If $(Y,\mathcal{F})$ is a martingale, show that $\mathbb{E} (Y_n)=\mathbb{E}(Y_0)$ for all n.

Question

I'm trying to show this by induction. So far I know by the properties of a martingale that $\mathbb{E}(Y_n)=\mathbb{E}(\mathbb{E}(Y_{n+1}|\mathcal{F}))=\mathbb{E}(Y_{n+1})$. Isn't this property just essentially showing $n=k+1$ holds and therefore $\mathbb{E}(Y_n)=\mathbb{E}(Y_0)$ is trivially true? I just want to know how to correctly show it by induction.

Thanks in advance! :)

Answer 1

Clearly $\mathbb E[Y_0]=\mathbb E[Y_0]$. Let $n$ be a nonnegative integer and assume that $\mathbb E[Y_n]=\mathbb E[Y_0]$. Then by the martingale property we have $$ \mathbb E[Y_{n+1}\mid\mathcal F_n] = Y_n. $$ Taking expectations on both sides yields $$ \mathbb E[\mathbb E[Y_{n+1}\mid \mathcal F_n]] = \mathbb E[Y_n] = \mathbb E[Y_0]. $$