Why is the Brownian Motion related to the Normal Distribution?

Question

I am trying to learn more about the Inverse Gaussian Distribution and its applications (e.g. First Passage Time).

First, define a Brownian Motion with Drift: If $X(t) = \mu t + W(t)$ represents a Brownian motion with drift $\mu$, where $W(t)$ is a standard Brownian motion (i.e. $W(t) \sim N(0,t)$). Here, the initial position of this Brownian Motion is 0, i.e. $W(t=0)=0$. The drift term is important because without drift, the Inverse Gaussian Distribution is not defined.

Next, define First Passage Time: The first passage time $T$ is the time it takes for the process $X(t)$ to reach a certain level $a > 0$ for the first time.

Finally, define the Inverse Gaussian Distribution: The first passage time $T$ follows an Inverse Gaussian distribution with parameters $\mu = \frac{a}{\mu}$ and $\lambda = a^2$, i.e., $T \sim IG(\frac{a}{\mu}, a^2)$ :

$$ f(x;\mu,\lambda) = \left(\frac{\lambda}{2\pi x^3}\right)^{1/2} e^{ -\frac{\lambda(x-\mu)^2}{2\mu^2 x} } $$

where:

• $x > 0$ is the variable
• $\mu > 0$ is the mean
• $\lambda > 0$ is the shape parameter.

Note: It's clear to see that if the drift term $\mu$ is 0, we would end up with division by zero when trying to define the mean of the Inverse Gaussian distribution, which makes it undefined: When $\mu = 0$, the term $\frac{\lambda(x-\mu)^2}{2\mu^2 x}$ in the exponent of the distribution becomes $\frac{\lambda x^2}{2 \cdot 0^2 \cdot x}$, which is undefined because we're dividing by zero.

Here is where I get confused: I can not understand the relationship between the Brownian Motion with Drift and the Inverse Gaussian Distribution. That is, (given some initial conditions) is it possible to mathematically manipulate a Brownian Motion $X(t)$ and mathematically demonstrate that the time at which $X(t)$ will reach a certain point is indeed given by the Inverse Gaussian Distribution?

For example, suppose $T$ is the first passage time of a Brownian motion with drift to reach a certain level $a > 0$.

$T$ follows an Inverse Gaussian distribution with parameters $\mu = \frac{a}{\mu}$ and $\lambda = a^2$, i.e., $T \sim IG(\frac{a}{\mu}, a^2)$.

The probability that $T$ falls in the interval $[t_1, t_2]$ can be written as:

$$ P(t_1 \leq T \leq t_2) = \int_{t_1}^{t_2} f(t;\mu,\lambda) dt $$

where $f(t;\mu,\lambda)$ is the probability density function of the Inverse Gaussian distribution:

$$ f(t;\mu,\lambda) = \left(\frac{\lambda}{2\pi t^3}\right)^{1/2} e^{ -\frac{\lambda(t-\mu)^2}{2\mu^2 t} } $$

The expected value of $T$ is equal to the mean parameter of the Inverse Gaussian distribution, which is $\mu = \frac{a}{\mu}$. This represents the average amount of time needed for the Brownian motion with drift to reach the level $a$.

But I am still confused: Why is the distribution of the first passage time an Inverse Gaussian distribution in the first place? I get it that a Brownian Motion is related to a Gaussian Distribution via the Wiener Process $W(t)$ ... but why is the distribution of the first passage time an Inverse Gaussian distribution?

Thanks!

PS: I tried to write the Inverse Gaussian Distribution for a Brownian Motion that does not start at 0.

If we define a Brownian Motion with drift such that the position at time =0 is given by $x_0$:

$$X(t) = x_0 + \mu t + W(t)$$

Then the Inverse Gaussian Distribution for this modified Brownian Motion becomes:

$$T' \sim IG\left(\frac{a - x_0}{\mu}, (a - x_0)^2\right)$$

$$ f(t;\mu',\lambda') = \left(\frac{\lambda'}{2\pi t^3}\right)^{1/2} e^{ -\frac{\lambda'(t-\mu')^2}{2\mu'^2 t} } $$

where $\mu' = \frac{a - x_0}{\mu}$ and $\lambda' = (a - x_0)^2$.

Answer 1

Let $W^{(\mu)}_t = \mu t + W_t$ denote the Brownian motion with drift starting at zero, and let $T^{(\mu)}_x=\inf\{t> 0: W^{(\mu)}_t\geq x\}$ denote the hitting time of $W^{(\mu)}_t$ to some $x>0$. Notice that the following two events are equivalent: $$ \{ T^{(\mu)}_x\leq t \} \Leftrightarrow \{\sup_{h\in[0,t]}W^{(\mu)}_h \geq x \},$$ since for $W^{(\mu)}_t$ to hit $x$ before time $t$, its maximum value on $[0,t]$ must exceed $x$. To simplify notation, let us write $M_t^{(\mu)} :=\sup_{h\in[0,t]}W^{(\mu)}_h $, and the two distributions satisfy the equality: $$ \Pr(T^{(\mu)}_x\leq t) = \Pr(M^{(\mu)}_t \geq x ).$$ Intuitively, this is where the Gaussian density gets 'inversed', as the above directly links the distribution of $M^{(\mu)}_t$, the behaviour of Brownian path up to time $t$, to that of the hitting time $T^{(\mu)}_x$.

This relation can be used to derive the density of $T^{(\mu)}_x$. The probability on the RHS has the form: $$ \Pr(M^{(\mu)}_t \geq x ) = \int_{-\infty}^{\infty}\Pr(M^{(\mu)}_t \geq x, W_t^{(\mu)}\in du ) = \Pr(T^{(\mu)}_x\leq t).$$ where the joint density of $(M^{(\mu)}_t, W_t^{(\mu)})$ can be found here, which can be derived by a standard Girsanov change of measure. Let us denote this joint density as $f_{M,W}(m,w;\mu,t)$, we have: $$ f_{M,W}(m,w;\mu,t) = e^{\mu w- \mu^2t/2}\frac{1}{\sqrt{2\pi t^3}}2(2m-w)e^{-\frac{(2m-w)^2}{2t}}\\ =-2e^{\mu w- \mu^2t/2}\phi'_{t}(2m-w), m>0, m>w,$$ where $\phi_{t}(x) = (2\pi t)^{-1/2}e^{-x^2/(2t)}$ is the Gaussian density function of $W_t$, and $\phi_t'(x)=-x\phi_{t}(x)/t$ is its derivative w.r.t. $x$. If we integrate out $m$ from $\max\{0,w\}$ to $\infty$ in the above density, we would get a Gaussian density for $W_t^{(\mu)}$ by construction.

Let us now derive the density of $T_x^{(\mu)}$. We start from: \begin{equation} \Pr(T^{(\mu)}_x\leq t) = \Pr(M^{(\mu)}_t \geq x ) =\int_{-\infty}^\infty \int_{x\vee w}^\infty f_{M,W}(m,w;\mu,t)dm dw, \end{equation} where $x\vee w$ means the larger of $x$ or $w$ (since $M_t^{(\mu)}$ must be larger than the endpoint $W_t^{(\mu)}$). For the inner integral: $$ \int_{x\vee w}^\infty -2e^{\mu w- \mu^2t/2}\phi'_{t}(2m-w) dm =-e^{\mu w- \mu^2t/2} \phi_{t}(2m-w)\Big|_{x\vee w}^\infty =e^{\mu w- \mu^2t/2} \phi_{t}(2(x\vee w)-w)\\ = \begin{cases} e^{\mu w- \mu^2t/2} \phi_{t}(2x-w), \quad x>w\\ e^{\mu w- \mu^2t/2} \phi_{t}(w), \quad x<w \end{cases}$$ The outer integral can thus be split into: $$ \int_{x}^\infty e^{\mu w- \mu^2t/2} \phi_{t}(w)dw+\int_{-\infty}^x e^{\mu w- \mu^2t/2} \phi_{t}(2x-w)dw$$ The first one is relatively easy: $$ \int_{x}^\infty e^{\mu w- \mu^2t/2} \phi_{t}(w)dw = \int_{x}^\infty\phi_{t}(w-\mu t) = 1-\Phi_t(x-\mu t)=\Phi_1(\mu\sqrt{t}-x/\sqrt{t}), $$ where $\Phi_{t}(x)=\int_{-\infty}^x\phi_t(x)dx$ is the CDF of a normal distribution with zero mean and variance $t$. For the second one: $$\int_{-\infty}^x e^{\mu w- \mu^2t/2} \phi_{t}(2x-w)dw=\int_{-\infty}^x e^{2\mu x}\frac{1}{\sqrt{2\pi t}}e^{-\frac{(w-\mu t-2x)^2}{2t}}= \int_{-\infty}^x e^{2\mu x}\phi_t(w-\mu t-2x)dw\\ = e^{2\mu x}\Phi_t(w-\mu t-2x)\Big|_{-\infty}^x = e^{2\mu x}\Phi_t(-\mu t-x) =e^{2\mu x}\Phi_1(-\mu\sqrt{t}-x/\sqrt{t}). $$ Finally, let $f_T(t;x,\mu)$ denote the density of $T_x^{(\mu)}$, we have: $$ f_T(t;x,\mu) = \frac{\partial \Pr(T^{(\mu)}_x\leq t)}{\partial t} = \frac{\partial (\Phi_1(\mu\sqrt{t}-x/\sqrt{t})+e^{2\mu x}\Phi_1\left(-\mu\sqrt{t}-x/\sqrt{t}\right))}{\partial t} \\ = \left(\frac{x}{2t^{3/2}}-\frac{\mu}{2t^{1/2}} \right) \phi_1(x/\sqrt{t}-\mu\sqrt{t}) + e^{2\mu x}\left(\frac{x}{2t^{3/2}}+\frac{\mu}{2t^{1/2}} \right) \phi_1(-x/\sqrt{t}-\mu\sqrt{t}) \\ = \left(\frac{x}{2t^{3/2}}-\frac{\mu}{2t^{1/2}} \right)\phi_1(x/\sqrt{t}-\mu\sqrt{t}) +\left(\frac{x}{2t^{3/2}}+\frac{\mu}{2t^{1/2}} \right) \phi_1(x/\sqrt{t}-\mu\sqrt{t})\\ = \frac{x}{t^{3/2}}\phi_1\left(\frac{x-\mu t}{\sqrt{t}} \right)=\frac{x}{\sqrt{2\pi t^3}}e^{\frac{-(x-\mu t)^2}{2t}}, $$ which is the desired inverse Gaussian distribution. Note that one can simply replace $ x= a - x_0$ in the above density to get the hitting time density of $W^{(\mu)}_t + x_0$ to $a$.