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I have some uncertainties about one of the requirements for martingale, i.e. showing that $\mathbb{E}|X_n|<\infty,n=0,1,\dots$ when $(X_n,n\geq 0)$ is some stochastic process. In particularly, in some solutions I find that lets say $\mathbb{E}|X_n|<n$ is accepted, for example here (2nd slide, 1.2 example). So my question is: what is the way of thinking if n goes to infinity? Why are we accepting n as a boundary or maybe I misunderstood something?

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The condition $\mathbb E|X_n|\lt n$ is odd. What is required for $(X_n)$ to be a martingale is, in particular, that each $X_n$ is integrable (if only to be able to consider its conditional expectation), but nothing is required about the growth of $\mathbb E|X_n|$.

Consider for example a sequence $(Z_n)$ of i.i.d. centered $\pm1$ Bernoulli random variables, and a real valued sequence $(a_n)$. Then $X_n=\sum\limits_{k=1}^na_kZ_k$ defines a martingale $(X_n)$ and $\mathbb E|X_n|\gt |a_n|$ by convexity. One sees that the growth of $n\mapsto\mathbb E|X_n|$ cannot be limited a priori.

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You misinterpreted the slide you link to: the inequality $\mathbb E|X_n|\leqslant n$ is there, only as a quick and easy argument showing that $\mathbb E|X_n|$ is finite. –  Did Mar 13 '12 at 18:11

For a definition of a martingale you only want each member $X_n$ of the sequence $\{X_n\}_{n\geq 0}$ to be integrable and you don't care about the growth of $\mathsf E|X_n|$, i.e. you only need to show that $\mathsf E|X_n|$ is finite - no matter how big is it or how fast does it grow with the growth of $n$.

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