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

143 Views Asked by At

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 $k=n$.

—————

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.

Thank you for your helps.

1

There are 1 best solutions below

0
On BEST ANSWER

\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*}