I have a big time trouble in evaluating the following probability. It is related to brownian motion and measure, so I am asking experts from both fields for help!
Denote $B_t$, $t\in [0, T]$ be univariate Brownian motion on $(\Omega, \Sigma, P)$ with filtration $\Sigma_t$ so that $\Sigma_T=\Sigma$. Now define a probability measure $M$ as follows. $$\forall K\in \Sigma:\;\; M(K)=\mathbb{E}[\exp(B_T-T/2)\mathbb{1}_K].$$ The question is to compute the probability $$M(\exp(B_T)-A\geq0)$$ where $A$ is a positive number.
This is what I did so far:
$$M(\exp(B_T)-A\geq0)=\mathbb{E}[\exp(B_T-T/2)\mathbb{1}_K] \geq\mathbb{E}[A\exp(-T/2)\mathbb{1}_K].$$ Then this implies $$M(\exp(B_T)-A\geq0) \geq A\exp(-T/2)\mathbb{E}[\mathbb{1}_K]\geq A\exp(-T/2)P(\exp(B_T)-A\geq0)$$ Then I found that since $\exp(B_T)$ follows log normal then follows the CDF, we have $$P(\exp(B_T)-A\geq0)=1-\Phi(\ln A/ \sqrt{T})$$ Therefore, $$M(\exp(B_T)-A\geq0) \geq A\exp(-T/2)[1-\Phi(\ln A/ \sqrt{T})]$$
Then I stop, since the question asks to find the exact answer, not the lower bound..but this is how I did and don't know how to proceed further..
Any help will be extremely helpful!
Let $K=[\mathrm e^{B_T}\gt A]$, then $M(K)=E[\mathrm e^{B_T-T/2}\mathbf 1_K]=\mathrm e^{-T/2}\cdot E[\mathrm e^{B_T};B_T\gt\log A]$. For every $x$, $$ E[\mathrm e^{B_T};B_T\gt x]=\int_x^{+\infty}\mathrm e^z\frac{\mathrm e^{-z^2/2T}}{\sqrt{2\pi T}}\mathrm dz. $$ One completes the square as usual, that is, one uses the change of variable $z=T+\sqrt{T}u$, which yields $$ E[\mathrm e^{B_T};B_T\gt x]=\mathrm e^{T/2}\int_{(x-T)/\sqrt{T}}^{+\infty}\frac{\mathrm e^{-u^2/2}}{\sqrt{2\pi}}\mathrm du=\mathrm e^{T/2}\,\Phi(\sqrt{T}-x/\sqrt{T}). $$ Finally, $$ M[\mathrm e^{B_T}\gt A]=\Phi(\sqrt{T}-(\log A)/\sqrt{T}). $$