Continuity of the stochastic process $X_t=\int_{0}^{t}(a+b\frac{u^n}{t^n} ) dW_{u} $

Question

I am wondering about the continuity of the stochastic process

$$X_t=\int_{0}^{t}(a+b\frac{u^n}{t^n} ) dW_{u} $$

Where n=1,2,.. At t=0, there seems to be a discontinuity except for b=0 . Is there a discontinuity?

Answer 1

This problem comes down to showing the continuity of

$$\frac{1} {t^n}\int_0^t u^n dW_u $$

From Itô's formula, we have

$$\frac{1} {t^n}\int_0^t u^n dW_u =W_t-\frac{n}{t^n}\int_{0}^{t}W_u u^{n-1}du$$

So essentially we have to evaluate the limit

$$\lim_{t\to 0} \frac{1}{t^n}\int_{0}^{t}W_u u^{n-1}du$$

Where n is any natural number Since $W_u u^{n-1}$ is continuous a.s thus from the fundamental theorem of calculus and L'Hôpital's rule, we have

$$\lim_{t\to 0} \frac{1}{t^n}\int_{0}^{t}W_u u^{n-1}du=\lim_{t\to 0}\frac{W_{t}t^{n-1}}{nt^{n-1}}=\lim_{t\to 0}\frac{W_{t}} {n} =0$$ a.s Thus $$X_t$$ is continuous a.s

More generally for the stochastic process

$$Y_t=\int_{0}^{t}f(u,t)dW_{u}$$ Where f is of the form $$f(u, t) =\frac{b(u)} {g(t)} $$ From Itô's formula, we have

$$f(t, t)W_{t} - f(0,t)W_{0}-\frac{1}{g(t) }\int_{0}^{t}\frac{db} {du} W_{u} du$$

If $\frac{db} {du}$ is continuous then

$$\lim_{t\to 0} \frac{1}{g(t) }\int_{0}^{t}\frac{db} {du} W_{u} du=\lim_{t\to 0}\frac{b' (t)} {g'(t)} W_{t} $$

Thus sufficient(but not necessary) conditions for continuity at t=0 are

$$f(t, t)$$ is continuous $$f(0,t)$$ is continuous $$\frac{db} {du}$$ is continuous $$\lim_{t\to 0}\frac{b' (t)} {g'(t)}$$ exists and is finite

Bonus:Check this for $$f(u, t) =\frac{\sin(u)} {t}$$

For the stochastic process $X_t$ given in the question, certain a,b can make it a Wiener process

We have already established $X_0=0$ a.s. Note that X_t has mean 0 and variance $(a^2+\frac{b^2}{2n+1}+\frac{2ab}{n+1})t$ and it is normally distributed since the limit of a sequence of normal random variables is also a normal random variable.

For its variance to be t $(a^2+\frac{b^2}{2n+1}+\frac{2ab}{n+1})=1$

Also note that since $X_t-X_s$ and $X_s$ are jointly normally distributed(any of their linear combination is individually normally distributed) and their covariance vanishes for $b=0$ and $\frac{a} {n+1}+\frac{b}{2n+1}=0$ These equations determine the a, b e.g two of the solutions are $a^2=1, b=0$ Hence $X_t-X_s$ is independent of $ X_s$ for these a, b.

Now since

$|X_t-X_s|=|t-s|^{\frac{1}{2}}Z$

Where Z is the standard normal random variable and the equality is in distribution. Then

$E|X_t-X_s|^{2m}=C_{m} |t-s|^{m} $

Thus from the Kolmogorov continuity theorem $X_{t} $ is a.s $<\frac{1}{2}$ holder continuous for those a, b.

These conditions make $X_t$ a Wiener process for those a, b.