$\mathbb E[(X_s-K)^+]\geq\mathbb E[(X_t-K)^+]$ for a martingale with $t<s$

Question

In a paper the authors use the fact that for a martingale $(X_t)_{t\geq0}$ with $K\geq0$ and $0\leq t<s$ we have the inequality $$\mathbb E[(X_s-K)^+]\geq\mathbb E[(X_t-K)^+]$$ I'm trying to prove this result using Jensen's inequality, since the $(\cdot)^+$ operator is convex, but could show the following inequality: \begin{align} \mathbb E[(X_s-K)^+]&\geq(\mathbb E[X_s]-K)^+ \\ &= (\mathbb E[\underbrace{\mathbb E[X_s|X_t]}_{=X_t\text{ due to martingality}}]-K)^+ \\ &=(\mathbb E[X_t]-K)^+ \quad(\leq\mathbb E[(X_t-K)^+]) \end{align} Therefore this approach fails. However, since the $\max$ operator is bounded from below by $0$ and it is increasing, we have that the difference between a driftless process and a costant must be increasing as well. Could somebody suggest how to formalize that, please?

Answer 1

For an integrable random variable $Y$ and a $\sigma$-algebra $\mathcal G$, $$ \mathbb E\left[Y^+\mid\mathcal G\right]=\mathbb E\left[\max\{Y,0\} \mid\mathcal G\right]\geqslant \max\{ \mathbb E\left[Y \mid\mathcal G\right] ,0\}. $$ Apply this to $Y=X_s-K$ and $\mathcal G=\mathcal F_t$ to get the wanted result after having taken the expectation. Indeed, we get the following (almost sure) pointwise inequality $$ \mathbb E\left[\left(X_s-K\right)^+\mid\mathcal F_t\right]\geqslant\max\{ \mathbb E\left[X_s-K\mid\mathcal F_t\right] ,0\} =\max\{ X_t-K\ ,0\}= \left(X_t-K\right)^+ $$ and then integrate.