Probability that $\int_0^tX_s\,dW_s$ lies within $1/t$ of $X_t$

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Consider the inequality $$f(x)-\frac1x\le f’(x)\le f(x)+\frac1x$$ on the positive axis. This tells us that $f(x)\sim e^x$ with infinitesimal deviation, and we can use identities such as Grönwall's inequality to establish bounds on $f(x)$. As there is no randomness in the function, this problem is deterministic.

In the random scenario, consider a stochastic process $X_t$ under Brownian motion (BM); that is, $$dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dW_t.$$ The problem thus becomes evaluating the value of $$F_{X_t}(t)=\operatorname P\left(\int_0^tX_s\,dW_s-\frac1t\le X_t\le\int_0^tX_sdW_s+\frac1t\right)\tag1$$ for any $t>0$. Note that the analogous version of $f(x)\sim e^x$ can be described through geometric Brownian motion (GBM).

If $X_t=W_t$, we expect $F(t)\to0$ as $t\to\infty$. Indeed, we have \begin{align}F_{W_t}(t)&=\operatorname P\left(-\frac12t-\frac1t+W_t^2\le W_t\le-\frac12t+\frac1t+W_t^2\right)\end{align} using Itô's lemma. This can be rewritten as $$F_{W_t}(t)=\operatorname P\left(\frac{2t^2+t-4}{4t}\le\left(W_t-\frac12\right)^2\le\frac{2t^2+t+4}{4t}\right)$$ to the form $\operatorname P\left(g(t)\le W_t\le h(t)\right)$, which can then be converted to a standard Normal.

But how about in the case of GBM (where $F(t)\to1$ with an appropriate choice of $\mu,\sigma$), or more generally any $X_t$?