square-root of a martingale

Question

A general result says that if $\{X_n\}$ is a martingale in $L^2$, then $\{X_n^2\}$ is also a martingale, and in general for a positive power $p>1$ this also holds. How about fractional powers, like $p=\frac{1}{2}$? I haven't been able to find a general result for this.

Answer 1

It is not true that $(X_n^{2})$ is martingale if $(X_n)$ is. If $(Y_i)$ is i.i.d. with standard normal distribution then $X_n=Y_1+Y_2+\cdots+Y_n$ defines a martingale. But $X_n^{2}$ is not a martingale because $EX_n^{2}=n$ which is not independent of $n$.

If $(X_n)$ is a nonnegative martingale then $(\sqrt X_n)$ is a supertmartingale. In fact for any increasing concave function $f: [0,\infty) \to \mathbb R$ the sequence $(f(X_n))$ is a supertmartingale. This follows immediately from Jensen's inequality for conditional expectations. [ Similarly for any increasing convex function $f: [0,\infty) \to \mathbb R$ the sequence $(f(X_n))$ is a submartingale].

Note that $x^{p}$ is convex on $[0,\infty)$ if $p \geq 1$ and concave if $0<p<1$.