Let $W={W_t}_{tin(0;T)}$ be a real-valued Brownian motion and let $sigmacolon (0;T) to mathbb R$ be deterministic and square-integrable.

For some constant $A>0$, I want to bound the probability

$$

mathbb Pbigg( max_{tin(0;T)} bigg| int^t_0 sigma(s) mathrm dW_s bigg| le Abigg)

$$

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 Pbigg( max_{tin(0;T)} bigg| int^t_0 sigma(s) mathrm dW_s bigg| ge Abigg)

le

frac 1 {A^2} mathbb Ebigg( max_{tin(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.