quality of approximation of Ito integral I am doing project in economics and I need to calculate the Ito integral:
$$\int_0^T f(X_t)dW_t$$
where $f$ is differentiable $\mathbb{R} \to \mathbb{R}$ and $X_t$ is a diffusion $dX_t = \mu(t,X_t)dt + \sigma(t, X_t)dW_t$
The problem is I don't know exactly the function $f$ but I have an approximation to it $f^*$. I use Euler discretization to calculate the integral.
Lets say $\sup|f - f^*| < \epsilon$. 
Is there any way to bound the error ?:
$$\int_0^T f(X_t)dW_t - \int_0^T (f^*)(X_t)dW_t$$
Do I need to impose some constraints on $f, \mu$ or $\sigma$? 
What I can do:
$$\int_0^T (f(X_t) - (f^*)(X_t))dW_t  = \sum_{i = 0}^N (f(X_{t_i}) - (f^*)(X_{t_i}))(W_{t_{i+1}}-W_{t_{i}})$$
$$\sum_{i = 0}^N (f(X_{t_i}) - (f^*)(X_{t_i}))(W_{t_{i+1}}-W_{t_{i}}) < \epsilon \sum_{i = 0}^N(W_{t_{i+1}}-W_{t_{i}})$$
Which is the sum of normal random variables and then I can write some inequality for the tails. Can I do anything better?
 A: We have 
$$\mathbb E\int_0^T |f(X_t) - f^*(X_t)|^p\,dt < T\epsilon^p < \infty, p\geq 2$$
Because of this $(\int_0^t f(X_t) - f^*(X_t)\,dW)_{t\geq 0}$ is a martingale and there are certain inequalities for moments of martingales (e.g. cf. Karatzas & Shreve "Brownian Motion and Stochastic Calculus" - 'Martingale Moment Inequalities [Millar (1968), Novikov
(1971)]).
Reformulated for a problem one of these inequalities says if $\mathbb E \int_0^T |X_t|^p \,dt < \infty$, then 
$$\| \int_0^T X_t\,dW_t \|_p \leq \sqrt{\frac{p(p-1)}{2}}T^{\frac 1 2 - \frac 1 p} \left(\mathbb E\int_0^T |X_t|^p\,dt\right)^{1/p}, p\geq 2.$$
I'll provide a sketch of the proof:
Of course the case $p = 2$ is just one half of the Ito isometry. For $p > 2$ with $M_t = \int_0^t X_t\,dW_t$,
$$\phi(M_t) = \phi(M_T) + \int_0^t \phi'(M_t)\,dM_t + \frac 1 2 \int_0^t \phi''(M_t)\,d\langle M \rangle_t,$$
by Itos formula, where $\phi(x) = |x|^p$.
Then,
$$|M_T|^p = p \int_0^T |M|^{p-2}X\,dW + \frac{p(p-2)}{2}\int_0^T |M_t|^{p-2} X_t^2\,dt + \int_0^T \frac p 2 |M_t|^{p-2} X_t^2\,dt.$$
While the first summand may not have expectation zero (because it is not necessarily a martingale), it does have expectation zero if stopped at appropiate stopping times, i.e.
$$\tau_n := \inf \{t\geq 0 : |M_t|\geq n\}.$$
For simplicity I'm going to assume the expectation is zero (for the real proof you have to carry along $\tau_n$ and then let $n\to \infty$ at the end).
Hölder's inequality then shows
$$\mathbb E|M_T|^p \leq \frac{p(p-1)}{2} \|\int_0^T|X_t|^2\|_{p/2} \|\int_0^T |M_t|^p\|_{\frac p {p-2}} = \frac{p(p-1)}{2} \left(\mathbb E\int_0^T |X_t|^p\,dt\right)^{2/p}\left(\mathbb E\int_0^T |M_t|^p\,dt\right)^{\frac {p-2} p}.$$
But then $(|M_t|^p)_{t\geq 0}$ is a submartingale, so $\mathbb E|M_T|^p \geq \mathbb E|M_t|^p$ and
$$\mathbb E\int_0^T |M_t|^p\,dt \leq T \mathbb E|M_T|^p.$$
Therefore,
$$\mathbb E|M_T|^p \leq \frac{p(p-1)}{2} \left(\mathbb E\int_0^T |X_t|^p\,dt\right)^{2/p}(T \mathbb E|M_T|^p)^{\frac {p-2} p},$$
and dividing by $(\mathbb E|M_T|^p)^{\frac {p-2} p}$ and then taking the square root yields
$$\|M_T\|_p = (\mathbb E|M_T|^p)^{(1 - \frac{p-2} p)/2} \leq \left(\frac{p(p-1)}{2}\right)^{1/2} \left(\mathbb E\int_0^T |X_t|^p\,dt\right)^{1/p}T^{\frac 1 2 - \frac 1 p}.$$
In your case
$$\|\int_0^T f(X_t) - f^*(X_t)\,dW_t\|_p \leq \sqrt{\frac{p(p-1)}{2}}T^{\frac 1 2 - \frac 1 p} \left(\mathbb E\int_0^T |f(X_t) - f^*(X_t)|^p\,dt\right)^{1/p} < \epsilon \sqrt{\frac{p(p-1)}{2}T}.$$
