I am trying to apply the Girsanov formula and Doobs optional sampling theorem to obtain an asymptotic form of first passage density of an fbm process with drift, but the answer i am getting seems strange, it does not match the brownian motion case result as well as i assumed it would. I am seeking help in debugging these computations.

I am also very much interested in learning about analytical ways of verifying this result.

Thank You

**Some known results**

The first passage density of Brownian motion case is given by the following theorem.

**Theorem:**
Let the arithmetic Brownian motion process $X \left(t\right)$ be defined by the following Brownian motion driven SDE
\begin{equation}
\mbox{d}X \left(t\right) = a \mbox{d}t + b \mbox{d}{W}\left(t\right).
\end{equation}
with initial value $X_0$. Let $\tau =\inf \left(u |X(u) \le B\right)$ denote the first passage time for the barrier $X_0 < B$. Then the first passage time $\tau$ is distributed as Inverse Gaussian Distribution
\begin{equation}
\tau \sim IG\left(\frac{B - X_0}{a}, \frac{\left(B - X_0\right)^2}{b^2}\right),\label{abmFirstPassageDist}
\end{equation}
and for $t > 0$ the pdf of $\tau$ is
\begin{equation}
f(t) = \sqrt{\frac{(B - X_0)^2}{2 \pi b^2 t^3}} \exp\left[-\frac{ \left(at - B + X_0\right)^2}{2 b^2 t}\right]\label{abmFirstPassageDensity}.
\end{equation}

**The Girsanov formula for fBm** Let $B^H \left(t\right)$ denote fractional brownian motion with mean $0$ and variance $t^{2H}$, for all $H \in \left(0, 1\right)$ defined on $\left(\Omega, \mathcal{F}, \mathbb{P}\right)$. Let $a$ be a scalar. Define a new probability measure $\mathbb{Q}$ on $\left(\Omega, \mathcal{F}\right)$ via the Radon Nikodym derivative with respect to $\mathbb{P}$ \begin{equation}
\frac{\mbox{d}{\mathbb{Q}}}{\mbox{d}{\mathbb{P}}} = \exp\left\{a M_T - \frac{1}{2}{a}^2 \langle M, M\rangle_T \right\}\label{girsanovRadonNikodymDerivativeWithInnerProduct}
\end{equation}
where
\begin{equation}
M_T = \frac{1}{2 H \Gamma \left(\frac{3}{2} - H\right) \Gamma \left(H + \frac{1}{2}\right)} \int_0^T \left(s\left(T - s\right)\right)^{\frac{1}{2} - H} \mbox{d}{{B_s}^H}.
\end{equation}
The process $M_T$ is a martingale with independent increments, zero mean and variance function $c^2 T^{2 - 2H}$ where
\begin{equation}\label{c}
c = \sqrt{\frac{\Gamma\left(\frac{3}{2} - H\right)}{2 H \left(2 - 2H\right)\Gamma\left(H + \frac{1}{2}\right)\Gamma\left(2 - 2H\right)}}.
\end{equation}
Then the process defined, for all $t \in \left[0, T\right]$, by $B^H \left(t\right) + a t$ is the standard $\mathbb{Q}$-fractional Brownian motion on $\left[0, T\right]$. In other words, under probability measure $\mathbb{Q}$,
$B^H \left(t\right)$ restricted to $t \in \left[0, T\right]$ is distributed as an arithmetic fractional Brownian motion with drift $a$.

A proof of the Girsanov formula for fractional Brownian motion can be found in Norros's paper, where the term the fundamental martingale is also coined for the process $M_t$. It's noteworthy that using the variance of $M_t$, Radon Nikodym Derivative can also be re-written as \begin{equation} \frac{\mbox{d}{\mathbb{Q}}}{\mbox{d}{\mathbb{P}}} = \exp\left\{a M_T - \frac{1}{2}{a}^2 c^2 T^{2 - 2H} \right\}.\label{girsanovRadonNikodymDerivative} \end{equation}

**corollary**
Let $X\left(t\right) = b B^H \left(t\right)$ be an arithmetic fractional brownian motion with volatility $b$, for all $H \in \left(0, 1\right)$ defined on $\left(\Omega, \mathcal{F}, \mathbb{P}\right)$. Let $a$ be a scalar. Define a new probability measure $\mathbb{Q}$ on $\left(\Omega, \mathcal{F}\right)$ via the Radon Nikodym derivative with respect to $\mathbb{P}$
\begin{equation}
\frac{\mbox{d}{\mathbb{Q}}}{\mbox{d}{\mathbb{P}}} = \exp\left\{\frac{a}{b} M_T - \frac{1}{2}\frac{a^2}{b^2} c^2 T^{2 - 2H}\right\}.\label{scaledGirsanovRadonNikodymDerivative}
\end{equation}
Then the process defined, for all $t \in \left[0, T\right]$, by $Z \left(t\right) = X \left(t\right) + a t$ is an arithmetic $\mathbb{Q}$-fractional Brownian motion process on $\left[0, T\right]$ with volatility $b$. In other words, under probability measure $\mathbb{Q}$, $X\left(t\right)$ restricted to $t \in \left[0, T\right]$ is distributed as an arithmetic fractional Brownian motion with drift $a$ and volatility $b$.

**Proposition**
Let $B^H \left(t\right)$ denote scaled fractional brownian motion with
mean $0$ and variance $b^2 t^{2H}$, for all $H \in \left(0, 1\right)$
with respect to measure $\mathbb{P}$. Define $\tau_k = \inf \left\{t \ge
0 : B^H \left(t\right) = k\right\}$ for $k > 0$. Then the conditional
mean and variance of $M_t$ given $B_t$ are
\begin{equation}
E\left(M_t | B^H \left(t\right) = k \right) = {t^{1-2H} k \over b}
\end{equation}
and
\begin{equation}
\hbox{Var}\left(M_t | B^H \left(t\right) = k\right) = t^{2-2H}\left(c^2-1\right).
\end{equation}

**Proof**
Both $B^H \left(t\right)$ and $M_t$ have mean zero, the variance
of $B^H \left(t\right)$ is $b^2 t^{2H}$, the variance of $M_t$ is $c^2 T^{2 - 2H}$,
and their covariance $bt$, can be derived
similarly to, as in Proposition 3.2, in Norros's paper. Hence
the correlation coefficient $\rho$ between $M_t$, $B^H \left(t\right)$
is $1/c$. Therefore, using elementary results for the bivariate normal
distribution, we find
\begin{eqnarray}
E\left(M_t | B^H \left(t\right) = k \right) &=& {\rho \sigma_{M_t}\over \sigma_{B^H \left(t\right)}}k\\\nonumber
&=& {t^{1-2H} k \over b}\nonumber
\end{eqnarray}
and
\begin{eqnarray}
\hbox{Var}\left(M_t | B^H \left(t\right) = k\right) &=& {\sigma^2}_{M_t}\left(1 - \rho^2\right) \\\nonumber
&=& t^{2-2H}\left(c^2-1\right).\nonumber
\end{eqnarray}

**The Derivation Attempt**
By Doob's optional sampling theorem (justified by the uniform integrability of the martingale
\begin{equation}
\exp\left\{\frac{a}{b} M_T - \frac{1}{2}\frac{a^2}{b^2} \langle M, M\rangle_T\right\}\nonumber
\end{equation}
on $\left[0, T\right]$
and the fact that $\left\{\tau_k \le T \right\} \in \mathcal{F}_{\tau_k}
\cap \mathcal{F}_{T} = \mathcal{F}_{\tau_k \bigwedge T} \subseteq \mathcal{F}_T$ ),
\begin{equation}
\mathrm{E}\left[\left.\exp\left\{\frac{a}{b} M_T - \frac{1}{2}\frac{a^2}{b^2} \langle M, M\rangle_T\right\} \right| \mathcal{F}_{\tau_k \bigwedge T}\right]
= \exp\left\{\frac{a}{b} M_{\tau_k \bigwedge T} - \frac{1}{2}\frac{a^2}{b^2} \langle M, M\rangle_{\tau_k \bigwedge T}\right\}.
\end{equation}
Therefore,
\begin{eqnarray}
\mathbb{P}^{a,T}\left[\tau_k \in (t,t+dt)\right] &=& \mathbb{E}^{a,T} \left[ 1_{\left\{\tau_k \in (t,t+dt)\right\}} \right]\\
\nonumber
&=& \mathbb{E}\left[\exp\left\{\frac{a}{b} M_T - \frac{1}{2}\frac{a^2}{b^2} \langle M, M\rangle_T\right\}1_{\left\{\tau_k \in (t,t+dt)\right\}}\right]
\end{eqnarray}
by Corollary above
\begin{eqnarray}
\nonumber
&=& \mathbb{E}\left[\left.\mathbb{E}\left[\exp\left\{\frac{a}{b} M_{\tau_k \bigwedge T}
- \frac{1}{2}\frac{a^2}{b^2} \langle M, M\rangle_{\tau_k \bigwedge T}\right\} \right|{\cal F}^{B^H}_{\tau_k\wedge T}\right]1_{\left\{\tau_k \le T\right\}}\right]\\
\nonumber
&=& \mathbb{E}\left[\exp\left\{\frac{a}{b} M_{\tau_k} - \frac{1}{2}\frac{a^2}{b^2} \langle M, M\rangle_{\tau_k}\right\}1_{\left\{\tau_k \le T\right\}}\right]\\
\end{eqnarray}
by Doob's sampling theorem
\begin{eqnarray}
\nonumber
&=& \mathbb{E}\left[\exp\left\{\frac{a k}{b^2}{\tau_k}^{1-2H} - \frac{1}{2}\frac{a^2}{b^2} {\tau_k}^{2-2H}\left(c^2 - 1\right)\right\}1_{\left\{\tau_k \le T\right\}}\right]\\
\end{eqnarray}
by propostion above
\begin{eqnarray}
\nonumber
&=& \int_0^T \exp\left[\frac{a k t^{1-2H}}{b^2} - \frac{1}{2}\frac{a^2 t^{2 - 2H}}{b^2} \left(c^2 - 1\right)\right] \mathrm{P}\left[\tau_k \in \mbox{d}{t}\right].
\end{eqnarray}
On the other hand, $\left\{B^H \left(t\right) + a t\right\}_{t \in \left[0, T\right]}$ is a scaled fractional Brownian motion under $\mathbb{P}^{a,T}$, so
\begin{equation}
\mathbb{P}^{a,T} \left[\tau_k \le T\right] = \mathbb{P}\left[\hat{\tau_k} \le T\right],
\end{equation}
where $\hat{\tau_k}$ is the first hitting time of the level $k$ of the scaled fractional Brownian motion
with drift $a$. Using the formula for asymptote of the first passage density of fBM without drift due to Molchan, it follows immediately that the long-time form of the first passage time density for fBm with drift is given by
\begin{eqnarray}
f\left(t\right) &=& \exp\left[\frac{a k t^{1-2H}}{b^2} - \frac{1}{2}\frac{a^2 t^{2 - 2H}}{b^2} \left(c^2 - 1\right)\right] t^{H-2}.\label{firstPassageDensityArithmeticFbm}
\end{eqnarray}

It is noteworthy that $C^2$ has a minimum at $H =\frac{1}{2}$, where $c^2 = 1$.

**Verification Against Brownian motion case**
Upon substituting $H = \frac{1}{2}$ in $c^2$ evaluates to $1$, and the function reduces to
\begin{eqnarray}
f_{H = \frac{1}{2}}\left(t\right) &=& \exp\left[\frac{a k}{b^2}\right] t^{-\frac{3}{2}}.
\end{eqnarray}

This seems a bit off as according to "Brownian motion and stochastic calculus" books, it should be \begin{eqnarray} f'_{H = \frac{1}{2}}\left(t\right) &=& \exp\left[\frac{a k}{b^2} - \frac{a^2}{2b^2}t \}\right] t^{-\frac{3}{2}}. \end{eqnarray} The presence of $c^2 - 1$ seems to prevents a perfect match with the Brownian motion case.

asymptotic form of the density? $\endgroup$ – Nawaf Bou-Rabee Dec 27 '17 at 22:04