I am having difficulty in understanding the solution for a GBM given the SDE:

$$dY(t)=\mu \ Y(t) \ dt + \sigma \ Y(t) \ dZ(t)$$ or $$\frac{dY(t)}{Y(t)}=\mu \ dt + \sigma \ dZ(t)$$

The solution for the above SDE is:

$$\int_0^t\frac{dY(t)}{Y(t)} = \int_0^t\mu \ ds + \int_0^t\sigma \ dZ(s)$$ $$Y(t) = Y(0)e^{(\mu-\frac{\sigma^2}{2})t+\sigma\ Z(t)}$$

What I can't understand is, how does this,

$$\int_0^t\frac{dY(t)}{Y(t)}=\ lnY(t) - \ lnY(0)$$

become this,

$$Y(t) = Y(0)e^{(\mu-\frac{\sigma^2}{2})t+\sigma \ Z(t)}$$

This part, $-\frac{\sigma^2}{2}$ is extra in the exponent. How?

Thank you in advance!


2 Answers 2


Your integration step is wrong. $Z$ here refers to Brownian motion, and thus you need to apply Ito Integration.

One way I like to see the extra drift is needed, is that we take derivative of the integration result, and see if they match the SDE or not, using Ito Lemma.

So if $Y_t =f(t, Z_t)$, with $f(t,z)=f(0, Z_0)e^{(\mu-\frac{\sigma^2}{2})t+\sigma z}$, where $Z_t$ is an adapted stochastic process, we have:

$$dY_t=df(t, Z_t)=\frac{\partial f(t, Z_t)}{\partial t} dt+ \frac{\partial f(t, Z_t)}{\partial z}dZ_t + \frac{1}{2} \cdot\frac{\partial^2 f(t, Z_t)}{\partial z^2}(dZ_t)^2$$

And for $Z_t$ being Brownian motion, we also have $(dZ_t)^2=dt$

Now let's plug in for the question here: $$Y_t= Y_t(0)e^{(\mu-\frac{\sigma^2}{2})t+\sigma Z_t}$$

$$dY_t=(\mu-\frac{\sigma^2}{2})Y_tdt + \sigma Y_tdZ_t + \frac{1}{2}\sigma\sigma Y_t(dZ_t)^2$$ $$=(\mu-\frac{\sigma^2}{2})Y_tdt + \sigma Y_tdZ_t + \frac{1}{2}\sigma^2 Y_tdt$$ Thus $$dY_t=\mu Y_tdt + \sigma Y_tdZ_t$$

Which means that the $Y$ we get is the solution for the SDE.


So OP wants to solve instead of see why that extra drift exists. Basically integration and derivative are the same thing, just notations:

So the following integration is Ito, $Y$ is not smooth enough for you to apply the calculus rule you are familiar with. $$\int \frac{dY}{Y} \color{red}\ne \ln(Y)$$

Notice $$d(\ln Y)=\frac{dY}{Y} -\frac{1}{2}\frac{1}{Y^2}(dY )^2$$

$$(dY)^2=(\mu Y_tdt + \sigma Y_tdZ_t)^2=\sigma^2Y^2 dt$$ We also know $\frac{dY}{Y}$ from our original SDE.

Plug them in and we get:

$$d(\ln Y)=\mu dt + \sigma dZ-\frac{\sigma^2}{2} dt$$

Now the right-hand-side do not have $Y$, you'll be able to solve it, i.e.

$$\ln Y_t - \ln Y_0 = \int_0^t d(\ln Y) = \int_0^t (\mu -\frac{\sigma^2}{2})dt + \int_0^t \sigma dZ_t=(\mu -\frac{\sigma^2}{2})t + \sigma Z_t$$

  • $\begingroup$ I understand the Ito integration but I have to work it backwards. If I have only this part, which is the SDE $$dY_t=\mu Y_tdt + \sigma Y_tdZ_t$$ then how can I come to this, $$Y(t) = Y(0)e^{(\mu-\frac{\sigma^2}{2})t+\sigma\ Z(t)}$$ What are the steps involved here? $\endgroup$
    – Nehal
    May 20, 2017 at 0:19
  • $\begingroup$ @Nehal I thought you wanted to understand. I'll update my answer. $\endgroup$
    – Jay Zha
    May 20, 2017 at 0:33
  • $\begingroup$ @Nehal no problem, I've updated. So difficult to layout the formula over the phone.. $\endgroup$
    – Jay Zha
    May 20, 2017 at 0:51
  • $\begingroup$ So what you mean is that $$\int_0^t\frac{dY(t)}{Y(t)} = d[lnY(t)] $\endgroup$
    – Nehal
    May 20, 2017 at 1:15
  • 1
    $\begingroup$ Another recommendation is P. Forsyth "An Introduction to Computational Finance. Without Agonizing Pain". The chapters discussing numerical approximations to the geometric BM should greatly help your understanding. $\endgroup$ May 20, 2017 at 8:08

I believe the answer by @Yujie Zha can be simplified substantially. Thanks to @Dr. Lutz Lehmann for providing a link to this, my solution is the same as the solution on page 15, but with more intermediate steps. I decided to write this as this helped me to figure out why the solution to the Geometric Brownian Motion SDE is the way it is. If I am wrong, please correct me.


Let $$dY(t) = \mu Y(t)dt + \sigma Y(t)dZ(t) ~~~~\text{(1)}$$ be our geometric brownian motion (GBM). Now rewrite the above equation as $$dY(t) = a(Y(t), t)dt + b(Y(t), t)dZ(t)~~~~\text{(2)}$$ where $a = \mu Y(t)$, $b = \sigma Y(t)$. Both are functions of $Y(t)$ and $t$ (albeit simple ones). Now also let $f = \ln(Y(t))$. We can now apply Ito's lemma to equation $(2)$ under the function $f = \ln(Y(t))$. This leads to $$ df = d(\ln(Y(t)) = \left(\frac{\partial f}{\partial t} + \frac{\partial f}{\partial Y}a + \frac{1}{2}\frac{\partial^2f}{\partial Y^2}b^2\right)dt + b\frac{\partial f}{\partial Y}dZ(t)~~~~\text{(3)}$$ Now we substitute all the derivatives in $(3)$ and the functions $a$ and $b$. Note that $\frac{\partial f}{\partial t} = \frac{\partial \ln(Y(t))}{\partial t} = 0 $ (partial derivative w.r.t. $t$ of a function of $Y$ is $0$), $\frac{\partial f}{\partial Y} = \frac{\partial \ln(Y)}{\partial Y} = \frac{1}{Y}$, $\frac{\partial^2 f}{\partial Y^2} = -\frac{1}{Y^2}$ (standard calculus). We finally have that $$ (3) = \left( 0 + \frac{1}{Y}Y\mu + \frac{1}{2}\left(-\frac{1}{Y^2}\right)\sigma^2 Y^2\right)dt + \sigma Y \frac{1}{Y}dZ(t) = \left(\mu - \frac{\sigma^2}{2}\right)dt + \sigma dZ(t)$$ i.e. $$d(\ln(Y(t)) = \left(\mu - \frac{\sigma^2}{2}\right)dt + \sigma dZ(t)$$ Integrating this from $0$ to $t$ gives $$\ln(Y(t)) - \ln(Y(0)) = \left(\mu - \frac{\sigma^2}{2}\right)t + \sigma (Z(t) - Z(0))$$ The integral $\int_0^{t}dZ(t)$ is, by the definition of the Ito integral, equal to $Z(t) - Z(0)$ as we are integrating the simple constant process $1$ w.r.t. Brownian motion.

If we rearrange and note that $Z(t) \sim N(0, t), Z(0) \sim N(0,0) = 0$ and are independent, we finnally get $$Y(t) = Y(0)\exp\left((\mu - \frac{\sigma^2}{2})t + \sigma Z(t)\right)$$

  • 2
    $\begingroup$ I provided two solutions, and each solution is not significantly more complicated than the solution you posted here. Plus, why would you downvote my answer - I think by all means there is no reason to do so. $\endgroup$
    – Jay Zha
    Nov 28, 2019 at 5:08
  • $\begingroup$ Why is > (partial derivative w.r.t. t of a function of Y is 0 ), true? $\endgroup$
    – Mossa
    Feb 6 at 11:44

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