
$$\mathbb{E}(\langle X \rangle_n ) = \mathbb{E}(X_n^2)-\mathbb{E}(X_0^2) \leq \mathbb{E}(X_n^2) \leq C^2 \tag{2}$$
for all $n \in \mathbb{N}$.
Now let's start estimating the left-hand side of your inequality. Clearly,
$$\mathbb{E} \left( \left[ \sum_{j=1}^n D_j^2 \right]^2 \right) = S_1+S_2 \tag{3}$$
where \begin{align*} S_1 &:= \sum_{j=1}^n \mathbb{E}(D_j^4) \\ S_2 &:= 2 \sum_{j=1}^n \sum_{k>j} \mathbb{E}(D_j^2 D_k^2). \end{align*}
Since $D_j^2 \leq 4C^2$, we see from $(1)$ and $(2)$ that
$$S_1 \leq 4C^2 \sum_{j=1}^n \mathbb{E}(D_j^2) = 4C^2 \mathbb{E}(\langle X \rangle_n) \leq 4 C^4. $$
To estimate $S_2$ we note that, by the tower property of conditional expectation,
\begin{align*} \mathbb{E}(D_j^2 D_k^2) &= \mathbb{E} \big[ \mathbb{E}(D_j^2 D_k^2 \mid \mathcal{F}_{k-1}) \big] \\ &= \mathbb{E}\big[ D_j^2 \mathbb{E}(D_k^2 \mid \mathcal{F}_{k-1}) \big] \end{align*}
for any $j<k$. Summing over $k=j+1,\ldots,n$ yields, by $(1)$,
\begin{align*} \sum_{k=j+1}^n \mathbb{E}(D_j^2 D_k^2) &= \mathbb{E} \left[ D_j^2 \sum_{k=j+1}^n \mathbb{E}(D_k^2 \mid \mathcal{F}_{k-1}) \right] \\ &\stackrel{\text{(1)}}{=} \mathbb{E}(D_j^2 (\langle X \rangle_n-\langle X \rangle_j)). \end{align*}
If we set $M_n := \langle X \rangle_n-X_n^2$, then $M$ is a martingale and we can write
\begin{align*} \sum_{k=j+1}^n \mathbb{E}(D_j^2 D_k^2) &= \mathbb{E}(D_j^2 (M_n-M_j)) + \mathbb{E}(D_j^2 (X_n^2-X_j^2)). \tag{4} \end{align*}
Since $(M_n)_{n \in \mathbb{N}}$ is a martingale, an application of the tower property yields
\begin{align*} \mathbb{E}(D_j^2 (M_n-M_j)) &= \mathbb{E} \bigg[ \mathbb{E}(D_j^2 (M_n-M_j)\mid \mathcal{F}_{j}) \bigg] \\ &= \mathbb{E} \bigg[ D_j^2 \mathbb{E}(M_n-M_j \mid \mathcal{F}_j) \bigg] = 0. \end{align*}
For the second term on the right-hand side of $(4)$ we note that
$$\mathbb{E}(D_j^2 (X_n^2-X_j^2)) \leq \mathbb{E}(D_j^2 X_n^2) \leq C^2 \mathbb{E}(D_j^2).$$
Summig over $j=1,\ldots,n$ in $(4)$, we conclude that
$$\sum_{j=1}^n \sum_{k=j+1}^n \mathbb{E}(D_j^2 D_k^2) \leq C^2 \sum_{j=1}^n \mathbb{E}(D_j^2) \stackrel{(2)}{\leq} C^4.$$
Combining all the estimates shows, by $(3)$, that
$$\mathbb{E} \left( \left[ \sum_{j=1}^n D_j^2 \right]^2 \right) \leq 6 C^4.$$
