Let me define the following SDE: $$\begin{cases} dX_{t}=X_{t}dW_{t}\\ X_{0}=1 \end{cases},$$ where $W$ is an appropriate Wiener process. My question is how we can compute
$$f\left(z\right)=\mathbb{E}\left(X_{2}\mid X_{1}>z\right)?$$
Is there any “analytic” solution? (If I find a solution where the $\varPhi$ standard normal distribution function appears, I will consider it “analytic”, or at least a very “nice” solution.)
It is not necessary to compute it with the $\left\{ X_{1}>z\right\}$ condition, I would be glad if at least I could compute $f\left(y\right)=\mathbb{E}\left(X_{2}\mid X_{1}=y\right)$.
The condition is an $\mathcal{F}_{1}$ measurable event, therefore $f\left(z\right)$ and $f\left(y\right)$ are real functions (and not “random variables”).
It is easy to verify, that the SDE has the following solution: $$X_{t}=\exp\left\{ W_{t}-\frac{1}{2}t\right\} ,$$ so $X_{t}\overset{d}{=}\exp\left\{ N\left(-\frac{1}{2}t,t\right)\right\}$, otherwise: $X_{t}$ is lognormal distributed.
My calculation is the following so far: $$\mathbb{E}\left(X_{2}\mid X_{1}>z\right) =\frac{\mathbb{E}\left(X_{2}\cdot\chi_{\left\{ X_{1}>z\right\} }\right)}{\mathbb{E}\left(\chi_{\left\{ X_{1}>z\right\} }\right)}=\frac{\mathbb{E}\left(\left(X_{2}-X_{1}+X_{1}\right)\cdot\chi_{\left\{ X_{1}>z\right\} }\right)}{\mathbb{P}\left(X_{1}>z\right)}= =\frac{\mathbb{E}\left(\left(X_{2}-X_{1}\right)\cdot\chi_{\left\{ X_{1}>z\right\} }\right)+\mathbb{E}\left(X_{1}\cdot\chi_{\left\{ X_{1}>z\right\} }\right)}{\mathbb{P}\left(X_{1}>z\right)}=\ldots$$
I could calculate the denominator and the second term of the numerator. The problem is $X$ doesn't have independent increments, therefore $X_{2}-X_{1}$ and $\chi_{\left\{ X_{1}>z\right\} }$ are not independent, so I can't factor the expectation: $$\mathbb{E}\left(\left(X_{2}-X_{1}\right)\cdot\chi_{\left\{ X_{1}>z\right\} }\right)\neq\mathbb{E}\left(X_{2}-X_{1}\right)\cdot\mathbb{E}\left(\chi_{\left\{ X_{1}>z\right\} }\right).$$ I know if they were uncorrelated, then it would be enough, but I don't think this is the case. I can't calculate $\mathbb{E}\left(X_{2}\mid X_{1}>z\right)$ as $$\mathbb{E}\left(X_{2}\mid X_{1}>z\right)=\mathbb{E}\left.\left(X_{2}-X_{1}+x\right)\right|_{x>z},$$ since $X_{2}-X_{1}$ is not independent from the condition.
I thought a lognormal “model” would yield a “nice” solution, but I don't know if there is any way to calculate the conditional expectation above analytically (and not with simulation).
I had another idea below. Because of the distribution of $X_{t}$, I can write the following: $$\mathbb{E}\left(X_{2}\mid X_{1}>z\right) =\mathbb{E}\left(\exp\left\{ N\left(0,2\right)-1\right\} \mid\exp\left\{ N\left(0,1\right)-\frac{1}{2}\right\} >z\right).$$
For first glance, I thought it would have had the same result if I had calculated $$\mathbb{E}\left(N\left(0,2\right)-1\mid N\left(0,1\right)-\frac{1}{2}>\ln z\right)$$ before (and as a side note I could have calculated it, because the distribution behind the expression is a multidimensional normal distribution,) and then I wanted to take the exponential because the exponential function is a monotonic transformation. That's right, but because the Jensen inequality $$\exp\left\{ \mathbb{E}\left(N\left(\mu,\sigma^{2}\right)\right)\right\} \leq\mathbb{E}\left(\exp\left\{ N\left(\mu,\sigma^{2}\right)\right\} \right).$$
So it doesn't work either. Anyway, is there any case when $$\exp\left\{ \mathbb{E}\left(\xi\right)\right\} =\mathbb{E}\left(\exp\left\{ \xi\right\} \right),$$ where $\xi$ is a random variable?