# Prove that Ito integral is large with large probability

Let $$W=\{W_t\}_{t\in[0;T]}$$ be a real-valued Brownian motion and let $$\sigma\colon [0;T] \to \mathbb R$$ be deterministic and square-integrable.

For some constant $$A>0$$, I want to bound the probability $$\mathbb P\bigg[ \max_{t\in[0;T]} \bigg| \int^t_0 \sigma(s) \mathrm dW_s \bigg| \le A\bigg]$$ from above, i.e. prove that the probability that this integral is large is not too small.

Well, if it was the other direction, I would just use Markov and Burkholder-Davis-Gundy: $$\mathbb P\bigg[ \max_{t\in[0;T]} \bigg| \int^t_0 \sigma(s) \mathrm dW_s \bigg| \ge A\bigg] \le \frac 1 {A^2} \mathbb E\bigg[ \max_{t\in[0;T]} \bigg| \int^t_0 \sigma(s) \mathrm dW_s \bigg|^2 \bigg] \le \frac {C_2} {A^2} \int^T_0 \sigma(t)^2 \mathrm dt,$$ i.e. if $$A$$ is large, the Ito integral is smaller with large probability.

But I have no idea how to do this in the other direction (prove that the Ito integral is larger than something with large probability) because the Markov inequality does not work that way.

• This seems to be an anti-concentration phenomena. Markov, Chebyshev, Chernoff etc. are all concentration inequalities, but are you aware of any non-concentration inequalities? Note that BDG can go either way, so I'm not really worried about that particular usage as much. It's the other way that sounds super interesting. Jul 14, 2021 at 11:15
• Kolodez, is my answer acceptable ? Jul 19, 2021 at 15:17

When $$\sigma$$ is deterministic then the quadratic variation of the stochastic integral is $$\int_0^t\sigma^2(s)\,ds\,,$$ and this is deterministic. Write $$I_t$$ for the stochastic integral and $$V_t$$ for this quadratic variation. Then take the inverse function of $$t\mapsto V_t$$ and call it $$V^{-1}_t\,.$$ The time-changed process $$I_{V^{-1}_t}$$ is a Brownian motion because it is a martingale with quadratic variation $$t\,$$ (see also Karatzas-Shreve, Brownian Motion and Stochastic Calculus, Thm. 3.4.6). The function $$t\mapsto V^{-1}_t$$ is a non decreasing one-to-one mapping of $$[0,V_T]$$ to $$[0,T]\,.$$ Therefore, $$\max_{0\le t\le T}\Bigg|\int_0^t\sigma(s)\,dW_s\Bigg|=\max_{0\le t\le V_T}\Bigg|\int_0^{V^{-1}_T}\sigma(s)\,dW_s\Bigg|=\max_{0\le t\le V_T}|I_{V^{-1}_t}|\,.$$ Then, $$\mathbb P\Bigg[\max_{0\le t\le T}\Bigg|\int_0^t\sigma(s)\,dW_s\Bigg|\ge A\Bigg]=\mathbb P\Big[\max_{0\le t\le V_T}|I_{V^{-1}_t}|\ge A\Big]\ge \mathbb P\Big[\max_{0\le t\le V_T}I_{V^{-1}_t}\ge A\Big]\,.$$ From K&S (2.8.3) it follows now that the last term equals $$\int_A^\infty\frac{2}{\sqrt{2\pi V^{-1}_T}}e^{-b^2/(2V^{-1}_T)}\,db=2-2\Phi\Bigg(\frac{A}{\sqrt{V_T^{-1}}}\Bigg)\,.$$ Where $$\Phi$$ is the normal CDF.
Maybe a bit care in the above derivation is needed when $$t\mapsto V_t^{-1}$$ is not continuous. This is the case if and only if $$t\mapsto V_t=\int_0^t\sigma^2(s)\,ds$$ has intervals on which it is constant.
• Thanks! When you write $V^{-1}_T$, you mean $T$, don't you? Jul 21, 2021 at 19:14
• By $V_t^{-1}$ I mean the inverse function of $t\mapsto V_t=\int_0^t\sigma^2(s)\,ds\,.$ The $V_T^{-1}$ is that inverse function evaluated at $T\,.$ Please let me know if anything further is unclear. Jul 22, 2021 at 6:05