Calculate the expectation and variance of $Y_T$. Is $Y_T$ a martingale?
The expectation is always zero. A sufficient condition for the integral $\int_0^t f(\omega, s)\, dB_s$ to be a martingale on $[0,T]$ is that
- $f(\omega,s)$ is adapted, measurable in s, and
- $\mathbb{E}\left(\int_0^T f^2(\omega,s)\,ds\right) < \infty$.
In this case, indeed, $\mathsf{E} \left(\int_0^T f(\omega,s)\, dB_s\right)=0$. For a proof see theorem 5 in Martingales and Elementary Integrals
. For the variance, as you mentioned one can just use Itô isometry
$$E[(\int s dB_{s})^{2}]=\int s^{2} ds.$$
A proof of Itô isometry is covered in most SDE textbooks eg. Revuz-Yor or many online notes eg "Notes on the Itô Calculus". The main idea is using the approximation
$$\int_0^T f(\omega,s)\, dB_s\approx \sum f(\omega, s_{i})(B_{s_{i+1}}-B_{s_{i}}),$$
(which to be clear, this approximation is only meant in L2; the almost-sure doesn't work see Convergence of approximating integral sum in case of stochastic integrals)
and so
$$E(\int_0^T f(\omega,s)\, dB_s)^{2}\approx E\left(\sum f(\omega, s_{i})f(\omega, s_{k})(B_{s_{i+1}}-B_{s_{i}})(B_{s_{k+1}}-B_{s_{k}})\right).$$
This is zero if we don't have k=i and so
$$=E\left(\sum f^{2}(\omega, s_{i})(B_{s_{i+1}}-B_{s_{i}})^{2}\right)$$
and thus by Markov property we get
$$=\sum Ef^{2}(\omega, s_{i})(s_{i+1}-s_{i}).$$
A) I have read somewhere that it was obvious that $Y_t$ is a martingale. First, why is it the case?
It is better to read an SDE book such as the Shreve-Karatzas textbook. But this follows from the above approximation we already mentioned. We have
$E[\sum^{T} f(\omega, s_{i})(B_{s_{i+1}}-B_{s_{i}})\mid \mathcal{F}_{t}]$
$=\sum^{t} f(\omega, s_{i})(B_{s_{i+1}}-B_{s_{i}})$
due to the independent increments having zero mean.
If $Y_t$ is a martingale, then we have that $\mathbb{E}(Y_t)=\mathbb{E}(Y_0)=0$. Other thank using the martingale property, is there a different way to conclude that the expectation is 0?
You can see it directly from the approximation as above.
