Can't see how implication on definition of Martingale was arrived at

Question

A Martingale is a discrete time stochastic process $Z_1, Z_2, ..., Z_n$ for any time $n$ that satisfies

$E[|Z_n|] < \infty$

$E[Z_{n+1}| Z_0, Z_1, ..., Z_n] = Z_n$

By the linearity of expectation the second statement is equivalent to

$E[Z_{n+1} - Z_n| Z_0, Z_1, ..., Z_n] = 0$

I understand linearity of expectation and how it lets us bring values "inside and outside " expected value functions but I don't see how that conclusion was arrived at...

$E[Z_{n+1}| Z_0, Z_1, ..., Z_n] = Z_n$

$\implies E[Z_{n+1}| Z_0, Z_1, ..., Z_n] - Z_n = 0$

Where do I go from here?

Answer 1

The random variable $\color{red}{Z_n}$ is measurable with respect to the $\sigma$-algebra $\sigma(Z_0,\dots,Z_{n-1},\color{red}{Z_n})$.

If $\mathcal G$ is a $\sigma$-algebra and $X$ is $\mathcal G$-measurable then $\mathbb E[X\mid\mathcal G]=X$.

Answer 2

You simply have that $$Z_n=E[Z_n|Z_n]$$ or in other words, given that at time $n$ you know $Z_n$ then the expected value of the random variable $Z_n$ will be $Z_n$. Note that the expectation on the right hand side is a random variable and not a number!

Obviously more knowledge does not change that so that also $$Z_n=E[Z_n|Z_0,Z_1,\ldots,Z_n]$$ The best (and most formal) explanation is already given (in another answer) and states that $$E[X|\mathcal G]=X$$ if $\mathcal G$ is $σ-$algebra $X$ is $\mathcal G$-miserable.