In preparation for my Options, Future's and Risk Management examination next week, I have been presented with a series of questions and their answers. Unfortunately, my lecturer, one of the less organised, does not respond to emails and attempts for consultation. I have resorted to these forums to relieve some stress.

My question is presented as follows:

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

Of course, I will display my attempted solution.

First, I assume that the change in stock price follows a geometric brownian motion (GBM). That is,

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

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

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

Substituting appropriate figures,

$$S \sim \left(58.21, \left(22.59\right)^2\right).$$

For probabilistic problems regarding normal distributions, I relate to standardised scores. I calculate that

$$z_{95} = 1.63.$$

Using Microsoft Excel, the probability that the z-score is greater than 1.63, and therefore, the price of the stock is greater than 95 is given by

$$1- \mathrm{NORMDIST(95,58.21,22.59,TRUE)}.$$

The answer I get is 5.17%. The answer states it is 8.23%.

I would be beyond thankful for any help and advice on how to properly solve this problem.

Thank you in advanced,


  • $\begingroup$ I do not see an error in your computation...Is the assumption of the geometric brownian motion for the return reliable? $\endgroup$ – muffin1974 Jun 18 '15 at 15:11
  • $\begingroup$ @muffin1974. Well, apparently my assumption that $$\frac{dS}{S_{0}} \sim \phi$$ is incorrect. Instead, I should be using $$\ln(S) \sim \phi.$$ Do you know why? $\endgroup$ – Gustavo Louis G. Montańo Jun 19 '15 at 0:33
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    $\begingroup$ Well, one standard argument against the assuming of normal distributed stock prices is of course that you cannot prevent the stock price to run negative. $\endgroup$ – muffin1974 Jun 19 '15 at 8:32

As the stock price process $S$ follows a geometric Brownian motion, we have that \begin{align*} S_T &= S_0 e^{(\mu-\frac{1}{2}\sigma^2)\, T + \sigma\, W_T}\\ &= S_0 e^{(\mu-\frac{1}{2}\sigma^2)\, T + \sigma\, \sqrt{T}\, \xi}, \end{align*} where $\xi$ is a standard normal random variable. Then, we have the probability \begin{align*} P(S_T > 95) &= P\Big( S_0 e^{(\mu-\frac{1}{2}\sigma^2)\, T + \sigma\, \sqrt{T}\, \xi} > 95\Big)\\ &= P\bigg(\xi > \frac{\ln \frac{95}{S_0} - (\mu-\frac{1}{2}\sigma^2)\, T}{\sigma\, \sqrt{T}} \bigg)\\ &= 1- NORMSDIST\left(\frac{\ln \frac{95}{S_0} - (\mu-\frac{1}{2}\sigma^2)\, T}{\sigma\, \sqrt{T}} \right)\\ &=NORMSDIST\left(\frac{\ln \frac{S_0}{95} + (\mu-\frac{1}{2}\sigma^2)\, T}{\sigma\, \sqrt{T}} \right). \end{align*}


Given its price today, the stock price at time T is lognormally distributed, whereas $lnS_T$ is normally distributed, that is

$lnS_T$ ~ $N \Bigr(lnS_0 + (\mu- \frac{\sigma^2}{2}T),\sigma^2T \Bigl)$

see for example Hull - Options, Futures, and other Derivatives.

Plugging in the numbers you get

$lnS_T$ ~ $N(3.981291519,0.16875)$

Then the probability you want is $P[lnS_T>ln95]=1-P[lnS_T<ln95]=1-P\Bigr[Z<z=\frac{ln95-3.981291519}{\sqrt{0.16875}}\Bigl] \\=1-P[Z<z=1.391]=1-0.9177=0.0823$

  • $\begingroup$ I see, indeed the solution you have provided is something I wanted to also address, however, I will do this in another question. I have one other question. Why does $$\mathcal{P}(\ln(S)>\ln(95)) = \mathcal{P}(S > 95)?$$ Thank you for your time and assistance. $\endgroup$ – Gustavo Louis G. Montańo Jun 19 '15 at 0:30
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    $\begingroup$ This Holds because $ln $ is a monotone transformation $\endgroup$ – muffin1974 Jun 19 '15 at 6:52

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