Economic Modelling · Stochastic models for security prices
Geometric Brownian Motion and the Lognormal Model for Share Prices
Updated 11 October 2026 · Fact-checked
The lognormal model says a share price follows geometric Brownian motion: dS = μS dt + σS dZ. Solving it gives S(t) = S(0) exp{(μ − σ²/2)t + σZ(t)}. So log returns are normal, prices are lognormal, and E[S(t)] = S(0)e^(μt). To solve questions, work with ln S(t) first.
Understand Geometric Brownian Motion and Lognormal Model
Start with a simple idea. Share prices cannot go below zero, and a ₹100 share moves by much larger rupee amounts than a ₹10 share. So it is more natural to model the percentage change in price than the rupee change. That is what geometric Brownian motion (GBM) does.
The model is written as a stochastic differential equation: dS(t) = μS(t) dt + σS(t) dZ(t). Here μ is the drift (expected growth rate), σ is the volatility, and Z(t) is standard Brownian motion. Over a short time dt, the relative change dS/S is normal with mean μ dt and variance σ² dt.
To solve it, put X(t) = ln S(t) and apply Ito's lemma. The second-derivative term brings in a correction of −σ²/2. This gives d(ln S) = (μ − σ²/2) dt + σ dZ. This is a plain Brownian motion with drift, so ln S(t) is normal. Integrating gives S(t) = S(0) exp{(μ − σ²/2)t + σZ(t)}.
The key consequence: ln(S(t)/S(0)) ~ N((μ − σ²/2)t, σ²t). The continuously compounded return over any period is normal, and returns over non-overlapping periods are independent. The price S(t) itself is lognormal, so it is always positive and right-skewed.
Watch the two different growth rates. The expected price grows at rate μ. The expected log return grows at rate μ − σ²/2, which is lower. The gap comes from the convexity of the exponential function. The model's limits are real: constant volatility, no jumps, no dividends, and thinner tails than real markets show.
Key rules to remember
- GBM stochastic differential equation
- dS(t) = μS(t) dt + σS(t) dZ(t)
- μ is the drift and σ the volatility, both constant. Z(t) is standard Brownian motion.
- Solution of the SDE
- S(t) = S(0) exp{(μ − σ²/2)t + σZ(t)}
- Obtained by applying Ito's lemma to ln S(t).
- Distribution of log price
- ln S(t) ~ N(ln S(0) + (μ − σ²/2)t, σ²t)
- Normal, so S(t) is lognormal.
- Distribution of log return over (t, t+s)
- ln[S(t+s)/S(t)] ~ N((μ − σ²/2)s, σ²s)
- Independent of the past and independent across non-overlapping intervals.
- Mean of the share price
- E[S(t)] = S(0) e^(μt)
- Uses E[e^Y] = exp(m + v/2) for Y ~ N(m, v).
- Variance of the share price
- Var[S(t)] = S(0)² e^(2μt) (e^(σ²t) − 1)
- Grows with σ² and t.
- Lognormal moments (general)
- If ln X ~ N(m, v): E[X] = e^(m + v/2), Var[X] = e^(2m + v)(e^v − 1)
- Use for any lognormal variable.
- Lognormal median
- Median of S(t) = S(0) exp{(μ − σ²/2)t}
- Lower than the mean.
- Probability statement
- P(S(t) > k) = 1 − Φ[(ln(k/S(0)) − (μ − σ²/2)t) ÷ (σ√t)]
- Standardise ln S(t) and use normal tables.
How to solve Geometric Brownian Motion and Lognormal Model questions
Almost every question on this topic reduces to a normal distribution for the log price. Follow this order.
- 1Write down the model parameters: S(0), μ, σ and the time t in years. Check whether σ is given as σ or σ².
- 2Define the log return: ln[S(t)/S(0)] ~ N((μ − σ²/2)t, σ²t). Compute the mean m and variance v numerically.
- 3If the question asks for a probability, convert the price condition to a log condition, standardise using (x − m) ÷ √v, and read Φ from tables.
- 4If it asks for the mean or variance of the price, use E[S] = S(0)e^(m + v/2) form, or the shortcut S(0)e^(μt) for the mean.
- 5If it asks to derive the solution, set X = ln S, apply Ito's lemma, show the −σ²/2 term, then integrate.
- 6For conditional questions, restart from the latest known price S(t) and use the time gap s only.
- 7State the result with units and one line on assumptions, such as constant μ and σ and no dividends.
Quickest way: Log-first shortcut
When to use it: Use for MCQs and for the first part of written questions on prices, probabilities and moments.
- Compute m = (μ − σ²/2)t and v = σ²t at once.
- Mean of price: S(0)e^(μt). You do not need m and v for this.
- Variance of price: mean² × (e^v − 1).
- Probability: standardise ln(k/S(0)) with m and √v, then use Φ.
- Sanity check: median < mean, and the price can never be negative.
Common mistakes in Geometric Brownian Motion and Lognormal Model
Using μt as the mean of the log return.
Students forget the −σ²/2 correction that comes from Ito's lemma.
Fix: Always write (μ − σ²/2)t for the log return mean. Use μt only for E[S(t)]/S(0) in the exponent.
Writing the variance of the log return as σt instead of σ²t.
Confusing standard deviation with variance in the normal parameters.
Fix: The N(m, v) notation uses variance. The standard deviation is σ√t.
Applying ordinary calculus to ln S and missing the Ito term.
Ordinary chain rule gives d(ln S) = dS/S, which looks natural.
Fix: Use Ito's lemma: d(ln S) = dS/S − ½σ² dt. Remember (dZ)² = dt.
Treating E[S(t)] as exp of the mean log price.
Thinking expectation passes through the exponential.
Fix: Use E[e^Y] = e^(m + v/2). The extra v/2 is the convexity adjustment.
Mixing time units, such as using σ per year with t in months.
Questions give t as 6 months or 90 days.
Fix: Convert t to years before any calculation.
Claiming the model is realistic because it fits the lognormal shape.
Memorising the assumptions without the limitations.
Fix: State the limits: constant volatility, no jumps, no dividends, thin tails, and no volatility clustering.
Worked examples
Example 1
A share has S(0) = ₹200, μ = 0.12 per year and σ = 0.20 per year under GBM. Find the mean and the distribution of the log return over 1 year, and E[S(1)].
Show the solution
- Mean of log return: (μ − σ²/2)t = (0.12 − 0.04 ÷ 2) × 1 = 0.12 − 0.02 = 0.10.
- Variance of log return: σ²t = 0.04.
- So ln[S(1)/S(0)] ~ N(0.10, 0.04), with standard deviation 0.20.
- E[S(1)] = 200 × e^0.12 = 200 × 1.127497 = ₹225.50 (to the nearest paisa).
Answer: The log return is N(0.10, 0.04) and E[S(1)] ≈ ₹225.50.
Example 2
For the share above, find the variance of S(1) and the probability that S(1) exceeds ₹200. Use Φ(0.5) = 0.6915.
Show the solution
- Variance: S(0)² e^(2μt)(e^(σ²t) − 1) = 40,000 × e^0.24 × (e^0.04 − 1).
- e^0.24 = 1.271249 and e^0.04 − 1 = 0.040811.
- Variance = 40,000 × 1.271249 × 0.040811 = 2,075.2 approximately.
- For the probability: P(S(1) > 200) = P(ln[S(1)/200] > 0).
- The log return is N(0.10, 0.04), so the standard deviation is 0.20.
- Standardise: P(Z > (0 − 0.10) ÷ 0.20) = P(Z > −0.5) = Φ(0.5) = 0.6915.
Answer: Var[S(1)] ≈ 2,075 (in ₹²) and P(S(1) > ₹200) ≈ 0.6915.
Exam tips
- Show the Ito's lemma step when asked to derive or verify the solution. Marks go to the −σ²/2 term and the integration.
- Quote the lognormal moment result in the form E[e^Y] = exp(m + v/2) before using it. This makes your method clear.
- Always state your units for t and check whether the question gives σ or σ². Examiners often change this between parts.
- In the computer-based paper, simulate with S(t+h) = S(t) exp{(μ − σ²/2)h + σ√h × N(0,1)}, and show the formula before the code.
- For discussion parts, give at least three limitations and link each to a real market feature, such as jumps, fat tails or changing volatility.
Practice questions from Stochastic models for security prices
- A stock follows dS = μS dt + σS dW with μ = 0.12 and σ = 0.20. Using Ito's lemma, what is the drift per year of ln S(t)?
- In the Ito multiplication rules used with Ito's lemma, which set of results is correct for small dt?
- A non-dividend-paying share trades at Rs 100. Under Black-Scholes, for a European call with strike Rs 100, maturity 1 year, risk-free force …
- A non-dividend-paying share trades at Rs 100. A European call with strike Rs 100 and one year to expiry is priced under Black-Scholes with a…
- A stock price S(t) follows geometric Brownian motion dS = μS dt + σS dW. Which statement about the process is correct?
Geometric Brownian Motion and Lognormal Model: frequently asked questions
Why does the −σ²/2 term appear in the GBM solution?
It comes from Ito's lemma. When you take ln S, the second-order term ½ × (−1/S²) × σ²S² dt equals −σ²/2 dt. It is the price of working with a non-smooth Brownian path.
What is the difference between the mean and median of a lognormal share price?
The mean is S(0)e^(μt) and the median is S(0)exp{(μ − σ²/2)t}. The median is lower by a factor e^(−σ²t/2). The gap grows with volatility and time, because the distribution is right-skewed.
What are the main assumptions of the lognormal model?
Drift and volatility are constant, and prices follow a continuous path with no jumps. Log returns over non-overlapping periods are independent and normal. There are no dividends or transaction costs in the basic version.
What are the main limitations of GBM for share prices?
Real returns show fat tails, jumps and volatility clustering, which GBM cannot produce. Volatility is not constant in practice, as the volatility smile shows. Dividends and trading costs also need extra modelling.