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Applications of Ito's Formula · Tutorial 186 of 600

The Lognormal Law of Geometric Brownian Motion

Use Ito’s formula to obtain the terminal lognormal distribution, then calculate its quantiles and threshold probabilities, including the degenerate deterministic case.

Advanced 11 min read

What You'll Learn

  • Derive the terminal logarithm and identify its mean and variance.
  • Explain why a deterministic Brownian integral is normally distributed.
  • Calculate lognormal quantiles from the standard normal quantile.
  • Evaluate strict and non-strict threshold probabilities.
  • Handle zero integrated variance as a point mass rather than a nondegenerate lognormal law.
  • Detect the missing Ito correction that changes the log-mean.

From Geometric Brownian Motion to a Lognormal Law

The distribution of geometric Brownian motion is easiest to identify after taking logarithms: the logarithm is a deterministic quantity plus a Brownian stochastic integral. That integral is Gaussian when its integrand is deterministic, so the terminal value is lognormal whenever the accumulated variance is positive. The case of zero accumulated variance must be handled separately: then the terminal value is deterministic.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) relative to the filtration. Fix a finite horizon \(T>0\). Let \(S_0>0\) be deterministic, and let the drift coefficient \(\mu(t)\) be integrable on \([0,T]\) and the diffusion coefficient \(\sigma(t)\) be deterministic and square-integrable there. Thus \(\int_0^T|\mu(s)|\,ds<\infty\) and \(\int_0^T\sigma(s)^2\,ds<\infty\). The drift is the coefficient of \(dt\); the diffusion coefficient is the coefficient of \(dB_t\).

Definition: Set \(M_T=\int_0^T\mu(s)\,ds\), \(V_T=\int_0^T\sigma(s)^2\,ds\), and \(m_T=\log S_0+M_T-\tfrac12V_T\). A lognormal random variable with parameters \(m\in\mathbb{R}\) and \(V>0\) is a positive random variable whose logarithm has distribution \(N(m,V)\). If \(V=0\), the corresponding exponential of a normal variable is instead the constant \(e^m\).

The integral \(\int_0^t\sigma(s)\,dB_s\) below is an Ito integral. Its deterministic integrand is progressively measurable and belongs to \(\mathcal{L}^2\), the class of progressively measurable processes \(H_s\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). The integrals in the SDE are Ito integrals as well. No Stratonovich integral is used here; in particular, no Stratonovich correction is being suppressed. The quadratic variation rule behind Ito’s formula is \([B]_t=t\), or \(dB_t\,dB_t=dt\); also \(dB_t\,dt=0\) and \(dt\,dt=0\). This is why Ito’s formula includes a second-derivative correction, unlike the ordinary chain rule.

The Terminal Distribution

For deterministic coefficients, the geometric Brownian solution can be written as \(S_t=S_0\exp\!\left(\int_0^t(\mu(s)-\tfrac12\sigma(s)^2)\,ds+\int_0^t\sigma(s)\,dB_s\right)\). The \(-\tfrac12\sigma^2\) term is essential. To verify the representation, define \(A_t\) to be the exponent and apply Ito’s formula to \(f(x)=e^x\). Its derivatives are \(f'(x)=e^x\) and \(f''(x)=e^x\), so the correction is explicitly \(\tfrac12 e^{A_t}\sigma(t)^2\,dt\). Since \(dA_t=(\mu(t)-\tfrac12\sigma(t)^2)\,dt+\sigma(t)\,dB_t\), the drift terms combine to \(\mu(t)S_t\,dt\), giving \(dS_t=\mu(t)S_t\,dt+\sigma(t)S_t\,dB_t\). This verification is local: after stopping when \(S_t\) reaches a fixed bound, the integrand \(\sigma(t)S_t\) is square-integrable, since its squared integral is bounded by that bound squared times \(\int_0^T\sigma(t)^2\,dt\). The representation is the geometric Brownian solution established by the earlier Ito-calculus material.

Theorem: Under the stated assumptions, \(\log S_T\) has distribution \(N(m_T,V_T)\) if \(V_T>0\). Consequently, \(S_T\) is lognormal with parameters \(m_T,V_T\). If \(V_T=0\), then \(S_T=e^{m_T}\) almost surely.

Proof. The representation gives \(\log S_T=m_T+\int_0^T\sigma(s)\,dB_s\). We justify the distribution of the integral. First suppose \(\sigma\) is a deterministic step function, with value \(c_j\) on each interval \((t_{j-1},t_j]\). The integral is then the sum \(\sum_j c_j(B_{t_j}-B_{t_{j-1}})\). The Brownian increments are independent normal random variables, so this sum is normal with mean zero and variance \(\sum_j c_j^2(t_j-t_{j-1})=\int_0^T\sigma(s)^2\,ds\). For a general deterministic square-integrable \(\sigma\), choose deterministic step functions \(\sigma_n\) converging to \(\sigma\) in \(L^2([0,T])\). The Ito isometry gives \(\mathbb{E}\left[\left(\int_0^T(\sigma_n-\sigma)\,dB_s\right)^2\right]=\int_0^T(\sigma_n(s)-\sigma(s))^2\,ds\to0\). Thus the integrals converge in \(L^2\), hence in probability. Their characteristic functions converge to that of the limit; those of the approximating integrals are \(\exp(-u^2\int_0^T\sigma_n(s)^2\,ds/2)\), which converge to \(\exp(-u^2V_T/2)\) for every real \(u\). This is the characteristic function of \(N(0,V_T)\) when \(V_T>0\), and of the constant zero when \(V_T=0\). Adding \(m_T\) proves the assertion for \(\log S_T\); exponentiating gives the stated law of \(S_T\). \(\square\)

Write \(\Phi\) for the standard normal cumulative distribution function and \(\Phi^{-1}\) for its inverse on \((0,1)\). When \(V_T>0\), the distribution has no atoms: for every \(K>0\), the event \(S_T=K\) is the event that a nondegenerate normal variable equals \(\log K\), which has probability zero. The formulas below therefore apply equally to strict or non-strict thresholds at positive \(K\).

Corollary: If \(V_T>0\), the \(p\)-quantile of \(S_T\), defined as \(\inf\{x:\mathbb{P}(S_T\le x)\ge p\}\), is \(q_p=\exp(m_T+\sqrt{V_T}\,\Phi^{-1}(p))\) for \(0<p<1\). For \(K>0\), \(\mathbb{P}(S_T>K)=1-\Phi((\log K-m_T)/\sqrt{V_T})\) and \(\mathbb{P}(S_T\le K)=\Phi((\log K-m_T)/\sqrt{V_T})\).

Proof. For \(x>0\), taking logarithms preserves order, so \(\mathbb{P}(S_T\le x)=\Phi((\log x-m_T)/\sqrt{V_T})\). Because \(\Phi\) is continuous and strictly increasing, the equation \(\mathbb{P}(S_T\le x)=p\) has the unique solution \(x=\exp(m_T+\sqrt{V_T}\Phi^{-1}(p))\), which is the defined quantile. Substituting \(x=K\) gives the non-strict probability. The absence of an atom at \(K\) means \(\mathbb{P}(S_T>K)=1-\mathbb{P}(S_T\le K)\), giving the strict-tail formula. If \(K\le0\), positivity of \(S_T\) instead gives \(\mathbb{P}(S_T>K)=1\) and \(\mathbb{P}(S_T\le K)=0\). \(\square\)

When \(V_T=0\), the distribution is a point mass at \(e^{m_T}\), not a nondegenerate lognormal distribution. Its \(p\)-quantile is \(e^{m_T}\) for every \(0<p<1\). For any real \(K\), \(\mathbb{P}(S_T>K)=1\) if \(e^{m_T}>K\), and is \(0\) otherwise; equality belongs to the non-strict event \(S_T\le K\).

Worked Quantiles and Thresholds

Worked Example: A Constant-Coefficient Model

Let \(S_0=100\), \(\mu=0.10\), \(\sigma=0.20\), and \(T=1\). Find the median, the 90th percentile, and \(\mathbb{P}(S_T>100)\).

Here \(M_T=0.10\cdot1=0.10\) and \(V_T=0.20^2\cdot1=0.04\), so \(\sqrt{V_T}=0.20\). The log-mean is \(m_T=\log100+0.10-\tfrac12(0.04)=\log100+0.08\). For the median, \(\Phi^{-1}(0.5)=0\), hence \(q_{0.5}=100e^{0.08}\), approximately \(108.3\). For the 90th percentile, \(\Phi^{-1}(0.9)\) is approximately \(1.2816\), so \(q_{0.9}=100\exp(0.08+0.20(1.2816))=100\exp(0.33632)\), approximately \(140.0\).

For the strict threshold \(K=100\), the standardized threshold is \((\log100-m_T)/0.20=-0.08/0.20=-0.4\). Thus \(\mathbb{P}(S_T>100)=1-\Phi(-0.4)=\Phi(0.4)\), approximately \(0.6554\). Check: the median \(108.3\) exceeds \(100\), and the probability of exceeding \(100\) is greater than one half, consistent with the log-mean lying \(0.08\) above \(\log100\). The strict and non-strict probabilities agree at this threshold because the law is continuous.

Worked Example: Piecewise Time-Dependent Coefficients

On \([0,2]\), let \(S_0=2\), let \(\mu(t)=\tfrac14\) on \([0,1]\) and \(\mu(t)=\tfrac12\) on \((1,2]\), and let \(\sigma(t)=\tfrac12\) on \([0,1]\) and \(\sigma(t)=1\) on \((1,2]\). Find the median and \(\mathbb{P}(S_2>2)\).

The drift integral is \(M_2=(\tfrac14)(1)+(\tfrac12)(1)=\tfrac34\). The variance integral is \(V_2=(\tfrac12)^2(1)+1^2(1)=\tfrac14+1=\tfrac54\). Therefore \(m_2=\log2+\tfrac34-\tfrac12(\tfrac54)=\log2+\tfrac18\), and \(\sqrt{V_2}=\sqrt5/2\). The median is \(e^{m_2}=2e^{1/8}\), approximately \(2.266\).

At the strict threshold \(K=2\), the standardized value is \((\log2-m_2)/\sqrt{V_2}=(-\tfrac18)/(\sqrt5/2)=-1/(4\sqrt5)\). It follows that \(\mathbb{P}(S_2>2)=\Phi(1/(4\sqrt5))\), approximately \(0.5445\). Check: the median \(2e^{1/8}\) is greater than \(2\), so a probability above one half of exceeding \(2\) is consistent. Both interval contributions were included in \(M_2\) and \(V_2\); using only the final coefficient on the whole interval would give a different, incorrect law.

Worked Example: A Tail Probability and an Upper Quantile

Let \(S_0=1\), \(\mu=\tfrac12\), \(\sigma=1\), and \(T=1\). Compute \(\mathbb{P}(S_T>e)\) and the 95th percentile.

We have \(M_T=\tfrac12\), \(V_T=1\), and \(m_T=\log1+\tfrac12-\tfrac12=0\). Thus \(\log S_T\) is standard normal. At \(K=e\), the standardized threshold is \((\log e-0)/1=1\), so \(\mathbb{P}(S_T>e)=1-\Phi(1)\), approximately \(0.1587\). The 95th percentile is \(q_{0.95}=\exp(\Phi^{-1}(0.95))\). Using \(\Phi^{-1}(0.95)\) approximately \(1.6449\), this is \(e^{1.6449}\), approximately \(5.180\).

Check: \(e\) is one standard deviation above the mean on the logarithmic scale, so the upper-tail probability is the standard normal upper tail beyond \(1\). The strict event \(S_T>e\) and the event \(S_T\ge e\) have the same probability because \(V_T=1>0\) gives a continuous distribution.

Worked Example: Zero Integrated Variance

Let \(S_0=3\), \(\mu(t)=-1\), \(\sigma(t)=0\), and \(T=2\). Find the terminal value, its median, and \(\mathbb{P}(S_T>1)\).

Here \(M_T=(-1)(2)=-2\) and \(V_T=0\). Thus \(m_T=\log3-2\) and \(S_T=e^{m_T}=3e^{-2}\) almost surely. The median is \(3e^{-2}\), approximately \(0.4060\). Since \(3e^{-2}<1\), the strict-threshold probability is \(\mathbb{P}(S_T>1)=0\).

Check: with zero diffusion, the SDE reduces to the ordinary differential equation \(dS_t=-S_t\,dt\), whose solution is \(S_t=3e^{-t}\); at \(t=2\) this gives exactly \(3e^{-2}\). This point mass has an atom at its terminal value, so strict and non-strict threshold events cannot be interchanged there.

A Common Error: Omitting the Ito Correction

A tempting but incorrect calculation is to write \(\log S_T=\log S_0+\int_0^T\mu(s)\,ds+\int_0^T\sigma(s)\,dB_s\). This treats the logarithm as though the ordinary chain rule applied. For constant coefficients in the first example, it would give log-mean \(\log100+0.10\) and median \(100e^{0.10}\), approximately \(110.5\). The correct Ito calculation gives log-mean \(\log100+0.08\) and median \(100e^{0.08}\), approximately \(108.3\). The discrepancy is the missing correction \(-\tfrac12V_T=-0.02\) in the log-mean.

The correction follows directly from Ito’s formula for \(f(x)=\log x\), whose derivatives are \(f'(x)=1/x\) and \(f''(x)=-1/x^2\). For \(dS_t=\mu(t)S_t\,dt+\sigma(t)S_t\,dB_t\), the second-order term is explicitly \(\tfrac12 f''(S_t)\sigma(t)^2S_t^2\,dt=-\tfrac12\sigma(t)^2\,dt\). The resulting logarithmic equation is \(d\log S_t=(\mu(t)-\tfrac12\sigma(t)^2)\,dt+\sigma(t)\,dB_t\). This is the source of the lognormal parameters, not an adjustment made after computing a normal probability.

All probabilities in these formulas are under the stated measure \(\mathbb{P}\). A change to the risk-neutral measure \(\mathbb{Q}\), when relevant in a financial model, may change the drift and therefore the log-mean; the distribution formulas must use coefficients specified under the measure being used. The results here identify a terminal distribution and its quantiles. They do not assert that \(S_t\) is a martingale: that is a separate property requiring its own conditions.

Check Your Understanding

Use the log-mean \(m_T\) and integrated variance \(V_T\), and distinguish the positive-variance and zero-variance cases.

  1. For deterministic \(\mu(t)\) and \(\sigma(t)\), derive \(m_T\) and \(V_T\) from the geometric Brownian representation.
  2. When \(V_T>0\), express the \(p\)-quantile of \(S_T\) using \(\Phi^{-1}(p)\), for \(0<p<1\).
  3. For \(K>0\), write formulas for \(\mathbb{P}(S_T>K)\) and \(\mathbb{P}(S_T\le K)\), and explain why no atom at \(K\) is present.
  4. If \(V_T=0\), what is the \(p\)-quantile, and how does \(\mathbb{P}(S_T>K)\) depend on the comparison between \(e^{m_T}\) and \(K\)?
  5. Identify the Ito correction when applying Ito’s formula to \(\log S_t\), and state the incorrect log-mean obtained if it is omitted.