Using the Correct Variance in a Geometric Brownian Motion Probability

Summary

The document describes a calculation of the probability that a price following geometric Brownian motion exceeds a threshold. The questioner reports a probability of one where the supplied solution gives a lower value, then identifies the source of the discrepancy: confusing a normal variable’s variance with its standard deviation when calculating a standardized score. Using volatility, rather than its square, as the denominator produces the stated solution.

The exchange illustrates a basic but important distinction in probability calculations: variance is the square of standard deviation, while volatility commonly denotes standard deviation. The correction is presented as the explanation for the mismatch, but the full model setup, derivation, and solution manual are not included in the excerpt. Readers would need those details to reproduce the probability or assess assumptions such as the time horizon and drift. It is a focused clarification of a calculation error, rather than a general treatment of geometric Brownian motion.

Key ideas

• A normal variable is parameterized by its mean and variance, while its standard deviation is the square root of variance.
• A probability calculation can be substantially wrong if volatility is confused with variance when standardizing a variable.
• The document attributes the reported mismatch to using the variance where the standard deviation belonged.
• The excerpt does not provide enough of the model setup to independently reproduce the probability.

Tags

# How to compute Pr(S>100) when S follows Geometric Brownian Motion?


# How to compute Pr(S>100) when S follows Geometric Brownian Motion?












I have been trying to resolve this problem, under (b), but I cannot find the correct answer. For i=1, my ultimate answer (P=1) deviates from the correct answer (P=0.7580). Please let me know whether my computations are wrong or not.

In blue is the solution according to the solution manual; my computations start at For i=1.

## Answer by AltTabsen (score 0)

https://quant.stackexchange.com/a/33146

I wrongly assumed that that a random normal variable's distribution was described by its mean and volatility. It is, however, described by its mean and variance.

So for the denominator in the Z computation I now use 0.1 (volatility) instead of 0.01 (the variance, which I had mistaken to be volatility), which results in the right answer.

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.