Continuity of the stochastic process $X_t=\int_0^t(a+b\frac{u}{t}) \, dW_u$

Question

I am wondering about the continuity of the stochastic process $$X_t=\int_0^t \left(a+b\frac{u}{t}\right)\,dW_u$$

which has variance $t$ and normally distributed for $a^2+\frac{b^2}{3}+ab=1$

The process seems to be discontinuous at $t=0$ except for $b=0.$ Is this stochastic process really discontinuous or does the limit $t\to 0$ have a finite value

Answer 1

We have $$\begin{align*}X_t&=\int_0^t \left(a+b\frac{u}{t}\right)\,dW_u \\\\ &=aW_t + \frac{b}{t}\int_0^t u\, dW_u\end{align*}$$

Using Itô's formula we get: $$\int_0^t u\, dW_u = tW_t - \int_0^t W_s\, ds$$ and so $$X_t = (a+b)W_t - \frac{b}{t}\int_0^t W_s \, ds$$

We know: $W_s$ is continous a.s. hence from the Fundamental theorem of calculus it follows that $$\int_0^t W_s \, ds$$ is differentiable a.s. in $t=0$ which means nothing else that $$\lim_{t\to 0} \frac{1}{t}\int_0^t W_s \, ds$$ exists a.s.

All together we get: $$\lim_{t \to 0} X_t = \lim_{t \to 0}\frac{b}{t}\int_0^t W_s \, ds$$ exists a.s. and so $X_t$ is continous in $0$.