# 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 $$\mbox{d}X \left(t\right) = a \mbox{d}t + b \mbox{d}{W}\left(t\right).$$ 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 $$\tau \sim IG\left(\frac{B - X_0}{a}, \frac{\left(B - X_0\right)^2}{b^2}\right),\label{abmFirstPassageDist}$$ and for $t > 0$ the pdf of $\tau$ is $$f(t) = \sqrt{\frac{(B - X_0)^2}{2 \pi b^2 t^3}} \exp\left[-\frac{ \left(at - B + X_0\right)^2}{2 b^2 t}\right]\label{abmFirstPassageDensity}.$$

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 $\mathbb{P}$ $$\frac{\mbox{d}{\mathbb{Q}}}{\mbox{d}{\mathbb{P}}} = \exp\left\{a M_T - \frac{1}{2}{a}^2 \langle M, M\rangle_T \right\}\label{girsanovRadonNikodymDerivativeWithInnerProduct}$$ where $$M_T = \frac{1}{2 H \Gamma \left(\frac{3}{2} - H\right) \Gamma \left(H + \frac{1}{2}\right)} \int_0^T \left(s\left(T - s\right)\right)^{\frac{1}{2} - H} \mbox{d}{{B_s}^H}.$$ The process $M_T$ is a martingale with independent increments, zero mean and variance function $c^2 T^{2 - 2H}$ where $$\label{c} c = \sqrt{\frac{\Gamma\left(\frac{3}{2} - H\right)}{2 H \left(2 - 2H\right)\Gamma\left(H + \frac{1}{2}\right)\Gamma\left(2 - 2H\right)}}.$$ 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 $$\frac{\mbox{d}{\mathbb{Q}}}{\mbox{d}{\mathbb{P}}} = \exp\left\{a M_T - \frac{1}{2}{a}^2 c^2 T^{2 - 2H} \right\}.\label{girsanovRadonNikodymDerivative}$$

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 $\mathbb{P}$ $$\frac{\mbox{d}{\mathbb{Q}}}{\mbox{d}{\mathbb{P}}} = \exp\left\{\frac{a}{b} M_T - \frac{1}{2}\frac{a^2}{b^2} c^2 T^{2 - 2H}\right\}.\label{scaledGirsanovRadonNikodymDerivative}$$ 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 $$E\left(M_t | B^H \left(t\right) = k \right) = {t^{1-2H} k \over b}$$ and $$\hbox{Var}\left(M_t | B^H \left(t\right) = k\right) = t^{2-2H}\left(c^2-1\right).$$

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 $$\exp\left\{\frac{a}{b} M_T - \frac{1}{2}\frac{a^2}{b^2} \langle M, M\rangle_T\right\}\nonumber$$ on $\left[0, T\right]$ and the fact that $\left\{\tau_k \le T \right\} \in \mathcal{F}_{\tau_k} \cap \mathcal{F}_{T} = \mathcal{F}_{\tau_k \bigwedge T} \subseteq \mathcal{F}_T$ ), $$\mathrm{E}\left[\left.\exp\left\{\frac{a}{b} M_T - \frac{1}{2}\frac{a^2}{b^2} \langle M, M\rangle_T\right\} \right| \mathcal{F}_{\tau_k \bigwedge T}\right] = \exp\left\{\frac{a}{b} M_{\tau_k \bigwedge T} - \frac{1}{2}\frac{a^2}{b^2} \langle M, M\rangle_{\tau_k \bigwedge T}\right\}.$$ 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 $$\mathbb{P}^{a,T} \left[\tau_k \le T\right] = \mathbb{P}\left[\hat{\tau_k} \le T\right],$$ 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.

• The density $f'_{H=1/2}$ is not a PDF: it is neither integrable nor does it integrate to $1$. Dec 27, 2017 at 20:26
• It is an asymptotic form of the density not the actual density Dec 27, 2017 at 21:55
• Could you define what you mean by asymptotic form of the density? Dec 27, 2017 at 22:04
• @NawafBou-Rabee , I do n't have a good precise definition of the Asymptotic form handy, I will search for it, but in essence it is the function obtained as an approximation of pdf, which captures how the density scales as function of $t$, for large $t$. I think it is also known as "long time form". Dec 27, 2017 at 23:23
• @NawafBou-Rabee I have added a theorem which precisely states the pdf result for arithmetic Bm case, which is well known all ready, pardon the variables used in the two results may be a little different, so in case of arithmetic BM. i guess $f'_{H=\frac{1}{2}}$ gives the asymptotic form of the pdf given in the theorem. Dec 27, 2017 at 23:27

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).

• Thank you so much, i see the mistake in my approach. I had gotten confused, if i needed to use the conditional variance as you can see. It makes sense. I will like to submit this derivation to a "letters" like Journal, it will be awesome, if you will be a co-author. I'll write it up. you just have to review Thoughts ? Dec 29, 2017 at 21:21
• Sounds good --- email me details when you're ready. Dec 29, 2017 at 23:41
• See arxiv.org/pdf/math/0606086.pdf for related work. Dec 30, 2017 at 19:09
• Thank your for sharing the link. I was familiar with that work, i will try cite it in the literature review section. I have sent you an email just now, please let me know, if you got it. Dec 30, 2017 at 20:20