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Is $\frac{W^{2}(t)}{t}$ a martingale w.r.t. the usual filtration? $t>0$ and $W(t)$ is the Wiener process.

What I have so far: Define $Z(t)=\frac{W^2(t)}{t}$.

By Ito we get


Now we find


Taking conditional expectations gives

$$\mathbb{E}[Z(t)|F_{u}]=\mathbb{E}[\int_{0}^{u}\frac{s-W^2(s)}{s^2}ds|F_{u}]+\mathbb{E}[\int_{u}^{t}\frac{s-W^2(s)}{s^2}ds|F_{u}]+2\mathbb{E}[\int_{0}^t\frac{W(s)}{s}dW(s)|F_{u}]$$ $$=\int_{0}^{u}\frac{s-W^2(s)}{s^2}ds+2\int_0^u\frac{W(s)}{s}dW(s)+\mathbb{E}[\int_{u}^{t}\frac{s-W^2(s)}{s^2}ds|F_{u}]$$


So if $\mathbb{E}[\int_{u}^{t}\frac{s-W^2(s)}{s^2}ds|F_{u}]=0$ then $Z(t)$ is a martingale.

I know that $\mathbb{E}[W^2(s)]=s$. Can I just say $$\mathbb{E}[\int_{s}^{t}\frac{s-W^2(s)}{s^2}ds|F_{u}]=\int_{u}^{t}\frac{s-\mathbb{E}[W^2(s)|F_{u}]}{s^2}ds$$?? And then does $\mathbb{E}[W^2(s)]=s$ imply $\mathbb{E}[W^2(s)|F_{u}]=s$??

Is Z a martingale?

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About the revised version: ?? And then does $\mathbb{E}[W^2(s)]=s$ imply $\mathbb{E}[W^2(s)|F_{u}]=s$?? Well, why should it? It does not (and the decomposition in my post shows what $\mathbb{E}[W^2(s)|F_{u}]$ is). – Did Nov 7 '12 at 19:18
up vote 2 down vote accepted


For every $s\geqslant0$, call $\mathcal W_s=\sigma(W_u;u\leqslant s)$, then, for every $t\geqslant s$, $W_t^2=W_s^2+2W_sY+Y^2$ with $Y=W_t-W_s$ centered and independent of $\mathcal W_s$ with variance $t-s$.

In particular, $\mathbb E(W_t^2\mid\mathcal W_s)=W_s^2+t-s$ hence $\mathbb E(Z_t\mid\mathcal W_s)=1+(Z_s-1)(s/t)$. Conditioning this on $\mathcal Z_s=\sigma(Z_u;u\leqslant s)\subseteq\mathcal W_s$ yields $$ \mathbb E(Z_t\mid\mathcal Z_s)=\mathbb E(\mathbb E(Z_t\mid\mathcal W_s)\mid\mathcal Z_s)=1+(Z_s-1)(s/t), $$ hence $(Z_t)$ is not a $(\mathcal Z_t)$-martingale.

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Ok I see what you've done. Can I ask where I went wrong with my approach? – Math Girl Nov 7 '12 at 17:06
Dunno... To begin with, you could correct those awful $\int_s^t(s-W^2(s))ds/s^2$ into $\int_s^t(u-W^2(u))du/u^2$, and see what happens. – Did Nov 7 '12 at 18:29

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