I don't know how to deal with the following stochastic integral:
$\int_0^t \frac{1}{\sqrt{t-s+1}} d W(s)$
As you can see, the variable $t$ appears both as an endpoint of the interval of integration and in the integrand. And that's exactly the problem. In general, the stochastic integral of a deterministic function is a normal random variable, but here? What can we say?
What about its stochastic differential? Is it possible to apply Ito formula to that process?
Could you find a way to, somehow, take the variable $t$ outside the integral? In this way we could apply Ito formula without problems
Thank you very much for any help!! :)
For fixed $t$, the function $f:s\mapsto 1/\sqrt{t-s+1}$ is an element of $L^2[0,t]$, so the stochastic integral $X_t:=\int_0^t{1\over\sqrt{t-s+1}}\,dW(s)=\int_0^t f(s)\,dW(s)$ is defined and is normally distributed with mean $0$ and variance $$ \int_0^t{1\over t-s+1}\,ds=\log(t+1). $$ If $0<t<u$ then the covariance of $X_t$ and $X_u$ is $$ \int_0^t{1\over\sqrt{(u-s+1)(t-s+1)}}\,ds=\log\left({t+u+2+2\sqrt{(t+1)(u+1)}\over u-t+2+2\sqrt{u-t+1}}\right). $$ (Integration courtesy of WolframAlpha.) This means that $$ \Bbb E[(X_{t+h}-X_t)^2] ={t\over 2(t+1)}h+o(1),\qquad h\to 0 $$ (if my calculation is correct) from which it follows that $t\mapsto X_t$ has a modification that is Holder continuous of order $\alpha$ for any $\alpha\in(0,1/2)$.
A formal integration by parts reveals that $$ X_t=W(t)+\int_0^t {W(s)\over 2(t-s+1)^{3/2}}\,ds, $$ and so (the continuous modification of) $X_t$ is a continuous semimartingale with martingale part $W(t)$.