Gaussian vs Inverse Gaussian Distribution?

Question

In our Finance class, we learned that if you have a Brownian Motion - then the time at which this Brownian Motion first passes a certain point ... this time is a Random Variable that follows an Inverse Gaussian Distribution. We were just told the final result - but I am interested and want to try and derive this myself... but I got stuck.

Here is my work:

Start with a Brownian Motion $X_t$ with drift $\mu$ and variance $\sigma$. Define $W_t$ as a random variable such that $W_t \sim N(0, t)$. We can write $X_t$ like this:

$$X_t = \mu \cdot t + \sigma \cdot W_t$$

Now, we define the first passage time $T_a$ as the time when the Brownian motion $X_t$ first passes the point $a$. We can update the earlier equation to contain this idea:

$$ X_{T_a} = a = \mu \cdot T_a + \sigma W_{T_a}$$

We can solve for $W_{T_a}$ :

$$W_{T_a} = \frac{a - \mu T_a}{\sigma}$$

Since $W_{T_a}$ is originally based on a Normal Distribution, we can write:

$$ W_{T_a} \sim N(0, T_a)$$

Now we have to start working towards analyzing $T_a$. For a second, lets define a new variable $G$ as:

$$G = \frac{a - \mu T_a}{\sigma}$$ $$G \sim N(0, T_a) $$

Using the properties of a Standard Normal Distribution (i.e. Normal Distribution with mean=0 and sigma=1), we say that:

$$Z = \frac{G - E(G)}{\sqrt{(Var(G)}} \sim N(0,1)$$

$$E(G) = 0$$ $$Var(G) = T_a$$

A standard Normal Distribution squared is equal to a Chi-Squared Distribution with 1 degree of freedom. Thus:

$$[\frac{G-0}{\sqrt{T_a}}]^2 \sim \chi^2_1 $$

Doing some manipulations, we get:

$$ \left[ \frac{a - \mu T_a}{\sigma^2 \sqrt{T_a}} \right]^2 = \frac{(a - \mu T_a)^2}{\sigma^2 T_a} \sim \chi^2_1 $$

We know that $E(\chi^2_k) = k$, thus $E(\chi^2_1) = 1$. We can use to write:

$$ E \left[ \frac{(a - \mu T_a)^2}{\sigma^2 T_a} \right] = E[\chi^2_1] = 1 $$ $$ (a - \mu T_a)^2 = \sigma^2 T_a $$

Using the Quadratic Formula, we can write:

$$ T_a = \frac{(2a \mu + \sigma^2) \pm \sqrt{ (2a \mu + \sigma^2)^2 - 4\mu^2 a^2}}{2\mu^2} $$

Since $T_a$ refers to a time and time can only be positive, we only consider the positive part of the above expression. Therefore (after some more simplification), we can write:

$$ T_a = \frac{2a\mu + \sigma^2 + \sigma \sqrt{4a\mu + \sigma^2}}{2\mu^2} $$

This is where I got stuck : I am confused how I can relate $T_a$ to an Inverse Gaussian Distribution.

I know that an Inverse Gaussian Distribution can be written like this:

$$ f(x;\mu,\lambda) = \sqrt\frac{\lambda}{2 \pi x^3} \exp\biggl(-\frac{\lambda (x-\mu)^2}{2 \mu^2 x}\biggr)$$

$$E(X) = \mu$$ $$Var(X) = \frac{\mu^3}{\lambda}$$

But I am confused how to relate to this $T_a$ .

My guess is that I can take the formula for a Inverse Gaussian, make some arbitrary modifications to it (e.g. introducing some multiplicative scaling factors) and show that somehow after these modifications, the Expected Value and Variance of this modified Inverse Gaussian Distribution somehow relate to $T_a$ ... but I am not sure if this is a valid/legitimate way of deriving something in math.

Can someone please help me finish this derivation? I am thinking perhaps something involving Moment Generating Functions?

Answer 1

You almost had it, but things went south after evaluating the expected values. You should have kept $Z$ instead of the illicit extraction of $T_a$ out of the expected value. From: $$ Z = \sqrt{\frac{(a-\mu T_a)^2}{\sigma^2T_a}} $$ you can directly deduce the pdf of $T_a$: $$ \begin{align} dP &= \frac{\exp(-z^2/2)}{\sqrt{2\pi}}dz \\ &= \frac1{\sqrt{2\pi}}\exp\left(-\frac{(a-\mu t)^2}{2\sigma^2t}\right)\frac{dz}{dt}dt \end{align} $$ I will let you calculate the derivative to complete the pdf.

However, this only gives you the distribution of all times crossing $a$, since you just imposed $a=X_T$. This includes the first crossing but all the subsequent crossings of your random walker on the line. To find the correct distribution, you also need to impose: $$ \forall t<T, X_t<a $$ Computing the corresponding probability is not easy as this is an intersection of events that are not independent.

You can get the correct answer by thinking physically. Imagine you have many random walkers, so that you have a diffusion process. Your particles start at $0$ and are free to walk in $x<a$. At $a$ you have an absorbing boundary. You therefore have the Fokker-Planck equation: $$ \partial_t\rho+\mu\partial_x\rho = \frac\sigma2\partial_x^2\rho $$ with the boundary condition $\rho(x=a) = 0$ and initial condition $\rho(t=0)=\delta$. By the method of images or mode exampnsion, you can solve it to be: $$ \rho = \frac1{\sqrt{2\pi\sigma t}}\exp\left(-\frac{(x-\mu t)^2}{2\sigma t}\right)-\frac1{\sqrt{2\pi\sigma t}}\exp\left(-\frac{(2a-x+\mu t)^2}{2\sigma t}\right) $$ The current of particles is: $$ j = \mu \rho-\frac\sigma2\partial_x\rho $$ so $j(x=a,t)$ is precisely the pdf of $T_a$, the number of particles who cross $x=a$ at time $t$. Due to the absorbing boundary, the particles cannot cross again, and this does give you the first crossing time. This is why the expression is different from the first method. Note that this method generalises quite well in higher dimensions.