Why an SDE Needs an Initial Condition for Uniqueness

# Baxter and Rennie: A question on Notation


# Baxter and Rennie: A question on Notation












On page 56 of Baxter and Rennie (Financial Calculus), we have

- The definition of a continuous stochastic process, in terms of the drift $\mu_s$ and volatality $\sigma_s$. Its important to keep in mind that the drift and volatality can be processes themselves.

- The uniquness result, a part of which tells us that for a given $\mu_s$ , $\sigma_s$ and $X_0$, we can construct only a single process $X_t$.

However, the bottom part of the page introduces SDEs, which may have multiple solutions. The definition of an SDE is given in terms of drift and volatality which may depend upon the current value of the process - the notation being $\mu(X_t,t)$ and $\sigma(X_t,t)$.

I think $\mu(X_t,t)$ and $\sigma(X_t,t)$ are special cases of $\mu_s$ and $\sigma_s$, which would imply that an SDE cannot have multiple solutions. This is obviously wrong, and I would like to know why - hopefully in terms of how those two notations differ. Thank you in advance!

## Answer by Kevin (score 2, accepted)

https://quant.stackexchange.com/a/46955

The notation $\mu(X_t,t)$ and $\sigma(X_t,t)$ is indeed a special case of the more general notation $\mu_t$ and $\sigma_t$. The latter may be any stochastic processes (with the conditions given in the defintion). Note firstly that if $\sigma_t=0$, the SDE reduces to an ODE and we already know that not all ODEs are solvable (explicitly and uniquely).

However, as you said, $\mu_t$ and $\sigma_t$ (or $\mu(X_t)$ and $\sigma(t,X_t)$) do not uniquely define a stochastic process. You also need the initial value $X_0$. In the definition of the SDE on the bottom of the page, the authors do not include the initial condition $X_0$ which is crucial to have a unique solution.

- Look at two geometric Brownian motions with two different starting points, you will get two different processes with different distributions. Remember that for every $t\geq0$, $$ \ln(X_t) \sim N\left( \ln(X_0) + \left( \mu-\frac{1}{2}\sigma^2\right)t, \sigma^2t\right).$$

One can show: If you fix an initial value (and if $\mu$ and $\sigma$ satisfy some regulatory (Lipschitz) conditions), then there exists indeed a unique solution to the SDE.

- For instance, $$\mathrm{d}X_t =3X_t^{3/2}\mathrm{d}t$$ with $X_0=0$. This is an ODE where $3x^{3/2}$ fails the Lipschitz condition, hence you can’t find a unique solution: for any $a>0$, \begin{align*} X_t = \begin{cases} 0 & t\leq a, \\ (t-a)^3 & t>a \end{cases} \end{align*} is a solution to the ODE.