Martingale: Whay $\mathbb E[S_n]=\mathbb E[S_1]$.

Question

I've got a theorem (without proof) that say:

If $(S_n)$ is a martingale refer to $(X_n)$, then $\mathbb E[S_n]=\mathbb E[S_1]$.

I don't really understand why. Is there an intuitive why to see it ?

Answer 1

$$\mathbb{E}\left[S_n\right]=\mathbb{E}\left[\mathbb{E}\left[S_n\mid X_{n-1}\right]\right]=\mathbb{E}\left[S_{n-1}\right]=\mathbb{E}\left[\mathbb{E}\left[S_{n-1}\mid X_{n-2}\right]\right]$$ $$=\mathbb{E}\left[S_{n-2}\right]=\ldots=\mathbb{E}\left[\mathbb{E}\left[S_{2}\mid X_{1}\right]\right]=\mathbb{E}\left[S_{1}\right].$$

To show the first equality, we use the definition of the conditional expectation : for all random variable $Y$ which is $\sigma\left(X_1,\ldots X_{n-1}\right)$-measurable and such that $\mathbb{E}\left[S_nY\right]$ exists, we have $$\mathbb{E}\left[S_nY\right]=\mathbb{E}\left[\mathbb{E}\left[S_n\mid X_1,\ldots X_{n-1}\right]Y\right]$$ and for $Y=1$, we get $$\mathbb{E}\left[S_n\right]=\mathbb{E}\left[\mathbb{E}\left[S_n\mid X_1,\ldots X_{n-1}\right]\right].$$ Then we use the martingale property $\mathbb{E}\left[S_n\mid X_1,\ldots X_{n-1}\right]=S_{n-1}$. As $1$ is constant, it is measurable for any $\left(\sigma\left(X_1,\ldots,X_k\right)\right)_{k\geq1}$ and then the above argument holds for each $k\geq1$.