# Martingale Difference Concentration for bounded moments

Let $$X_1,\dots X_n$$ be real random variables such that \begin{align} E[X_i\mid X_{i-1}, \dots, X_1] &= 0\\ \text{and}\\ E[|X_i|^p\mid X_{i-1}, \dots, X_1] &\le \sqrt{\nu_i p}^{p} \end{align} for some values $$\nu_i\ge0$$ and all $$2\le p\le q$$ for some $$q\ge2$$.

We would like to show \begin{align} E\big[\big(\sum_i X_i\big)^p\big]^{1/p} \le K \sqrt{\big(\sum_i \nu_i\big)\, p} \end{align} for $$p\le q$$ and some universal constant $$K$$ not dependent of $$n$$.

Note that this follows easily when the moments above are defined for all $$p$$, since we can then bound the moment generating function. See e.g. for sub-gaussians and sub-exponentials.

Other classic Martingale inequalities include bounded moments and Martingale-Bernstein, however these are both less tight than the linked moment-generating approaches and what we're looking for.

Unfortunately, only being able to bound the first moments, we don't necessarily have the moment generating function, and thus have to prove the inequalities some other way.

Question: Do you know of a proof of sub-Gaussian Martingale concentration that doesn't rely on the moment generating function?

We introduce independent copies of the $$X_i$$ named $$Y_i$$, such that
1. $$E[Y_i \mid X_{i-1}, \dots, X_1] = E[X_i \mid X_{i-1}, \dots, X_1]$$.
2. The $$Y_i$$ are conditionally independent given $$X_{n}, \dots, X_1$$.
3. $$E[Y_i \mid X_{i-1}, \dots, X_1] = E[Y_i \mid X_{n}, \dots, X_1]$$.
The theorem then states that $$\|\sum_i X_i\|_p \lesssim \|\sum_i Y_i\|_p$$, and we can bound
\begin{align} \big\|\sum_i Y_i\big\|_p &= E\big[E\big[\big|\sum_i Y_i\big|^p \mid X_{n}, \dots, X_1\big]\big]^{1/p} \\&\le E\left[K^p \sqrt{\big(\sum_i \nu_i\big)p}^p\right]^{1/p} \\&= K \sqrt{\big(\sum_i \nu_i\big)p} \end{align} as we wanted. Note that this just used the normal bound on a sum of fully independent sub-gaussian variables.