Probability that the price of stock following a brownian motion goes under a certain value

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:

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

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

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

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

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

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

• 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 $0.07$. – KeSchn Jul 19 at 20:21
• Thanks! I shall buy a new calculator! – Magic is in the chain Jul 19 at 20:24