# Ito's lemma and Lognormal Property

What would be the difference between:

\begin{align} dS = udt + \sigma dz \end{align}

and

\begin{align} dS=u*S*dt + \sigma*S*dzdS \end{align} Is that the former is in absolute terms and the latter is in relative terms with the stock price?

Therefore, if I want to derive the lognormal property for ( $$G =\ln S => dG = (u-\sigma^2/2)*dt + \sigma*dz$$) pricing an option, can the first equation be used and how? In John Hull book is done by using the second one.

Thank you.

• The first SDE is arithmetic Brownian motion and can become negative. Stocks have limited liability, so a negative stock price is not very sensible for a model. – kurtosis Aug 21 '20 at 7:42
• Does this mean that the first SDE cannot be used to come up to the lognormal property? I have to use the second one? – S_Star Aug 21 '20 at 8:03
• The first process is just normally distributed with mean $ut$ and variance $\sigma^2 t$ - no $\log$s at all – StackG Aug 21 '20 at 8:04
• To be a little more careful: $dS_t$ in the first equation is arithmetic Brownian motion -- so $dS_t$ and thus changes in $S$ are normally-distributed. The second equation should be $dS_t=uS_tdt+\sigma S_tdz$ and is geometric Brownian motion. Then, $dS_t$ is log-normal or (equivalently) $d(\log(S_t))=\frac{dS_t}{S_t}$ is normally-distributed as are changes in $\log(S)$. – kurtosis Aug 21 '20 at 14:26

\begin{align} dS = \mu dt + \sigma dW_t \end{align} is simply

\begin{align} S(t) - S(0) = \mu t + \sigma W_t \end{align}

(as discussed here in the first page, for example)

• I understand this, thank you, but it doesn't fully answer my questions above, as I am a bit confused. – S_Star Aug 21 '20 at 8:04
• The first equation is a normal process, not a log-normal process – StackG Aug 21 '20 at 8:05
• The first one it says is the instantaneous rate of return and the second one is the Brownian motion with drift followed by the stock price. Does it mean that the first one is in absolute terms and the second one in relative terms? And how do I come to the lognormal property with the first equation if possible? – S_Star Aug 21 '20 at 8:08
• The first equation says that the change in $dS$ has a drift that is linear in time, and a variance that is growing linearly with time. It is NOT log-normally distributed. Lognormal behaviour comes from geometric brownian motion (ie. the second equation) when the drift and variance terms have a factor of $S$ in them as well – StackG Aug 21 '20 at 8:19
• $dS/S=\mu dt$. dS/S is the relative performance. – dm63 Aug 21 '20 at 10:56

Perhaps it might help if we define the difference between Brownian Motion (BM) and Geometric Brownian Motion (GBM). BM has independent, identically distributed increments while GBM has independent, identically distributed ratios between successive factors. The definition is inherited from that of arithmetic random walks, which are modelled as sums of random terms, and geometric random walks, modelled as products of random factors.

Let's look at them a bit more in detail.

The BM differential equation is:

$$dS_{t} = \mu dt + \sigma dW_{t}$$

where the first term, $$\mu dt$$, is the drift term and the second term $$\sigma dW_{t}$$ is the diffusion term characterised by the Wiener process $$W_{t}$$.

To resolve it, we add integrals on both sides:

$$\int_{t=0}^T dS_{t} =\mu \int_{t=0}^T dt + \sigma \int_{t=0}^T dW_{t}$$

Here, the last term $$\int_{t=0}^T dW_{t}$$ is your random variable, i.e. shock.

Let us now look at the GBM. As we said earlier, the GBM is characterised by i.i.d
ratios between successive factors. We define it as

$$\frac{dS_{t}}{S_{t}} = \mu dt + \sigma dW_{t}$$

Here, $$\frac{dS_{t}}{S_{t}}$$ is the finite-time price. To resolve, we take logs and after applying ito's lemma we obtain

$$d(logS_{t}) = (\mu - \frac{1}{2} \sigma^2)dt+\sigma dW_{t}$$

now we can add integrals, as we have a normal diffusion:

$$\int_{t=0}^Td(logS_{t}) = (\mu - \frac{1}{2} \sigma^2)\int_{t=0}^Tdt+\sigma \int_{t=0}^TdW_{t}$$

hence

$$log S_{T}-log S_{0} = (\mu - \frac{1}{2} \sigma^2)T + \sigma \int_{t=0}^TdW_{t}$$

Here, $$(\mu - \frac{1}{2} \sigma^2)T$$ is the mean of the log price after T years, and $$\sigma \int_{t=0}^TdW_{t}$$ is the shock, i.e. the variance after T years (normally distributed with mean 0 and variance 1).

Finally, we have the rate of return equal to

$$\frac{S_{t}}{S_{0}} = exp((\mu - \frac{1}{2} \sigma^2)T + \sigma \int_{t=0}^TdW_{t})$$

which is log-normally distributed.

I hope this helps!