Suppose we have a process $(X_t)$ such that under a measure $Q$ we know that
$$X_t=\mathcal{E}\left(\int_0^t\lambda(s) \, dW(s)\right)$$
for a deterministic function $\lambda(s)$ and a Brownian Motion $W$ under $Q$. Hence $X_t$ is lognormal distributed with mean $\mu:=-\frac{1}{2}\int_0^t |\lambda(s)|^2ds $ and variance $\sigma^2:=\int_0^t |\lambda(s)|^2ds$. Now, the process $(X_t)$ is adapted w.r.t to a filtration $(\mathcal{F}_t)$. Let $K$ be a positive constant, $t>s$ how do I calculate the following:
$$E_Q[\mathbf1_{\{X_t>K\}}|\mathcal{F}_s]$$
If we need, we can assume that $\mathcal{F}$ is generated by the Brownian motion $W$.
Call $X_t^\lambda$ the exponential martingale defined in the post. Note that $X^\lambda_t=X^\lambda_sY$ where $Y$ is independent of $\mathcal F_s$ and distributed like $X^{\lambda_s}_{t-s}$ where $\lambda_s(\cdot)=\lambda(s+\cdot)$. Thus, $$ Q(X^\lambda_t\gt K\mid\mathcal F_s)=G(K(X^\lambda_s)^{-1}),\qquad G:x\mapsto Q(X^{\lambda_s}_{t-s}\gt x). $$ Note that $X^{\lambda_s}_{t-s}=\exp(\mu+\sigma Z)$ where $\mu=\mu_t-\mu_s$, $\sigma^2=\sigma^2_t-\sigma^2_s$ and $Z$ is standard normal. Thus, for every $x$, $G(x)=Q(\sigma Z\gt\log x-\mu)$, that is, $G(x)=\Phi((\mu-\log x)/\sigma)$. Finally, $$ Q(X_t\gt K\mid\mathcal F_s)=\Phi\left(\frac{\mu_t-\mu_s-\log K+\log X_s}{\sqrt{\sigma^2_t-\sigma^2_s}}\right). $$