Let $X_i, i \geq 1$ be a $(\mathcal{F}_i)_{i \geq 1}$ measurable sequence of Random variables.
It is a martingale if
$$E[X_{i+1} \mid \mathcal{F}_i] = X_i.$$
But how do we conclude from this that $E[X_i]=E[X_j]$ for all $i,j$?
MY IDEA:
show it by induction on $i$. Let $\mu:=E[X_1]$.
Then $$E[X_2] = E[E[X_2 \mid \mathcal{F}_1]]=E[X_1]=\mu,$$
and $$E[X_i]=E[E[X_i \mid \mathcal{F}_{i-1}]]= E[X_{i-1}]=\mu$$
by induction, using the property that the conditional expectation and the random variable itself have the same expectation.
Is this correct?
In the same way one can show that for a submartingale one has $E[X_i] \geq E[X_j]$ for $i \geq j$ and for a supermartingale $E[X_i] \leq E[X_j]$ for $i \geq j$. Is this correct?
Best Answer
Note that an equivalent definition is $E[X_n \mid \mathcal F_m] = X_m$ if $n > m$.
The fact that they are equivalent can be proven in a very similar way as you did. In any case it's a little more intuitive now, as you can just take the expectation to both sides to conclude that $E[X_n] = E[X_m]$ for $n > m$ which is what we wanted