Distribuiton of stochastic integral

Question

If $(W_t)_{t\geq 0}$ is a Wiener process, $X_0=0$ and for all $t$, $t>0$ and $\alpha>0$. $X_t=\int_0^t\frac{u^\alpha}{t}dW_u$.

I have want to answer 2 questions:

What is the distribution of $X_t$? Is $X_t$ a martingale in $W_t$'s filtration?

I tried to write the Integral in sum form:

$X_t=\frac{1}{t}\sum t_i^\alpha(W_{t_{i+1}}-W_{t_i})$

So it is just the sum of normal distributions, so it is normal distribution isn't it?

Answer 1

It is the L2 limit of a family of variables of the form $\frac{1}{t}\sum t_i^\alpha(W_{t_{i+1}}-W_{t_i})$ hence it is measurable according to the filtration of $W$.

Answer 2

This is a Wiener integral and so it has a Gaussian law with first two moments of the form :

Mean $$m_t=E[\int_0^t \frac{u^\alpha}{t}dW_u]=0 $$ Variance $$v_t=E[\int_0^t \frac{u^{2.\alpha}}{t^2}du]=\int_0^t \frac{u^{2.\alpha}}{t^2}du=...$$

Check the web for properties on Wiener Integrals there is plenty.

Best regards