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Let $X_t=\int_0^t e^{W_s^2}dW_s$ for $0\leq t\leq 1$ and show that is not a martingale.

I guess the reason is that the expectation is not finite, but I'm not sure how to show it precisely. In fact $Ee^{2W_s^2}=(1-4s)^{-1/2}$ for $0<t<1/4$ and infinity otherwise, but I don't think I can use Ito's isometry directly since this expectation is not finite. If I could that would give me $E X_t=\int_0^t Ee^{2W_s^2} ds=\infty$ for $t>1/4$.

So here is what I tried:

As in the construction of the integral, let $$t_n=\inf \left\{ t\, |\, \int_0^te^{2W_s^2}ds \geq n \right\} \wedge t$$ which is a sequence of stopping times increasingly converging to $t$. Now we can use Ito' s isometry and write $$E ( \int_0^{t_n} e^{W_s^2}dW_s)=\int_0^{t} E(e^{2W_s^2}1_{s\leq t_n}) ds.$$ but I don't know how to compute $E\left(e^{2W_s^2}1_{s\leq t_n}\right)$ because I would need to know the joint distibution first.. Thank you

bibo
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  • Note that Itô's isometry gives (at best) $X_t \notin L^2$ - but you want to prove $X_t \notin L^1$. – saz May 18 '15 at 18:57
  • You are right, what I did was actually for $E X_t^2$. But then how to proceed to show it's not in $L^1$? – bibo May 19 '15 at 02:13
  • Sorry but Itô's isometry does not yield $$E \left( \int_0^{t_n} e^{W_s^2}dW_s\right)=\int_0^{t} E(e^{2W_s^2}1_{s\leq t_n}) ds.$$ – Did May 19 '15 at 08:08
  • Yes, I should have put a square. I was computing EX_t^2 and forgot to type it – bibo May 19 '15 at 14:30

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