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I am reading Stochastic Calculus by Shreve and am a bit confused by the notation when he first introduces a Martingale with the definition: $E_n(X_{n+1})=X_n $ What I don't understand is why the $X_n$ is capitalized. I thought that when we refer to a specific value a random variable takes we would write $x$ as opposed to $X$. Doesn't the $X_n$ here refer to a known value at time $n$?

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    $\begingroup$ Usually capital letters are used to refer to the random variables, and non-capitalized the possible specific values it can take. So, for example for a positive random variable $E[X] = \int_0^\infty x p(x)$ where $p(x)$ is the probability density of $X$. And $X_n$ indeed refers to a known unique value (today's value) of the random variable. $X_{n+1}$ refers to the future (unknown) values of the random variable. $\endgroup$
    – user34971
    Oct 12, 2020 at 12:24

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You've got to make clear for yourself what the notation here means. The operator $\mathbb{E}_{n}$ is an abbreviation for a conditional expectation, given some sigma algebra say $\mathcal{F}_{n}$ of a filtration $\lbrace \mathcal{F}_{n}\rbrace_{n\ge 1}$, i.e. \begin{equation} \mathbb{E}_{n}[X]:=\mathbb{E}[X|\mathcal{F}_{n}]. \end{equation} And this guy is not deterministic but random! Namely it's defined to be the random variable which integrates against all $\mathcal{F}_{n}$-measurable random variables in the same way as $X$ does. Now $\mathbb{E}_{n}[X_{n+1}]$ being random, you should be less surprised about $X_{n}$ (which is random) being capitalized on the right hand side of your equation. It is an identity between two random variables (which btw therefore is only asked to be true $\mathbb{P}$-almost surely).

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