# Sinh equation
\begin{align*}
dE&= i \frac{\lambda}{2} E dt+ \frac{2}{\sqrt{\beta t}}\frac{E-E^*}{2i}(dw_1+i dw_2)
\\
dE^*&=-i \frac{\lambda}{2}E^* dt+ \frac{2}{\sqrt{\beta t}} \frac{E-E^*}{2i}(dw_1-i dw_2)
\end{align*}
Let $R=\log E^*/E=l+i \alpha$

\begin{align}
dR =& -i \lambda dt+ \frac{2}{\sqrt{\beta t}} \frac{e^{-R}-1}{2i}(dw_1-i dw_2)- \frac{2}{\sqrt{\beta t}}\frac{1-e^R}{2i}(dw_1+i dw_2)\\
&=-i \lambda dt-i \frac{2}{\sqrt{\beta t}}(\cosh(R)-1)dw_1+ \frac{2}{\sqrt{\beta t}} \sinh(R) dw_2
\end{align}
\begin{align}
d l=&\Im \lambda dt+\frac{2}{\sqrt{\beta t}}(\sin (\alpha ) \sinh (l) dw_1+\cos (\alpha ) \sinh (l)dw_2)\\
=&\Im \lambda dt+\frac{2}{\sqrt{\beta t}}\sinh(l) dB
\end{align}
where
$$
dB=\sin \alpha\, dw_1+\cos \alpha \,dw_2.
$$
Note that $l(\bar \lambda)=-l(\lambda)$.

We consider the case $\Im \lambda>0$. Then $l(0)=0$ and $l(t)>0$ for $t>0$.
In the other timescale we have
\begin{align}
d l=\Im \lambda f(t) dt+\sinh(l) dB
\end{align}
We set $X=\log \tanh (l/2)$, then $X(0)=-\infty$ and $X< 0$.
$$
dX=-\Im\lambda f(t) \sinh X \, dt +\frac12 \coth X\,dt+ dB
$$

Guess: $e^{-X(t)} f(t)\sim O(1)$, which means $X(t)\sim \frac{\beta}{4}t$.

Then $l=2 \operatorname{arctanh}(e^X)\sim 2 e^{\frac{\beta}{4}t}$, then $e^{-l}\sim 1-2 e^{\frac{\beta}{4}t}$.

Q: how to use this for $\frac{E(z)}{E(w)}$
Use the tail bound for hitting $(\beta/4+\epsilon)t$

In the exponential timescale:
\begin{align*}
d \frac{E(z)}{E(w)}
=i \frac{z-w}{2} f(t)\frac{E(z)}{E(w)}dt+\frac{E(z)}{E(w)}\left(\frac{E^*(w)}{E(w)}-\frac{E^*(z)}{E(z)} \right)(-i dw_1+ dw_2)
\end{align*}

Note that
\begin{align}
\frac{E(z)}{E(w)}=e^{i \frac{z-w}{2}\int_{-\infty}^t f(s) ds} e^{\int_{-\infty}^t dZ}
\end{align}
where
\begin{align}
dZ=\left(\frac{E^*(w)}{E(w)}-\frac{E^*(z)}{E(z)} \right)(-i dw_1+ dw_2)
\end{align}
We can choose $w=0$ to get a representation for $E(z)$.

Check this: it seems that if we look at $|E(z)|$ then its distribution is the same as the distribution of $|E(z)|^{-1} g(\Im z)$ with a deterministic function $g$. This would imply that if the $p$th moment exists for $|E(z)|$ then the same is true for $-p$. (This might not be true...)

Conjecture: on one halfplane we have all moments, and on the other we have the moments that we have on the real line.

This has the same distribution as the solution of
\begin{align}
d\tilde Z=(e^{l(w)+i \alpha(w)}-e^{l(z)+i \alpha(z)})dW
\end{align}
with complex BM $dW$.

It seems that proving
\begin{align}
E[e^{p \int_{-\infty}^t |e^{l(w)+i \alpha(w)}-e^{l(z)+i \alpha(z)}|^2 ds}]<\infty
\end{align}
would imply the existence of all moments of $\frac{E(z)}{E(w)}$. In that case $e^{p \int_{-\infty}^t dZ}=\left(\frac{E(z)}{E(w)}\right)^p e^{- i p \frac{z-w}{2}\int_{-\infty}^t f(s) ds}$
is a proper martingale by Novikov's criterion.

This would follow from
\begin{align}
E[e^{p \int_{-\infty}^0 |\frac{E^*(z)}{E(z)}-1|^2 ds}]<\infty.
\end{align}

If we are in the region where $l<0$ then we can estimate this expectation by considering
\begin{align}
\tau_z=\inf_{t\le 0}\left\{ l(z,t)\ge \min(e^{(\frac{\beta}{4}+\varepsilon)t},1/2) \right\}
\end{align}
(or something similar) and that use $\tau=\min(\tau_z,\tau_w)$ to write
\begin{align}
E[e^{p \int_{-\infty}^0 |e^{l(w)+i \alpha(w)}-e^{l(z)+i \alpha(z)}|^2 ds}]&\le E[e^{2 p \int_{-\infty}^\tau |e^{l(z)}-1|^2+|e^{l(w)-1|^2 ds}} e^{2p |\tau|}]\\
&\le c E[e^{2 p |\tau|}]
\end{align}