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Let $(X_t)$ be a non-negative supermartingale. Then for $s<t$, $$ E[X_t\mid \mathcal{F}_s] \leq X_s,$$ which can be written as $$ -X_s \leq E[-X_t\mid \mathcal{F}_s],$$ hence $(-X_t)$ is a submartingale, and thus because the absolute value is convex, also $(|-X_t|)$ is a submartingale. But because $X$ was non-negative, we have $|-X_t| = X_t$, so then we conclude that $(|-X_t|) = (X_t)$ is both a submartingale and a supermartingale, hence a martingale.

This is obviously wrong. But right now I don't see where I'm missing something.

Protawn
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2 Answers2

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There is the following result:

a) Let $(X_t)_{t \geq 0}$ be a martingale and $F$ a convex function such that $F(X_t) \in L^1$ for all $t \geq 0$. Then $(F(X_t))_{t \geq 0}$ is a submartingale.

b) Let $(X_t)_{t \geq 0}$ be a submartingale and $F$ an increasing convex function such that $F(X_t) \in L^1$ for all $t \geq 0$. Then $(F(X_t))_{t \geq 0}$ is a submartingale.

Note that b) requires weaker assumption on the process (martingale property is replaced by submartingale property) but the price which we have to pay is that we need to assume additionally that $F$ is increasing.

Since the modulus $F(x)=|x|$ is clearly not increasing, this means that $(X_t)_{t \geq 0}$ being a submartingale does not imply that $(|X_t|)_{t \geq 0}$ is a submartingale. An example is for instance the deterministic submartingale $X_t := - \frac{1}{1+t^2}$.

Community
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saz
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It is not true that $|-X_t|$ is a submartingale. Think of constant random variables. In that case $(-X_t)$ is a submartingale iff it is increasing. But that does not imply that $|-X_t|$ is increasing. In fact it is decreasing!.

An increasing convex function of a submartingale is a submartingale. 'Increasing' is essential.

Kavi Rama Murthy
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