Consider the following SDE:

$$ dX_t = X_t (\sigma dW_t + \mu_tdt), \text{with } \mu_t \text{ a bounded integrable function of time} $$

The way I would solve this would be to use that:

$$ X_t = \exp(\sigma W_t + \nu t) \quad \Leftrightarrow \quad dX_t = X_t \left(\sigma dW_t + (\nu + \frac{1}{2}\sigma^2 )dt\right) $$

I assume that this would also hold for:

$$ X_t = \exp\left(\sigma W_t + \int_0^t \nu_s ds \right) \quad \Leftrightarrow \quad dX_t = X_t \left(\sigma dW_t + (\nu_t + \frac{1}{2}\sigma^2 )dt\right) $$

So I guess that $ \mu_t = \nu_t + \frac{1}{2}\sigma^2 $, such that $\nu_t = \mu_t - \frac{1}{2}\sigma^2 $ yielding:

$$ X_t = \exp\left(\sigma W_t + \int_0^t \mu_s ds - \frac{1}{2}\sigma^2 t \right) $$

I can use this in Itô's lemma to check that it is correct:

$$ dX_t = X_t \left( \left[ \mu_t - \frac{1}{2}\sigma^2 + \frac{1}{2}\sigma^2 \right] dt + \sigma dW_t \right) = X_t \left( \mu_t dt + \sigma dW_t \right) $$

so Itô's lemma suggests that it is indeed correct. However, my book (Financial Calculus: An introduction to derivatives pricing, Martin Baxter & Andrew Rennie, 21st printing 2014), suggests the solution to be:

$$ X_t = X_0 \exp \left( \sigma W_t + \int_0^t \mu_s d_s - \frac{1}{2}\sigma^2 t \right) $$

I don't get where the $X_0$ comes from. Sure $\int_0^tdX_t = X_t - X_0$, but then why is $X_0$ multiplied with the right-hand-side? I can't make any sense of this.

Does someone see what I did wrong? Or is my solution correct?


1 Answer 1


Notice in your SDE, scalilng $X_t$ by any fixed constant still keeps the result as a solution of the SDE.

You can think of it as the integration constant in the exponent, i.e. $$ e^{f+c} = Ke^f $$


In your solution, you assume $$ X_t = \exp(\sigma W_t + \nu t),$$ thus forcing $X_0 = \exp(\sigma W_0 + \nu \cdot 0) = 1$, which is restrictive. You should assume $$ X_t = K \exp(\sigma W_t + \nu t)$$ instead, and in this case note that $K = X_0$. Now $X$ can take any value at time $0$, retaining its other properties

  • $\begingroup$ I don't get what you mean. Could you elaborate? $\endgroup$
    – Jean-Paul
    Oct 12, 2015 at 16:42
  • $\begingroup$ @Jean-Paul please see the update $\endgroup$
    – gt6989b
    Oct 12, 2015 at 16:45
  • $\begingroup$ That makes sense. However, the book uses my assumption on the previous page to prove that exact SDE. I don't get why I would need to add the $K$ if the example is exactly the same. Is it because my $\mu_t$ is a function instead of a constant? (See the assumption I make from the example with $\nu$ a constant and the version with $\nu_t$ a function) $\endgroup$
    – Jean-Paul
    Oct 12, 2015 at 16:54
  • $\begingroup$ @Jean-Paul you want to make as generic of an assumption as possible; forcing $X_0=1$ is very restrictive... $\endgroup$
    – gt6989b
    Oct 12, 2015 at 16:59

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