Let $\{W_t\}_{0 \le t \le h}$ be Brownian motion in $\mathbb{R}$. If $\{g_t\}_{0 \le t \le h} \subset \mathbb{R}$ is a fixed scalar field (independent of the Brownian motion) parameterized by time, then I know how to obtain concentration of the following integral: $$\int_0^h g_t dW_t$$ because it is equivalent via time-change to a draw from the distribution $$\mathcal{N}\left(0, \int_0^h g_t^2 dt \right). $$ My question is what to do in the case when we are dealing with a scalar field that depends on the Brownian motion. In particular, suppose $g: \mathbb{R} \rightarrow \mathbb{R}$ is some scalar field (say it is $1$-Lipschitz, so it can grow if the Brownian motion gets too large), and I want to understand concentration of $$\int_0^h g(W_t)dW_t.$$ Is there a standard way of doing this? My apologies if this is a naive question.
Edit: I am trying to be competitive with the following strategy -- first, apply some tail bound on how large $W_t$ ever gets, and then using this tail bound obtain a bound on the variance in $$\mathcal{N}\left(0, \int_0^h g(W_t)^2 dt\right). $$ Then, apply a standard Gaussian tail bound. However, I feel this doesn't make any sense since if I tell you the trajectory of the Brownian motion then the integral is no longer a random variable, so it doesn't seem right to treat the result as a draw from an appropriate Gaussian.
Answer for linear g
By definition for partitions $\pi=\{0=t_0<...<t_n=h\}$, $|\pi| = \sup_{i} |t_{i+1} - t_i |$
$$\int_0^h g(W_t)dW_t = \lim_{|\pi|\to 0} \sum_{i=1}^{n-1} g(W_{t_i})(W_{t_{i+1}}-W_{t_i}) $$
So for $g(x) = ax +b$ we can do the following:
$$\begin{align} \sum_{i=1}^{n-1} (aW_{t_i} + b)(W_{t_{i+1}}-W_{t_i}) &= a \sum_{i=1}^{n-1} W_{t_i} (W_{t_{i+1}}-W_{t_i}) + b\sum_{i=1}^{n-1} (W_{t_{i+1}}-W_{t_i}) \\ &= a \sum_{i=1}^{n-1} W_{t_i} (W_{t_{i+1}}-W_{t_i}) + b(W_h - W_0) \end{align}$$
So we only really need $g(x) = x$, which we can get using the following argument: $$x(y-x) = xy - x^2 =\frac{x^2 + y^2 -(x-y)^2}{2} - x^2 = \frac{1}{2}((y^2-x^2) -(y-x)^2)$$
Which means
$$\begin{align} \sum_{i=1}^{n-1} W_{t_i} (W_{t_{i+1}}-W_{t_i}) &= \frac{1}{2}\left(\underbrace{\sum_{i=1}^{n-1} (W_{t_{i+1}}^2-W_{t_i}^2)}_{=W^2_h - W^2_0} + \underbrace{\sum_{i=1}^{n-1} (W_{t_{i+1}}-W_{t_i})^2}_{\to [W]_h-[W]_0 = h-0}\right)\\ &\to\frac{1}{2}(W_h^2 - h) \end{align}$$
where $[W]_h=h$ is the quadratic variation of the brownian motion. So for linear $g(x)=ax+b$, we get:
$$\int_0^h g(W_t)dW_t = \frac{a}{2}(W_h^2 - h) + bW_h$$
Possible approach for non linear, differentiable g
By the Ito Formula
$$F(W_h) - F(W_0) = \int_0^h \underbrace{\frac{\partial F}{\partial x}(W_t)}_{\stackrel{!}{=}g(W_t)} dW_t + \frac{1}{2}\int_0^h \underbrace{\frac{\partial^2 F}{\partial x^2}(W_t)}_{=g'(W_t)} d[W]_t$$
So
$$\int_0^h g(W_t)dW_t = \int_0^{W_h}g(x)dx - \frac{1}{2}\int_0^h g'(W_t) dt$$
The issue here of course is the calculation of the integral over $g'$. If the derivative is bounded this might provide you with some bounds though.