Asymptotic form of pdf of Escape Time of arithmetic fBm 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.
 A: Suppose $b=1$, for simplicity.  
According to the OP, $(B^H_t)_{t \ge 0}$ is a standard fractional Brownian motion with parameter $H \in (0,1)$ under probability measure $\mathbb{P}$, and the measure $\mathbb{Q}$ is defined by $$
\frac{d \mathbb{Q}}{d \mathbb{P}} (B^H_{[0,T]}) =\exp\left(a M_T -\frac{1}{2} a^2 c^2 T^{2-2H}\right) \;.
$$ 
By Girsanov's Theorem for fractional Brownian motion (see Thm 4.1 in the reference given below), \begin{align*}
\mathbb{Q}(A) = \mathbb{E}_{\mathbb{Q}}(I(A)) &= \mathbb{E}_{P}\left\{ I(A) \frac{d\mathbb{Q}}{d \mathbb{P}} \right\} \\
&=\mathbb{E}_{P}\left\{ I(A) \exp\left(a M_T -\frac{1}{2} a^2 c^2 T^{2-2H} \right)\right\} \\
&= \mathbb{E}_{P}\left\{ I(A) \exp\left(a M_T -\frac{1}{2} a^2 c^2 T^{2-2H} \right) \right\} \\
&= \mathbb{E}_{P}\left\{ \exp\left(a M_T -\frac{1}{2} a^2 c^2 T^{2-2H} \right) \mid A \right\} \mathbb{P}(A)
\end{align*}
If $A=\{ \tau_k \in (t, t+d t) \}$, then $$
\mathbb{E}_{P}\left\{ \exp\left(a M_t -\frac{1}{2} a^2 c^2 t^{2-2H}\right) \mid A \right\} = \exp\left( -\frac{1}{2} a t^{1- 2H} (-2 k+a t) \right) 
$$
which follows from the elementary lemma given below with $X = M_t$, $Y=Z_t$, $\sigma_X^2 = c^2 t^{2-2 H}$, $\sigma_Y^2 = t^{2 H}$, and $\rho = 1/c$.   

Lemma Let $X$ and $Y$ be centered, normal random variables with variances $\sigma_X^2$ and $\sigma_Y^2$ (respectively), and correlation coefficient $\rho$.  Then $$
\mathbb{E}\{ e^{a X} \mid Y=k \} = e^{\frac{a \sigma_X}{2 \sigma_Y} (2 k \rho + a (1-\rho^2) \sigma_X \sigma_Y )} \;.
$$

I surmise that this lemma was the main missing ingredient in the OP's derivation attempt.  It is easy to check that one recovers the standard Brownian motion result when $H=1/2$.  
Reference
Norros, Ilkka; Valkeila, Esko; Virtamo, Jorma, An elementary approach to a Girsanov formula and other analytical results on fractional Brownian motions, Bernoulli
   5, No.4, 571-587 (1999).
