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Let $W_{t} = (W_{t}^{1}, \ldots, W_{t}^{d}), t \geqslant 0$ be a $d$-dimensional Wiener process, e.i. $W^{1}, \ldots, W^{d}$ are independent 1-dimensional Wiener processes. Let $ d \geqslant 3 $.

Show that $(|W_{t}|^{2-d})_{t \in (0, \infty)}$ is a supermartingale.

I've got a hint that I have to show the existance of a positive constant $C_{d}$, dependent only of the dimension $d$, such that for every $x \in \mathbb{R}^{d}$ we have $|x|^{2-d} = C_{d} \int_{0}^{\infty}{p_{t}(x) dt}$, where $p_t$ is the density of $W_t$.

I also know that there is a theorem which says:

If $X_t$ is a supermartingale and $f$ is a nondecreasing concave function such that $\mathbb{E} |f(X_t)| < \infty$, then also $f(X_t)$ is a supermartingale.

This theorem is true because of Jensen's inequality.

Olga98
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1 Answers1

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The fact that $\frac 1 {|W_t|}$ is a super-martingale follows easily from the fact that $(W_t)$ is a martingale. Now use the fact that $x \to x^{d-2}$ is an increasing convex function on $[0,\infty)$.

Kavi Rama Murthy
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  • It's all right, but - probably a stupid question - how exactly one can show that $\frac{1}{|W_t|}$ is a super-martingale? I was trying to use the triangle inequality, but it wasn't working. – Olga98 Sep 09 '19 at 17:53
  • @Kavi Rama Murthy Are you sure about this? To me, it s already non obvious how to define $\frac{1}{|W_t|}$.. – Tom Sep 10 '19 at 10:02
  • @Tom Well $\frac 1 {W_t}$ is defined almost everywhere but $E\frac 1 {|W_t|}=\infty$. So we don't have a supermartingale in the traditional sense. But $E(X\mathcal G)$ can be defined for non-negative $X$ even when it does not have finite mean. The supermartingales here are in this extended sense – Kavi Rama Murthy Sep 10 '19 at 10:07
  • @KaviRamaMurthy Ok but dont you have then a hard time in defining the process $\frac{1}{W}$ for every $(t,\omega)$, since the almost sure set will depend on $t$? – Tom Sep 10 '19 at 10:14
  • @Tom The point here we h ave to go beyond the usual real valued random variables. If you allow $\infty$ as a value then \frac 1 {W(t,\omega)}$ is defined and measurable. – Kavi Rama Murthy Sep 10 '19 at 10:17
  • Ok I see. But (dont know if it matters) $\frac{1}{W}$ defined in such a way cannot be continuous. Right? – Tom Sep 10 '19 at 10:19
  • @Tom Continuity does not matter for this question. – Kavi Rama Murthy Sep 10 '19 at 10:23
  • I think I understand now - Thx a lot – Tom Sep 10 '19 at 10:23