Finding a Brownian Stock’s Probability of Falling Below a Threshold
Summary
The note works through the probability that a stock modeled by arithmetic Brownian motion will be at or below a specified level after six months. With a starting price of 10, zero drift, and annual volatility of 20, the price at the horizon is normally distributed with mean equal to the starting price and variance equal to volatility squared times elapsed time. Standardizing the threshold of 8 converts the question into a standard normal cumulative probability, which can be read from a normal table. The response gives an approximate probability of 0.444.
A second answer presents the same standardization directly, with the resulting z-score rounded to about -0.14. The calculation is useful as a hand method under the stated arithmetic Brownian model. Its main limitation is the model itself: an arithmetic normal price process can assign positive probability to negative stock prices. The note identifies this as a Bachelier-style modeling assumption and does not compare it with a positive-price model such as geometric Brownian motion.
Key ideas
- Under arithmetic Brownian motion, the stock price at a fixed horizon is normally distributed.
- The horizon mean is the starting price plus drift times elapsed time.
- The horizon variance equals volatility squared times elapsed time.
- Standardize the threshold to a z-score and use the normal cumulative distribution table.
- The normal price model can assign positive probability to negative stock prices.
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Full text
# Probability that the price of stock following a brownian motion goes under a certain value
# 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)
## Answer by Kevin (score 4)
https://quant.stackexchange.com/a/46707
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.
## Answer by Magic is in the chain (score 4)
https://quant.stackexchange.com/a/46714
I think there is a typo in the previous answer- assuming arithmetic brownian is meant- here is my working:
$P\left[S_t \le 8\right]=P\left[S_0+\mu t+\sigma B_t \le 8\right]$
$=P\left[S_0+\mu t+\sigma \sqrt{t}Z \le 8\right]$
$=P\left[Z\le \frac{8-S_0-\mu t}{\sigma \sqrt{t}}\right]$
$=P\left[Z \le \frac{8-10}{20 \sqrt{0.5}}\right]$
$=P\left[Z \le \frac{-1}{10 \sqrt{0.5}}\right]$
$=P\left[Z\le -0.14\right]$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.