Stochastic Differential Equation solution for Geometric Brownian Motion

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?

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.

EDIT

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.

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

• 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? – Nehal May 20 '17 at 0:19
• @Nehal I thought you wanted to understand. I'll update my answer. – Yujie Zha May 20 '17 at 0:33
• @Nehal no problem, I've updated. So difficult to layout the formula over the phone.. – Yujie Zha May 20 '17 at 0:51
• So what you mean is that \int_0^t\frac{dY(t)}{Y(t)} = d[lnY(t)] – Nehal May 20 '17 at 1:15
• 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. – LutzL May 20 '17 at 8:08