Brownian Motion: Covariance, Independent Increments, and Conditional Probability
Let $\{B_t\}_{t \geq 0}$ be a standard Brownian motion (i.e., $B_0 = 0$, continuous paths, stationary independent increments, and $B_t \sim N(0, t)$).
- Show that $\text{Cov}(B_s, B_t) = \min(s, t)$ for any $s, t \geq 0$.
- Prove that increments over disjoint intervals are independent. That is, if $0 \leq a < b \leq c < d$, show that $B_b - B_a$ and $B_d - B_c$ are independent.
- Given $0 < s < t$ and $B_s = b$, compute $P(B_t > 0 \mid B_s = b)$ in closed form and interpret what the result tells you about the future path of Brownian motion.
Hints
- For part (i), try writing $B_t$ as $B_s + (B_t - B_s)$ and use the fact that increments over non-overlapping intervals are independent.
- For part (ii), think about how any collection of disjoint-interval increments can be embedded in a partition of consecutive increments, which are independent by definition.
- For part (iii), use the Markov property: $B_t - B_s$ is independent of $B_s$ and distributed as $N(0, t-s)$, so $B_t \mid B_s = b$ is just a shifted normal.
Worked Solution
How to Think About It: These three parts test whether you really understand what Brownian motion *is*, not just its definition. Part (i) is the workhorse identity -- it falls straight out of the independent increments property and the variance formula. Part (ii) is essentially baked into the definition for most constructions, but the interviewer wants to see you handle it cleanly. Part (iii) is where you show you understand the Markov property: given $B_s = b$, the future increment $B_t - B_s$ is independent of the past and normally distributed, so $B_t \mid B_s = b$ is just $N(b, t - s)$. Before any computation, your gut should say: the conditional probability is just a normal CDF evaluated at a shifted, rescaled argument.
Quick Estimate: For part (iii), suppose $s = 1$, $t = 4$, $b = 1$. Then $B_t \mid B_s = 1 \sim N(1, 3)$, so $P(B_t > 0 \mid B_s = 1) = P\left(Z > \frac{-1}{\sqrt{3}}\right) = \Phi\left(\frac{1}{\sqrt{3}}\right) \approx \Phi(0.577) \approx 0.72$. Makes sense -- you are starting at $b = 1$ with standard deviation $\sqrt{3} \approx 1.73$, so being above zero is comfortably likely but not certain.
Approach: We use the defining properties of Brownian motion (independent, stationary, Gaussian increments) to derive each result.
Formal Solution:
Part (i): Covariance
Assume without loss of generality that $s \leq t$. Write $B_t = B_s + (B_t - B_s)$. Then:
$\text{Cov}(B_s, B_t) = \text{Cov}(B_s, B_s + (B_t - B_s))$
$= \text{Cov}(B_s, B_s) + \text{Cov}(B_s, B_t - B_s)$
Now $B_s = B_s - B_0$ is the increment over $[0, s]$, and $B_t - B_s$ is the increment over $[s, t]$. Since $s \leq t$, these are increments over disjoint intervals, so they are independent, giving $\text{Cov}(B_s, B_t - B_s) = 0$. Therefore:
$\text{Cov}(B_s, B_t) = \text{Var}(B_s) = s = \min(s, t)$
By symmetry in $s$ and $t$, the general result is $\text{Cov}(B_s, B_t) = \min(s, t)$. $\square$
Part (ii): Independence of increments over disjoint intervals
Let $0 \leq a < b \leq c < d$. We want to show $B_b - B_a$ and $B_d - B_c$ are independent.
Decompose the increments using the partition $0 \leq a < b \leq c < d$:
- $B_b - B_a$ is the increment over $[a, b]$
- $B_d - B_c$ is the increment over $[c, d]$
By the definition of Brownian motion, the increments $B_a - B_0$, $B_b - B_a$, $B_c - B_b$, and $B_d - B_c$ (over the non-overlapping intervals $[0,a]$, $[a,b]$, $[b,c]$, $[c,d]$) are mutually independent. In particular, $B_b - B_a$ and $B_d - B_c$ are independent as they are components of this mutually independent collection. $\square$
Note: if the interviewer defines Brownian motion only with *consecutive* independent increments $B_{t_1} - B_{t_0}, B_{t_2} - B_{t_1}, \ldots$, then you get the disjoint case by noting that $B_b - B_a$ and $B_d - B_c$ are each functions of disjoint subsets of these consecutive increments. Independence of functions of independent random variables gives the result.
Part (iii): Conditional probability
By the Markov property, $B_t - B_s$ is independent of $\mathcal{F}_s$ (and in particular of $B_s$), and $B_t - B_s \sim N(0, t - s)$. Therefore:
$B_t \mid B_s = b \sim N(b, \, t - s)$
So:
$P(B_t > 0 \mid B_s = b) = P\left(\frac{B_t - b}{\sqrt{t - s}} > \frac{-b}{\sqrt{t - s}}\right) = \Phi\left(\frac{b}{\sqrt{t - s}}\right)$
where $\Phi$ is the standard normal CDF.
Interpretation: The conditional probability depends on $b$ and $t - s$ only through the ratio $b / \sqrt{t-s}$. This makes sense:
- If $b > 0$ (you are currently above zero), the probability exceeds /2$ and grows as $b$ increases or $t - s$ shrinks.
- If $b = 0$, the probability is exactly
/2$ by symmetry.- If $b < 0$, the probability is below
/2$.- As $t - s \to 0^{+}$, the probability goes to
$ if $b > 0$ (continuity of paths -- the motion has not had time to diffuse away from $b$).- As $t - s \to \infty$, the probability approaches
/2$ regardless of $b$ (the diffusion "forgets" its starting point).Answer:
- (i) $\text{Cov}(B_s, B_t) = \min(s, t)$
- (ii) Follows directly from the defining independent increments property of Brownian motion applied to disjoint intervals.
- (iii) $P(B_t > 0 \mid B_s = b) = \Phi\!\left(\dfrac{b}{\sqrt{t - s}}\right)$
Intuition
The covariance formula $\text{Cov}(B_s, B_t) = \min(s, t)$ is the single most important identity for Brownian motion. It tells you that the correlation between $B_s$ and $B_t$ is $\sqrt{\min(s,t) / \max(s,t)}$ -- high when the times are close, and decaying as they separate. Every Gaussian process is determined by its mean and covariance function, so this one formula pins down the entire distribution of the process. In practice, whenever you need to simulate, condition on, or regress against Brownian paths, you are implicitly using this identity.
Part (iii) illustrates the Markov property in its most concrete form: the conditional distribution of $B_t$ given $B_s = b$ is $N(b, t-s)$, which depends only on the current value and the remaining time -- not on how you got to $b$. This is the foundation of option pricing (Black-Scholes starts here), filtering, and stochastic control. The ratio $b / \sqrt{t-s}$ shows up constantly in finance as a "signal-to-noise" measure: how far the current level is from a threshold, measured in units of future diffusion. A common mistake is to overthink part (iii) by trying to use Bayes' rule on densities rather than simply exploiting the independence of the increment $B_t - B_s$ from $B_s$.
- If $b = 0$, the probability is exactly