Covariances Across Times for Geometric Brownian Motion

Why Values at Different Times Are Related

Geometric Brownian motion at a later time contains the Brownian history accumulated by the earlier time. To calculate how strongly the two values are related, we need to separate that shared history from the Brownian increment that occurs afterward. This gives the product moment $\mathbb{E}[S_sS_t]$, from which covariance and correlation follow.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Fix a deterministic initial value $S_0>0$, a drift $\mu\in\mathbb{R}$, and a diffusion coefficient $\sigma\in\mathbb{R}$. The geometric Brownian motion considered here is the solution of the Ito SDE $dS_t=\mu S_t\,dt+\sigma S_t\,dB_t$. The first coefficient is the drift; the second is the diffusion coefficient. The stochastic integral is an Ito integral with progressively measurable integrand $\sigma S_t$ in $\mathcal{L}^2$ on every finite time interval: indeed, the moment calculation below gives $\mathbb{E}\int_0^T \sigma^2S_u^2\,du<\infty$. The stochastic integral is not a Stratonovich integral.

Definition: For square-integrable random variables $X,Y$, their covariance is $\operatorname{Cov}(X,Y)=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]$. If both variances are positive, their correlation is $\operatorname{Corr}(X,Y)=\operatorname{Cov}(X,Y)/\sqrt{\operatorname{Var}(X)\operatorname{Var}(Y)}$. Correlation is undefined if either variance is zero.

The Solution and Its Moments

For $t\ge0$, the explicit solution is $S_t=S_0\exp((\mu-\tfrac12\sigma^2)t+\sigma B_t)$. To verify the formula, take $f(t,x)=S_0\exp((\mu-\tfrac12\sigma^2)t+\sigma x)$. Its derivatives are $f_t=(\mu-\tfrac12\sigma^2)f$, $f_x=\sigma f$, and $f_{xx}=\sigma^2 f$. Ito's formula, including its second-derivative correction, gives

df(t,B_t)=\left(f_t+\tfrac12 f_{xx}\right)dt+f_x\,dB_t =\mu f\,dt+\sigma f\,dB_t,

where the correction is $\tfrac12 f_{xx}\,dt=\tfrac12\sigma^2f\,dt$, using $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$. Also $f(0,B_0)=S_0$. Thus the candidate satisfies the SDE. The ordinary chain rule would omit the $\tfrac12 f_{xx}\,dt$ term and would not give this Ito solution.

Since $B_u$ has the normal distribution $N(0,u)$, its exponential moment is $\mathbb{E}[e^{aB_u}]=e^{a^2u/2}$ for real $a$. Substitution into the explicit solution yields, in particular,

\mathbb{E}[S_u]=S_0e^{\mu u}, \qquad \mathbb{E}[S_u^2]=S_0^2e^{(2\mu+\sigma^2)u}.

The second expression is finite for every finite $u$. Consequently, $\mathbb{E}\int_0^T\sigma^2S_u^2\,du=\int_0^T\sigma^2S_0^2e^{(2\mu+\sigma^2)u}\,du<\infty$ for finite $T$, as required for the Ito integrand. These formulas also give $\operatorname{Var}(S_u)=S_0^2e^{2\mu u}(e^{\sigma^2u}-1)$.

Product Moment and Covariance Across Times

Theorem: For $s,t\ge0$, geometric Brownian motion with the parameters above satisfies $$ \mathbb{E}[S_sS_t]=S_0^2e^{\mu(s+t)+\sigma^2\min(s,t)} $$ and $$ \operatorname{Cov}(S_s,S_t)=S_0^2e^{\mu(s+t)}\left(e^{\sigma^2\min(s,t)}-1\right). $$

Proof. First suppose $0\le s\le t$. From the explicit solution, factor the later value into the earlier value and the increment after time $s$: $S_t=S_s\exp((\mu-\tfrac12\sigma^2)(t-s)+\sigma(B_t-B_s))$. The increment $B_t-B_s$ is independent of $\mathcal{F}_s$ and has distribution $N(0,t-s)$. Since $S_s$ is $\mathcal{F}_s$-measurable, the normal exponential moment gives $\mathbb{E}[S_t\mid\mathcal{F}_s]=S_se^{\mu(t-s)}$. The variables involved are integrable; in particular $S_sS_t$ is integrable by the finite second moments and the Cauchy–Schwarz inequality. Therefore $\mathbb{E}[S_sS_t]=e^{\mu(t-s)}\mathbb{E}[S_s^2]$. Using the second-moment formula above, this is $S_0^2e^{\mu(t-s)+(2\mu+\sigma^2)s}=S_0^2e^{\mu(s+t)+\sigma^2s}$. Subtracting $\mathbb{E}[S_s]\mathbb{E}[S_t]=S_0^2e^{\mu(s+t)}$ gives the covariance formula for $s\le t$. If $t<s$, interchange the labels $s$ and $t$: both the product and covariance are symmetric in their arguments, so $s$ in the exponent is replaced by $\min(s,t)$. This proves both formulas for all $s,t\ge0$. $\square$

The term $\sigma^2\min(s,t)$ records how much Brownian history is common to both values. If $s\le t$, the random value at time $s$ depends on $B_s$, and the later value contains that same $B_s$ in addition to the independent increment $B_t-B_s$. Only the later increment is new after time $s$.

Corollary: If $s,t>0$ and $\sigma\ne0$, then $$ \operatorname{Corr}(S_s,S_t)= \frac{e^{\sigma^2\min(s,t)}-1} {\sqrt{(e^{\sigma^2s}-1)(e^{\sigma^2t}-1)}}. $$ For $0<s\le t$, this simplifies to $\sqrt{(e^{\sigma^2s}-1)/(e^{\sigma^2t}-1)}$. If $s=0$ or $t=0$, or if $\sigma=0$, at least one value is deterministic and its variance is zero, so the correlation is undefined.

Proof. The variance formula gives $\operatorname{Var}(S_u)=S_0^2e^{2\mu u}(e^{\sigma^2u}-1)$. It is positive exactly when $u>0$ and $\sigma\ne0$, since $S_0>0$. Under those conditions at both times, insert the covariance from the Theorem and these two variances into the definition of correlation. The positive factors $S_0^2e^{\mu(s+t)}$ cancel, giving the displayed expression. If $s\le t$, then $\min(s,t)=s$, and cancelling the positive factor $e^{\sigma^2s}-1$ between numerator and denominator gives the simplified formula. In the stated degenerate cases the variance definition shows directly why correlation is not defined. $\square$

Worked Calculations

Worked Example: Two Positive Times

Let $S_0=2$, $\mu=\tfrac14$, and $\sigma=\tfrac12$. Compute the product moment, covariance, and correlation at $s=1$ and $t=3$.

Here $\sigma^2=\tfrac14$, $s+t=4$, and $\min(s,t)=1$. The product moment is therefore $\mathbb{E}[S_1S_3]=2^2e^{(\frac14)(4)+(\frac14)(1)}=4e^{5/4}$. The means are $\mathbb{E}[S_1]=2e^{1/4}$ and $\mathbb{E}[S_3]=2e^{3/4}$, whose product is $4e$. Hence $\operatorname{Cov}(S_1,S_3)=4e^{5/4}-4e=4e(e^{1/4}-1)$.

For the correlation, the exponential factors from the means cancel, leaving $\operatorname{Corr}(S_1,S_3)=\sqrt{(e^{1/4}-1)/(e^{3/4}-1)}$. Check: multiplying the two means gives $4e^{1/4+3/4}=4e$; subtracting this from the product moment $4e^{5/4}$ recovers $4e(e^{1/4}-1)$. The correlation is positive and less than $1$, since $0<e^{1/4}-1<e^{3/4}-1$.

Worked Example: Zero Drift and a Shared Brownian Increment

Set $S_0=1$, $\mu=0$, and $\sigma=1$, and take $s=1,t=2$. Find the two-time moments and correlation, then identify the shared contribution.

The moments are $\mathbb{E}[S_1]=1$ and $\mathbb{E}[S_2]=1$. Since $\min(1,2)=1$, $\mathbb{E}[S_1S_2]=e$, so $\operatorname{Cov}(S_1,S_2)=e-1$. The variances are $\operatorname{Var}(S_1)=e-1$ and $\operatorname{Var}(S_2)=e^2-1$. Thus $\operatorname{Corr}(S_1,S_2)=(e-1)/\sqrt{(e-1)(e^2-1)}=1/\sqrt{e+1}$, because $e^2-1=(e-1)(e+1)$.

In this case $S_1=e^{-\frac12+B_1}$, and $S_2=e^{-\frac12+B_1}e^{-\frac12+(B_2-B_1)}$. The first factor is exactly $S_1$; the second factor uses the increment after time $1$, independent of the history through time $1$. Check: the product of the means is $1$, and subtracting it from $\mathbb{E}[S_1S_2]=e$ gives $e-1$, as required.

Worked Example: Equal Times and a Degenerate Case

First take $S_0=2,\mu=0,\sigma=1$ at $s=t=1$. The product moment is $\mathbb{E}[S_1^2]=4e$. Since $\mathbb{E}[S_1]=2$, the variance is $4e-2^2=4(e-1)$, and the covariance of $S_1$ with itself is the same variance. Its correlation with itself is $4(e-1)/\sqrt{4(e-1)\,4(e-1)}=1$. This agrees with the general formula at equal positive times.

Now take $S_0=5,\mu=-\tfrac13,\sigma=0$, at $s=1,t=2$. Then $S_u=5e^{-u/3}$ is deterministic. Thus $\mathbb{E}[S_1S_2]=(5e^{-1/3})(5e^{-2/3})=25e^{-1}$, and $\mathbb{E}[S_1]\mathbb{E}[S_2]=25e^{-1}$, so the covariance is $0$. Both variances are zero, so correlation is undefined. Check: the general product formula gives $25e^{-\frac13(1+2)+0}=25e^{-1}$, exactly the deterministic product.

Dependence, Independence, and a Common Mistake

For $0<s<t$ and $\sigma\ne0$, the covariance formula is strictly positive: $S_0^2e^{\mu(s+t)}>0$ and $e^{\sigma^2s}-1>0$. Since these variables have finite second moments, independence would imply $\mathbb{E}[S_sS_t]=\mathbb{E}[S_s]\mathbb{E}[S_t]$, and hence zero covariance. The positive covariance therefore proves that $S_s$ and $S_t$ are not independent in this case.

The qualification matters. If $\sigma=0$, the process is deterministic, so its values are independent in the usual probabilistic sense: events determined by deterministic random variables have probability zero or one, and their joint probabilities factor. If $s=0$, $S_s=S_0$ is deterministic and is independent of $S_t$, even when $\sigma\ne0$. In either deterministic case correlation is still undefined because the variance of at least one variable is zero.

Common mistake. Treating $S_s$ and $S_t$ as independent for $s<t$ gives the incorrect calculation $\mathbb{E}[S_sS_t]=\mathbb{E}[S_s]\mathbb{E}[S_t]$, and so incorrectly sets their covariance to zero. In the second worked example, this would give $1$ instead of the correct product moment $e$, and zero instead of covariance $e-1$. The error is that $B_t$ is not a fresh Brownian value independent of $B_s$: it decomposes as $B_s+(B_t-B_s)$. The increment $B_t-B_s$ is independent of the past, but $B_s$ is shared by the two process values.

The covariance records the magnitude of this shared contribution in the original units of $S$; correlation rescales it by the two standard deviations. Neither quantity should be confused with the conditional mean $\mathbb{E}[S_t\mid\mathcal{F}_s]=S_se^{\mu(t-s)}$: that conditional expectation depends on the realized value at time $s$, whereas covariance and correlation are unconditional summaries.

Check Your Understanding

Use the two-time formulas and check whether each correlation is defined before calculating it.

For $0<s\le t$, derive the simplified correlation formula from the general expression.
For $S_0=1,\mu=0,\sigma=2$, compute $\mathbb{E}[S_1S_2]$ and $\operatorname{Cov}(S_1,S_2)$.
Explain why positive covariance proves non-independence when $0<s<t$ and $\sigma\ne0$.
For $\sigma=0$, explain why the covariance is zero but the correlation is undefined.
At $s=t>0$ with $\sigma\ne0$, verify from the definitions that the correlation is $1$.