Proving a claim that $Z_n = \max\{(X_n^{(1)})^2, X_n^{(2)}\}$ is a submartingale

Question

Let $X_n^{(1)}, X_n^{(2)}, n \in \mathbb{N}$ be martingales. Let $\mathcal{F}_n$ be a filtration. I believe that $Z_n := \max\{X_n^{(1)}, X_n^{(2)}\}$ is a submartingale. I tried to show it by using Jensen's inequality since the mapping $(x,y) \mapsto \max\{x^2, y\}$ is convex. So, \begin{align*} \mathbb{E}[Z_n|\mathcal{F}_{n-1}] \geq \max\{\mathbb{E}[(X_n^{(1)})^2|\mathcal{F}_{n-1}], \mathbb{E}[X_n^{(2)}|\mathcal{F}_{n-1}]\} \end{align*} Well, $\mathbb{E}[X_n^{(2)}|\mathcal{F}_{n-1}] = X_{n-1}^{(2)}$ by assumption. But I am not sure how to justify $\mathbb{E}[(X_n^{(1)})^2|\mathcal{F}_{n-1}] = (X_{n-1}^{(1)})^2$.

Answer 1

I think that just another application of Jensen's inequality does the trick: \begin{align*} \mathbb{E}[(X_n^{(1)})^2|\mathcal F_{n-1}] &\ge (\mathbb{E}[X_n^{(1)}|\mathcal F_{n-1}])^2 \\ &=(X_{n-1}^{(1)})^2, \end{align*} so \begin{align*} \mathbb{E}[Z_n|\mathcal F_{n-1}] &\ge \max\{\mathbb{E}[(X_n^{(1)})^2|\mathcal F_{n-1}],\mathbb{E}[X_n^{(2)}|\mathcal F_{n-1}]\} \\ &\ge \max\{(X_{n-1}^{(1)})^2, X_{n-1}^{(2)}\} \\ &= Z_{n-1}. \end{align*}