The price of the stock XYZ follows a brownian motion pattern with starting price = 10, μ = 0 and σ = 20 (on annual basis). What's the probability that in 6 months the price is less or equal to 8? Also i must solve this with paper and pen (I can consult the Normal distribution tabel)


Let $(S_t)$ be the price process of your stock such that $S_t = S_0+ \mu t + \sigma B_t$ where $(B_t)$ is a standard Brownian motion. Then, since $B_t\sim N(0,t)$, we get $S_t\sim N(S_0+\mu t, \sigma^2 t)$. In six months, $t=\frac{1}{2}$, we have $S_{0.5}\sim N(10,200)$, i.e. $S_{0.5}=10+\sqrt{200}Z=10+10\sqrt{2}Z$ where $Z\sim N(0,1)$. Thus, \begin{align*} \mathbb{P}[\{S_{0.5}\leq8\}] &= \mathbb{P}[\{10+10\sqrt{2}Z\leq8\}] \\ &= \mathbb{P}\left[\left\{Z\leq-\frac{1}{5\sqrt{2}}\right\}\right] \\ &= \Phi\left(-\frac{1}{5\sqrt{2}}\right) \\ &= 1-\Phi\left(\frac{1}{5\sqrt{2}}\right) \\ &\approx 1-\Phi\left(0.141\right) \\ &\approx 0.444. \end{align*}

You get the last number from your normal table. Note that under your model, the stock price may be negative with positive probability. Furthermore, a model with a normal distributed stock price was originally proposed by Bachelier.


I think there is a typo in the previous answer- assuming arithmetic brownian is meant- here is my working:

$P\left[S_t \le 8\right]=P\left[S_0+\mu t+\sigma B_t \le 8\right]$

$=P\left[S_0+\mu t+\sigma \sqrt{t}Z \le 8\right]$

$=P\left[Z\le \frac{8-S_0-\mu t}{\sigma \sqrt{t}}\right]$

$=P\left[Z \le \frac{8-10}{20 \sqrt{0.5}}\right]$

$=P\left[Z \le \frac{-1}{10 \sqrt{0.5}}\right]$

$=P\left[Z\le -0.14\right]$

  • $\begingroup$ Yes, cheers mate. I corrected a mistake, just a little computation error. Happy to exchange the favour though: in your last line $\frac{1}{10\sqrt{0.5}}\approx 0.141$ and not $0.07$. $\endgroup$ – Kevin Jul 19 '19 at 20:21
  • $\begingroup$ Thanks! I shall buy a new calculator! $\endgroup$ – Magic is in the chain Jul 19 '19 at 20:24

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