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Explicit Solutions: Linear SDEs · Tutorial 240 of 600

Lognormal Transition Laws For Geometric Brownian Motion

Learn how to obtain and use the conditional distribution, density, interval probabilities and quantiles of geometric Brownian motion.

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What You'll Learn

  • Express the transition from time s to time t using a Brownian increment independent of the information at time s.
  • Derive the conditional CDF and density when the elapsed time and volatility are positive.
  • Handle zero elapsed time and zero volatility as degenerate transition laws.
  • Convert interval endpoints into standard-normal probabilities, including strict and non-strict endpoints.
  • Calculate conditional quantiles and check them by substitution into the CDF.

From the SDE to a Conditional Distribution

Geometric Brownian motion does more than give an explicit solution to a stochastic differential equation: it gives a transition law from any current value \(S_s\) to a later value \(S_t\). This law is useful when calculating the probability that a price, population or other positive quantity lies in a specified interval at a future time. Its form follows from two facts: the logarithm of the process is normal over a time interval, and Brownian increments after time \(s\) are independent of the information available at time \(s\).

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). Let \(S_0=s_0>0\) be deterministic, and let \(S_t\) solve the constant-coefficient geometric Brownian motion SDE \(dS_t=\mu S_t\,dt+\sigma S_t\,dB_t\), where \(\mu\in\mathbb{R}\) and \(\sigma>0\). The drift coefficient is \(\mu S_t\), and the diffusion coefficient is \(\sigma S_t\). The Ito integral is understood for progressively measurable integrands in \(\mathcal{L}^2\), meaning that \(\mathbb{E}\int_0^T H_u^2\,du<\infty\) for each finite \(T\). For this SDE, the integrand is \(H_u=\sigma S_u\); with deterministic \(s_0\), it is square-integrable on finite time intervals.

The explicit solution established for geometric Brownian motion is \(S_t=s_0\exp((\mu-\tfrac12\sigma^2)t+\sigma B_t)\). Dividing the expression at time \(t\) by the one at time \(s\) gives the transition representation. We use this established solution rather than repeating its SDE derivation.

$$ S_t=S_s\exp\!\left(\left(\mu-\frac12\sigma^2\right)(t-s)+\sigma(B_t-B_s)\right),\qquad 0\le s<t. $$

Here \(S_s\) is \(\mathcal{F}_s\)-measurable, while \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has distribution \(N(0,t-s)\). Thus the conditional distribution of \(S_t\) is determined by the current value \(S_s\) and the elapsed time \(t-s\).

The Conditional Lognormal Law

Theorem: Let \(h=t-s>0\), and set \(m=(\mu-\tfrac12\sigma^2)h\) and \(v=\sigma\sqrt{h}\). For \(\sigma>0\), the conditional distribution of \(S_t\) given \(\mathcal{F}_s\) is the distribution of \(S_s\exp(m+vZ)\), where \(Z\sim N(0,1)\) is independent of \(\mathcal{F}_s\). In particular, for \(y>0\), \(\mathbb{P}(S_t\le y\mid\mathcal{F}_s)=\Phi\big((\ln(y/S_s)-m)/v\big)\) almost surely. Its conditional density, for \(y>0\), is \(p_{s,t}(S_s,y)=\dfrac{1}{yv\sqrt{2\pi}}\exp\!\left(-\dfrac{(\ln(y/S_s)-m)^2}{2v^2}\right)\). The density is zero for \(y\le0\).

Proof. From the transition representation, \(S_t=S_s\exp(m+\sigma(B_t-B_s))\). The standardized increment \(Z=(B_t-B_s)/\sqrt{h}\) has distribution \(N(0,1)\) and is independent of \(\mathcal{F}_s\); hence \(\sigma(B_t-B_s)=vZ\). Since \(S_s>0\), taking logarithms shows, conditionally on \(\mathcal{F}_s\), that \(S_t\le y\) for \(y>0\) exactly when \(Z\le(\ln(y/S_s)-m)/v\). The standard normal CDF gives the stated conditional CDF. For \(y\le0\), \(S_t>0\), so the conditional probability is zero. To obtain the density, differentiate the CDF with respect to \(y>0\): the derivative of \(\ln(y/S_s)\) is \(1/y\), and the standard normal density at \(z\) is \(\exp(-z^2/2)/\sqrt{2\pi}\). The chain rule therefore gives the displayed expression. It integrates to one by the substitution \(z=(\ln(y/S_s)-m)/v\), which maps \(y\in(0,\infty)\) onto \(z\in\mathbb{R}\). \(\square\)

The formula is conditional on the information \(\mathcal{F}_s\): its current-value argument is the random variable \(S_s\). Equivalently, for a fixed current value \(x>0\), it gives the transition kernel conditional on \(S_s=x\), for values of \(x\) where a regular conditional distribution is specified. The same formula applies to either a strict or non-strict interval endpoint when \(h>0\) and \(\sigma>0\), because the conditional distribution has a density and hence assigns probability zero to any single endpoint.

Degenerate cases: If \(t=s\), then \(S_t=S_s\) and the conditional law is a point mass at \(S_s\), not a density. If \(\sigma=0\), the transition is deterministic: \(S_t=S_s e^{\mu(t-s)}\). The density formula and the normal-standardization formulas require both \(t-s>0\) and \(\sigma>0\).

Interval Probabilities

For \(0<a<b\), standardize both logarithmic endpoints. With \(m=(\mu-\tfrac12\sigma^2)(t-s)\) and \(v=\sigma\sqrt{t-s}>0\), the conditional probability of an interval is

$$ \mathbb{P}(a<S_t\le b\mid\mathcal{F}_s) = \Phi\!\left(\frac{\ln(b/S_s)-m}{v}\right) - \Phi\!\left(\frac{\ln(a/S_s)-m}{v}\right). $$

Replacing either strict inequality by a non-strict one does not change this probability under the nondegenerate assumptions. If the interval includes nonpositive values, use \(S_t>0\): for example, \(\mathbb{P}(S_t\le b\mid\mathcal{F}_s)=0\) when \(b\le0\). For a degenerate transition, endpoint inclusion can matter, since a point mass at an endpoint is counted by a non-strict inequality but not by the corresponding strict inequality.

Worked Example: Probability Between Two Logarithmic Thresholds

Suppose the current value is \(S_s=100\), the elapsed time is \(t-s=1\), the drift is \(\mu=0.10\), and the volatility is \(\sigma=0.20\). Find the conditional probability that \(100e^{-0.12}<S_t\le100e^{0.28}\).

First compute the log-drift and log-standard-deviation parameters: \((0.10-\tfrac12(0.20)^2)\cdot1=(0.10-0.02)\cdot1=0.08\), and \(0.20\sqrt{1}=0.20\). The lower standardized endpoint is \((-0.12-0.08)/0.20=-1\). The upper standardized endpoint is \((0.28-0.08)/0.20=1\). Therefore

$$ \mathbb{P}(100e^{-0.12}<S_t\le100e^{0.28}\mid S_s=100) =\Phi(1)-\Phi(-1) =2\Phi(1)-1. $$

Using \(\Phi(1)\approx0.8413\), the probability is approximately \(2(0.8413)-1=0.6826\). The endpoints are strict at the lower end and non-strict at the upper end as asked; changing either convention leaves the result unchanged here because the transition has a continuous density. As a check, the two standardized thresholds are symmetric about zero, so the probability must be the standard normal probability between \(-1\) and \(1\), as obtained.

Quantiles of the Transition Law

For a nondegenerate transition and a probability level \(p\) with \(0<p<1\), define the conditional \(p\)-quantile as the value \(q_p\) whose conditional CDF equals \(p\). The CDF is continuous and strictly increasing on \((0,\infty)\), so this value is unique. Inverting the CDF gives

$$ q_p(S_s)=S_s\exp\!\left((\mu-\tfrac12\sigma^2)(t-s)+\sigma\sqrt{t-s}\,\Phi^{-1}(p)\right). $$

Here \(\Phi^{-1}(p)\) is the unique real number \(z\) satisfying \(\Phi(z)=p\). The formula is conditional: when \(S_s\) is random, \(q_p(S_s)\) is an \(\mathcal{F}_s\)-measurable random threshold. A quantile for a degenerate transition is instead its deterministic transition value; the inverse-CDF formula above is not applicable to a point mass.

Justification of uniqueness and inversion. The standard normal CDF \(\Phi\) is continuous and strictly increasing from zero to one. The map \(y\mapsto(\ln(y/S_s)-m)/v\) is also continuous and strictly increasing from \(-\infty\) to \(+\infty\) for \(y>0\). Their composition, the conditional CDF, therefore has exactly one positive solution to \(F(y)=p\) for each \(0<p<1\). Setting its normal-CDF argument equal to \(\Phi^{-1}(p)\) and exponentiating yields the stated expression.

Worked Example: A Conditional Median and an Upper Quantile

Let \(S_s=80\), \(t-s=1\), \(\mu=0.10\), and \(\sigma=0.20\). Compute the conditional median and the quantile corresponding to probability level \(p=\Phi(1)\).

The log-drift is \((0.10-\tfrac12(0.20)^2)\cdot1=0.08\), and the log-standard deviation is \(0.20\sqrt{1}=0.20\). Since \(\Phi^{-1}(0.5)=0\), the median is \(q_{0.5}=80e^{0.08}\). For \(p=\Phi(1)\), the defining identity gives \(\Phi^{-1}(p)=1\), so the quantile is \(q_p=80e^{0.08+0.20}=80e^{0.28}\).

$$ q_{0.5}=80e^{0.08},\qquad q_{\Phi(1)}=80e^{0.28}. $$

To check the first value, substitute \(80e^{0.08}\) into the CDF: its standardized logarithm is \((\ln((80e^{0.08})/80)-0.08)/0.20=(0.08-0.08)/0.20=0\), giving \(\Phi(0)=0.5\). For the second value, the standardized logarithm is \((0.28-0.08)/0.20=1\), giving \(\Phi(1)=p\), as required. The second threshold is larger because its probability level is above the median.

Worked Example: Conditional Density at the Current Value

Suppose \(S_s=50\), \(t-s=0.25\), \(\mu=0.08\), and \(\sigma=0.40\). Find the conditional density at \(y=50\), and the probability that \(50<S_t\le50e^{0.08}\).

The log-drift is \((0.08-\tfrac12(0.40)^2)\cdot0.25=(0.08-0.08)\cdot0.25=0\). The log-standard deviation is \(0.40\sqrt{0.25}=0.20\). At \(y=50\), the standardized logarithm is \((\ln(50/50)-0)/0.20=0\), so the density is

$$ p_{s,t}(50,50)=\frac{1}{50(0.20)\sqrt{2\pi}} =\frac{1}{10\sqrt{2\pi}}. $$

For the interval, the standardized lower endpoint is zero and the upper endpoint is \((0.08-0)/0.20=0.4\). Hence its probability is \(\Phi(0.4)-\Phi(0)=\Phi(0.4)-0.5\), approximately \(0.6554-0.5=0.1554\). The density has units of inverse value, so it is not itself a probability. The interval calculation is the corresponding CDF difference and is between zero and one, as it must be.

Common Pitfall: Using the SDE Drift as the Log-Drift

The mean rate in the SDE and the drift in the logarithm are different. The correction \(-\tfrac12\sigma^2\) is essential in the transition law. For example, with \(\mu=0.10\), \(\sigma=0.20\), and elapsed time one, the log-drift is \(0.10-\tfrac12(0.20)^2=0.08\), not \(0.10\). A wrong calculation for the upper endpoint \(100e^{0.28}\) in the first example might use \((0.28-0.10)/0.20=0.9\), leading to \(\Phi(0.9)\) rather than \(\Phi(1)\). The correction is to standardize using \(m=0.08\): \((0.28-0.08)/0.20=1\). The log-drift follows from the distribution of \(\log S_t-\log S_s\), not by copying the SDE drift coefficient unchanged.

Keep the roles of the parameters separate: the drift \(\mu\) and the quadratic-variation correction determine the center of the log-transition, while \(\sigma\sqrt{t-s}\) determines its spread. The resulting conditional law is under \(\mathbb{P}\), the probability measure of the stated SDE; a change to a risk-neutral measure \(\mathbb{Q}\), when relevant, must be specified separately and can change the drift parameter.

Check Your Understanding

In each question involving a density or normal standardization, assume \(t>s\) and \(\sigma>0\). State separately what changes for a degenerate transition.

  1. For \(S_s=x>0\), write the conditional CDF of \(S_t\) and identify its log-drift and log-standard deviation.
  2. For \(0<a<b\), derive the conditional probability of \(a<S_t\le b\) by standardizing both logarithmic endpoints.
  3. Find the conditional median in terms of \(S_s,\mu,\sigma\), and \(t-s\).
  4. Explain why changing a strict endpoint to a non-strict endpoint does not alter a nondegenerate interval probability.
  5. For \(\sigma=0\), state the transition value and explain why the nondegenerate density formula cannot be used.