# If M is martingale, then $E[M_n] = E[ M_0]$ for all n.

I would like to show that

If M is martingale, then $$E[M_n] = E[ M_0]$$ for all n.

What I have done is

As we know the definition of martingale, $$E[ M_{n+1} | F_n] = M_n$$

By using this definition, I have firstly shown that $$M_{n+k}$$ replace $$M_{n+1}$$.

That’s, I have $$E[ M_{n+k} | F_n] = M_n$$

By the law of total expectation, I get

$$E[M_{n+k}]= E[ M_{n+k} | F_n] = E[M_n]$$

so I assume n=0, I have

$$E[M_{0+k}]= E[ M_{0+k} | F_0] = E[M_0]$$

i.e. $$E[M_k]=E[M_0]$$

where I may accept $$k=n$$.

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I did this proof in order to show the statement in the yellow box.

But, the proof which I did seems trivial to me. I would like to learn the actual proof of the statement.

\begin{align*} {\mathbb E}[M_{n}] ={\mathbb E}[{\mathbb E}[M_{n}|{\mathcal F}_{n-1}]] ={\mathbb E}[M_{n-1}] =\cdots ={\mathbb E}[M_{0}]. \end{align*}