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Let $X_1,X_2,\dotsc$ be a sequence of a.s. bounded, zero-mean random variables. For $\alpha \in (0,1)$ define $Z_t$ as the geometric series with $Z_t = \sum_{i=1}^t\alpha^{t-i}X_i$ and $\mathcal{F}_k = \sigma (X_1,..,X_k)$ to be the natural filtration.

I wanted to apply Azuma's inequality, but the process $\{Z_t\}$ does not seem to be a martingale:

\begin{align} \mathbb{E}[Z_t \mid \mathcal F_{t-1}] &= \sum_{i=1}^t\alpha^{t-i} \mathbb{E}[X_i \mid \mathcal{F}_{t-1}] \\ &= \mathbb{E}[X_t \mid \mathcal{F}_{t-1}] + \alpha \sum_{i=1}^{t-1}\alpha^{t-i-1} \mathbb{E}[X_i \mid \mathcal{F}_{t-1}] \\ &= \mathbb{E}[X_t \mid \mathcal{F_{t-1}}] + \alpha \mathbb{E}[Z_{t-1} \mid \mathcal{F}_{t-1}] \\ &= 0 + \alpha Z_{t-1} \end{align} and hence Azuma's inequality cannot be applied.

However, as in my previous question, for any fixed $t$, the random variables $Y_k = \alpha^{t-k}X_k$ are independent and \begin{align} Z_t & = \sum_{k=1}^t Y_k. \end{align}

It is now possible to use Hoeffding's inequality to bound $Z_t$ for any $t$.

Why can Hoeffding's inequality be applied, but not Azuma's, while the first is a special case of Azuma's inequality?

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For $i\leqslant t-1$, we have $\mathbb E\left[X_i\mid \mathcal F_{t-1}\right]=X_i$, hence the sequence $(Z'_t)_{t\geqslant 1}$, where $Z'_t=\sum_{i=1}^t\alpha^{-i}X_i$ is a martingale with respect to the filtration $(\mathcal F_t)_{t\geqslant 1}$ as long as $\mathbb E\left[X_t\mid\mathcal F_{t-1}\right]=0$ (which is seemed to be assumed in the first part of the opening post). Then you may apply Azuma's inequality to this martingale.

Indeed, $(Z_t)_{t \geqslant 1 }$ is not a martingale for $( \mathcal F_t)_ {t \geqslant 1 }$.

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  • $\begingroup$ Thx! I don't see where your implication "the sequence $(Z_t)$ is a martingale" comes from. $\endgroup$ – Manuel Schmidt Apr 16 '15 at 5:09
  • $\begingroup$ It was actually not quite correct. I've edited. $\endgroup$ – Davide Giraudo Apr 16 '15 at 16:32
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You do have a martingale!

\begin{align*} \mathbb{E} [Z_t | \mathcal{F}_{t-1}] &= \mathbb{E} [X_t | \mathcal{F}_{t-1}] + \mathbb{E}[Z_{t-1}|\mathcal{F}_{t-1}] \\ &= 0 + Z_{t-1} \end{align*}

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