In Hull, we are presented that

$$\frac{\Delta S}{S_{0}}=\mu \Delta t+\sigma\sqrt{\Delta t}\cdot \varepsilon.$$

Following some algebra,

$$ \begin{align*} \frac{\Delta S}{S_{0}} &=\mu \Delta t+\sigma\sqrt{\Delta t} \cdot \varepsilon \\ \frac{S-S_{0}}{S_{0}} &= \mu \Delta t+\sigma\sqrt{\Delta t} \cdot \varepsilon \\S &= \left(S_{0} + \mu S_{0} \Delta t\right) + \sigma S_{0} \sqrt{\Delta t} \cdot \varepsilon \end{align*} $$

Therefore the distribution of future stock price is given by

$$S \sim \phi\left(S_{0} + \mu S_{0} \Delta t,\left(\sigma S_{0} \sqrt{\Delta t}\right)^{2}\right).$$

That is, the future stock price follows a normal distribution.

We are then introduced to Itô's Lemma. By letting $G = \ln(S_{0})$, we derive that

$$dG = \left(\mu - \frac{1}{2}\sigma^{2}\right)dt+\sigma dz.$$

Since $G = \ln{S_{0}}$, in a discrete sense, it can be said that

$$dG = \ln{S_{T}} - \ln{S_{0}}.$$


$$\ln{S_{T}} - \ln{S_{0}} = \left(\mu - \frac{1}{2}\sigma^{2}\right)dt+\sigma dz \\ \implies \ln{S_{T}} = \ln{S_{0}} + \left(\mu - \frac{1}{2}\sigma^{2}\right)dt+\sigma dz. $$

It then follows that since $\ln{S_{T}}$ follows a normal distribution, the future stock price must follow a lognormal distribution

I am now confused, which process do I use to answer questions about the probabilistic nature of future stock prices?

I have one other question. Why does

$$\mathcal{P}(\ln{S_{T}} > \ln{X}) = \mathcal{P}(S_{T} > X)?$$

The context for my last question can be found here.

  • $\begingroup$ In Hull, most likely it assumes that $$\frac{\Delta S_{t+\Delta t}}{S_{t}}=\mu \Delta t+\sigma\sqrt{\Delta t}\cdot \varepsilon,$$ instead of $$\frac{\Delta S}{S_{0}}=\mu \Delta t+\sigma\sqrt{\Delta t}\cdot \varepsilon,$$ where $\Delta S_{t+\Delta t} = S_{t+\Delta t} - S_t$. That is, log-normal distribution for stock price. $\endgroup$
    – Gordon
    Jun 19, 2015 at 12:50

2 Answers 2


You ask 2 questions and I try to answer:

1) Why do we use geometric Brownian motion ($\ln S_t-\ln S_0$ is normally distributed)? In this case you have $$ S_t = S_0 \exp( (\mu-\sigma^2/2) t + \sigma B_t), $$ which means that you model positive prices. Furthermore the log-return $$ \ln(S_t/S_0) = (\mu-\sigma^2/2) t + \sigma B_t, $$ is normally distributed. As log returns can cover the whole real line $(-\infty,\infty)$ this is a nice model. Keep in mind that simple returns $S_t/S_0-1$ can only take values from $[-1,\infty)$. The best place to model a normal distribution is the whole real line.

If you use a model (the Bachelier model) $$ S_t = S_0 + \mu t + \sigma B_t, $$ then your returns $S_t-S_0$ are normally distributed. But there is the chance that prices get negative (if $B_t$ becomes very negative). You probably don't want this in your model. Some people use this model nevertheless to price options that are close to maturity as you don't need such large $\sigma$ to match (relatively high) prices of OTM options.

For 2) Why is $P(\ln S>\ln X)=P(S > X)$? because the logarithm is a monotonous transformation. We speak of the same events. If $S>X$ then always $\ln S > \ln X$. Thus the same events have the same probability. Another example $$ P ( S > X ) = P ( S+4 > X + 4). $$ Just the same events.

  • $\begingroup$ Hi @Richard, thank you for you time. What I have gathered from your response to my first question is that the model of choice depends on the analyst. However, log returns are preferred. When you therefore read a test question such as: "The share price of company XYC Inc. exhibits an instantaneous drift of 7% per year with return volatility of 45%. What is the probability that XYZ shares exceed 95 after 10 months when they cost 55 today" Notice how there is no explicit reference to the process. Would you assume lognormal returns, as my lecturer did? $\endgroup$ Jun 19, 2015 at 8:28
  • $\begingroup$ It is difficult to say what is meant if they do not state it explicetly. But the word "instantaneous drift" could point to the GBM setting - thus lognormal prices ( i.e. log returns are normal). $\endgroup$
    – Richi Wa
    Jun 19, 2015 at 8:39

You know that Brownian motion {W(t)} is a stochastic process with the following properties:

  1. (Independence of increments) W(t) − W(s) , for t > s , is independent of the past, that is, of W(u) , 0 ≤ u ≤ s, or of $F_s$ , the σ-field generated by W(u), u ≤ s.

  2. (Normal increments) W(t) − W(s) has Normal distribution with mean 0 and variance t − s. This implies (taking s = 0) that W(t) − W(0) has N(0, t) distribution.

  3. (Continuity of paths) W(t), t ≥ 0 are continuous functions of t.

Let ${{x}_{t}}=\ln ({{S}_{t}})$ . The Itˆo formula gives us \begin{align} & d{{x}_{t}}=(\mu -\frac{1}{2}\sigma ^{2})dt+{{\sigma }}d{W}(t) \\ \end{align} Then \begin{align} & {x}_{t}={x}_{0}+(\mu -\frac{1}{2}\sigma^{2})t+{{\sigma }_{t}}{W}(t) \\ \end{align} As we said W(t) has Normal distribution with mean 0 and variance t then $x_t$ is normal process such that the expected value is given by \begin{align} & E[{x}_{t}]=E[{x}_{0}+(\mu -\frac{1}{2}\sigma^{2})t]+E[\sigma{W}(t)]={x}_{0}+(\mu -\frac{1}{2}\sigma ^{2})t\\ \end{align} and \begin{align} & Var[{x}_{t}]=Var[{{\sigma }_{t}}{W}(t)]=\frac{1}{2}\sigma ^{2}t\\ \end{align} And your other question.fix t and let

\begin{align} & A_\omega =\{\omega|S_t,\omega>K\}=\{\omega|\ln S_t,\omega>\ln K\}\,\, (Almost\,Surely) \end{align}


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