Conditional Moments and Transition Laws of Geometric Brownian Motion

From the Explicit Solution to a Conditional Law

Let \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carry a standard Brownian motion \(B_t\) relative to the filtration: in particular, for \(0\le s<t\), \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has distribution \(N(0,t-s)\). Consider geometric Brownian motion with deterministic \(S_0>0\) and constant drift \(\mu\in\mathbb{R}\) and diffusion coefficient \(\sigma\ge0\). Under \(\mathbb{P}\), it satisfies \(dS_t=\mu S_t\,dt+\sigma S_t\,dB_t\).

For each finite \(T\), the stochastic integral in this equation is the Ito integral with integrand \(H_t=\sigma S_t\). This integrand is progressively measurable and belongs to \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_t^2\,dt<\infty\). The explicit solution, established by applying Ito's formula to an exponential transform, is \(S_t=S_0\exp((\mu-\tfrac12\sigma^2)t+\sigma B_t)\). The half-variance term is essential: it comes from Ito's \(\tfrac12 f''\,dt\) correction, not from the ordinary chain rule.

Fix \(0\le s<t\), and write \(h=t-s>0\). Dividing the explicit solution at time \(t\) by the one at time \(s\) gives \(S_t=S_s\exp((\mu-\tfrac12\sigma^2)h+\sigma(B_t-B_s))\). Thus, given the information \(\mathcal{F}_s\), the random factor multiplying \(S_s\) depends only on a Gaussian increment independent of that information. This yields the transition law and, by conditioning on \(S_s\), the conditional distribution given the current value.

Conditional Transition Theorem

Proof. Set \(Z=B_t-B_s\). By the Brownian increment property, \(Z\) is independent of \(\mathcal{F}_s\) and has distribution \(N(0,h)\). The explicit solution gives \(S_t=S_s\exp((\mu-\tfrac12\sigma^2)h+\sigma Z)\). For fixed \(x>0\), let \(g_A(x,z)=\mathbf{1}_A(x\exp((\mu-\tfrac12\sigma^2)h+\sigma z))\), a bounded measurable function. Independence implies \(\mathbb{E}[g_A(S_s,Z)\mid\mathcal{F}_s]=\int_{\mathbb{R}}g_A(S_s,z)\,\nu_h(dz)\), where \(\nu_h=N(0,h)\). The integral is \(K_h(S_s,A)\) by definition. This proves the conditional law given \(\mathcal{F}_s\). Since \(S_s\) is \(\mathcal{F}_s\)-measurable, conditioning that identity again on \(S_s\) gives \(\mathbb{P}(S_t\in A\mid S_s)=K_h(S_s,A)\) almost surely. For \(\sigma=0\), the displayed map of \(z\) is constant, giving the stated point mass. \(\square\)

Conditioning on a particular value \(S_s=x\) is shorthand for evaluating this regular conditional distribution at \(x\). If \(S_s\) has a continuous distribution, the event \(S_s=x\) may have probability zero; the transition kernel still specifies a conditional law for every \(x>0\). Because \(S_0>0\) and the exponential solution is positive, \(S_s>0\) almost surely.

Conditional Moments

Proof. When \(\sigma>0\), let \(Z=B_t-B_s\sim N(0,h)\). Raising the solution ratio to the power \(p\) gives \(S_t^p=S_s^p\exp(p(\mu-\tfrac12\sigma^2)h+p\sigma Z)\). For a centered normal variable with variance \(h\), completing the square in its density gives \(\mathbb{E}[e^{aZ}]=e^{a^2h/2}\) for every real \(a\). Independence of \(Z\) from \(\mathcal{F}_s\) therefore yields \[ \mathbb{E}[S_t^p\mid\mathcal{F}_s] =S_s^p\exp\!\left(p(\mu-\tfrac12\sigma^2)h+\tfrac12p^2\sigma^2h\right) =S_s^p\exp\!\left(p\mu h+\tfrac12p(p-1)\sigma^2h\right). \] The Gaussian exponential moment is finite for every real \(a\), establishing finiteness even for negative \(p\), since \(S_t>0\). When \(\sigma=0\), direct substitution of \(S_t=S_s e^{\mu h}\) gives the same formula. Conditioning once more on \(S_s\) gives the assertion conditional on \(S_s\). \(\square\)

Taking \(p=1\) gives the conditional mean; taking \(p=2\) and subtracting the square of that mean gives the conditional variance. With \(q=\sigma^2h\), the formulas are \[ \mathbb{E}[S_t\mid\mathcal{F}_s]=S_s e^{\mu h},\qquad \operatorname{Var}(S_t\mid\mathcal{F}_s)=S_s^2e^{2\mu h}(e^q-1). \]

For an integer \(n\ge1\), the raw conditional moment is \(S_s^n\exp(n\mu h+\tfrac12n(n-1)\sigma^2h)\). Higher central moments can be formed from these raw moments. For example, writing \(m=S_s e^{\mu h}\), the conditional third central moment is \(m^3(e^{3q}-3e^q+2)\): expand \(\mathbb{E}[(S_t-m)^3\mid\mathcal{F}_s]\) as the third raw moment minus \(3m\) times the second raw moment plus \(2m^3\).

Worked Conditional Calculations

Common Mistake: Using the Unconditional Time Horizon

A frequent error is to use an unconditional moment formula with \(t\) in place of the elapsed time \(h=t-s\), and to omit the current value \(S_s\). For example, take \(\mu=0\), \(\sigma=1\), \(S_0=1\), \(s=1\), and \(t=2\). The unconditional formula gives \(\mathbb{E}[S_2^2]=e^2\), but this is not the conditional second moment given \(\mathcal{F}_1\). The correct calculation uses \(h=2-1=1\): \(\mathbb{E}[S_2^2\mid\mathcal{F}_1]=S_1^2e^1\). The current value is random when conditioning on \(\mathcal{F}_1\); conditioning only on \(S_1\) gives the same expression with the observed current value substituted.

This distinction also identifies the role of the coefficients. The drift \(\mu\) affects the conditional mean through \(e^{\mu h}\), while the diffusion coefficient determines the log variance \(q=\sigma^2h\) and the factor \(e^q-1\) in the conditional variance. The formulas here are under \(\mathbb{P}\); no change to a risk-neutral measure \(\mathbb{Q}\) is being made. Nor do the conditional moment calculations assert that \(S_t\) is a martingale: the conditional mean equals \(S_s e^{\mu h}\), which need not equal \(S_s\).

The result is useful whenever a calculation starts from the current asset value rather than from time zero. The kernel gives transition probabilities, while its moments support conditional forecasting and moment calculations. In each use, the conditioning information and the elapsed interval must be kept explicit.