Show $(X_n+a)^2$ is a submartingale

Question

Let $(X_n)$ be a martingale, and let $EX_n^2 < \infty$ - then I am told to show $E(X_n+a)^2 $ is a sub martingale.

I wrote $$(X_n+a)^2 = ((X_{n-1} + a) + (X_n - X_{n-1}))^2 $$ then $$E((X_n+a)^2 | \mathcal{F_{n-1}}) = (X_{n-1}+a)^2 + 2(X_{n-1}+c)E(X_{n}-X_{n-1} | \mathcal{F}_{n-1}) + E((X_n-X_{n-1})^2| \mathcal{F_{n-1}})$$ and the second term vanishes, but I don't know hwat to do with the last term.

Answer 1

You just need to apply the direct definition of sub-martingale: i.e. we want to show $$E((X_{n+1}-a)^2 - (X_n -a)^2\vert \mathcal{F}_n) \geq 0$$ \begin{align} E((X_{n+1}-a)^2 - (X_n -a)^2 \vert \mathcal{F}_n) &= E(X_{n+1}^2 - X_n^2\vert \mathcal{F}_n)\\ & = E(X_{n+1}^2\vert \mathcal{F}_n) - E(X_{n}^2\vert \mathcal{F}_n)\\ &= E(X_{n+1}^2\vert \mathcal{F}_n) -E(X_{n}\vert \mathcal{F}_n)^2\\ &= E(X_{n+1}^2\vert \mathcal{F}_n) - E(X_{n+1}\vert \mathcal{F}_n)^2\\ &>= 0 \end{align} The last inequality is a result of Jensen inequality.

Answer 2

We need only note that \begin{align*} E((X_{n+1}-a)^2 - (X_n -a)^2 \mid\mathcal{F}_n) &= E(X_{n+1}^2 - X_n^2\mid\mathcal{F}_n)\\ &=E(X_{n+1}^2 -2X_{n+1}X_n + X_n^2\mid\mathcal{F}_n)\\ &=E\left((X_{n+1}-X_n)^2 \mid \mathcal{F}_n \right)\\ &\ge 0. \end{align*}