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Consider the following stochastic differential equation (SDE)

$$d X_s= \mu (X_s + b)ds + \sigma X_s d w_s $$

where constants $\mu, \sigma, b > 0$ and initial position $X_0$ are given.

If $b=0$, then the above equation is a geometric Brownian motion (GBM) and the distribution of $X_t$ at time $t$ is lognormally distributed. If $b>0$, can I say anything about the distribution of $X_t$ at a later time $t$? Is it possible to find the probability that $X_t \in (B, B+1)$?

Rodrigo de Azevedo
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S_R
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2 Answers2

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Simplifying the Drift

As will all linear SDEs, let $Y_t=e^{-\mu t}X_t$. Then, \begin{align*} \text{d}Y_t &=-\mu e^{-\mu t}X_t\text{d}t+e^{-\mu t}\text{d}X_t \\ &=\mu b e^{-\mu t}\text{d}t+\sigma Y_t\text{d}W_t. \end{align*}

Product Rule

Consider the geometric Brownian motion $Z_t$ with $\text{d}Z_t=\sigma^2Z_t\text{d}t-\sigma Z_t\text{d} W_t$ and $Z_0=1$ such that $Z_t=\exp\left(\frac{1}{2}\sigma^2 t-\sigma W_t\right)$.

Then, \begin{align*} \text{d}Y_tZ_t&= Y_t\text{d}Z_t+Z_t\text{d}Y_t+\text{d}Y_t\text{d}Z_t \\ &=\sigma^2Y_tZ_t\text{d}t-\sigma Y_tZ_t\text{d} W_t+\mu b e^{-\mu t}Z_t\text{d}t+\sigma Y_tZ_t\text{d}W_t-\sigma^2Y_tZ_t\text{d}t \\ &=\mu b e^{-\mu t}Z_t\text{d}t. \end{align*} Thus, \begin{align*} Y_tZ_t-Y_0Z_0=\mu b\int_0^te^{-\mu s}Z_s\text{d}s. \end{align*} Finally, \begin{align*} X_t&=X_0e^{\mu t}Z_t^{-1}+\mu be^{\mu t}Z_t^{-1}\int_0^te^{-\mu s}Z_s\text{d}s\\ &=e^{\mu t}Z_t^{-1}\left(X_0+\mu b\int_0^te^{-\mu s}Z_s\text{d}s\right). \end{align*} However, I do not think the distribution of $X_t$, which includes an integrated geometric Brownian motion, is known? This is the entire struggle of pricing Asian options.

Special Cases

We can recover two special cases:

  • If $\mu=0$, we get $X_t=X_0\exp\left(-\frac{1}{2}\sigma^2 t+\sigma W_t\right)$.
  • If $b=0$, we get $X_t=X_0\exp\left(\left(\mu-\frac{1}{2}\sigma^2\right) t+\sigma W_t\right)$.
Kevin
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  • Thanks. A follow up question. In this case, for any finite B>0 and X_o > 0, can I say that P(X_t ∈ (B,B+1) ) > 0 for any t>0? I think in case of GBM we can say that because CCDF of lognormal distribution has a error function which goes to 1 only if B tends to infinity. – S_R Feb 25 '21 at 11:40
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    Yes, I think you can say this. $e^{\mu t}Z_t^{-1}$ is a geometric Brownian motion. Thus, $X_t$ is the product of a GBM with another factor which is positive (assuming $X_0,b,\mu>0$). Because the support of a log-normal distribution is $(0,\infty)$, I'd guess the same applies $X_t$ and the process should be able to attain values in any given interval $(B,B+1)$, i.e. $\mathbb{P}[{X_t\in(B,B+1)}]>0$ for all time points $t>0$ and constants $B>0$. – Kevin Feb 25 '21 at 12:11
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If b>0, can I say anything about the distribution of at a later time t?

Yes - The solution is in Kloeden and Platen

You want to refer to section 4.4 of Numerical solutions of stochastic differential equations by Kloeden and Platen (which is my go-to book for SDEs). Under the section on linear SDEs with multiplicative noise on page 119, an SDE of this form is explicitly listed in (4.9), which I repeat here: $$ \mathrm{d}X_t = (aX_t + c) \,\mathrm{d}t + (bX_t + d)\, \,\mathrm{d}W_t $$ has the solution $$ X_t = \Phi_t \left(X_0 + (c-bd)\int_0^t \Phi^{-1}_s\,\mathrm{d}s + d\int_0^t \Phi^{-1}_s\,\mathrm{d}W_s\right) $$ with the fundamental solution $$ \Phi_t = \exp\left( \left(a-\tfrac{1}{2}b^2\right)t + bW_t\right). $$

A few equations later they give the solution for when the coefficients are all functions of time too, but it's a bit too much for me to type up here.

oliversm
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