Brownian motion W(t) has stationary increments (W(t+h) - W(s+h) has the same distribution as W(t) - W(s)). Is Brownian motion itself a stationary process?
AYes — stationary increments implies stationarity
BNo — Var(W(t)) = t grows with time, so the distribution of W(t) depends on t, violating stationarity
CYes, but only in the wide-sense (second-order) meaning
DIt depends on the initial condition W(0)
Stationarity requires the marginal distribution of X(t) to be time-invariant. Since W(t) ~ N(0,t), the variance grows linearly — the distribution at time 2 is N(0,2), different from the distribution at time 1 which is N(0,1). Brownian motion has stationary increments but is not itself stationary. The Ornstein-Uhlenbeck process, which adds mean reversion, is the canonical modification that produces a stationary Gaussian process.
Question 2 Multiple Choice
A wide-sense stationary process has autocovariance R(τ). The power spectral density S(ω) is defined as the Fourier transform of R(τ). If R(τ) = σ²e^{-α|τ|} (as in the Ornstein-Uhlenbeck process), what is S(ω)?
AS(ω) = σ²/(α² + ω²) · (2α), a Lorentzian (Cauchy) spectral density
BS(ω) = σ²e^{-ω²/(2α²)}, a Gaussian spectral density
CS(ω) = σ²δ(ω), concentrated at frequency zero
DS(ω) = σ²/(2π) for all ω, flat (white noise)
The Fourier transform of e^{-α|τ|} is 2α/(α² + ω²). Multiplying by σ² gives S(ω) = 2ασ²/(α² + ω²), a Lorentzian. This decays as 1/ω² for large |ω|, meaning the OU process has less power at high frequencies — it is 'smoother' than white noise (which has flat S(ω)) but 'rougher' than processes with Gaussian spectral density. The parameter α determines the cutoff frequency: below α, the spectrum is approximately flat; above α, it decays. This characterizes the OU process as colored noise with a specific bandwidth.
Question 3 True / False
Strict stationarity implies wide-sense stationarity whenever the first two moments exist.
TTrue
FFalse
Answer: True
Strict stationarity means all finite-dimensional distributions are time-invariant. If the first two moments exist, this implies E[X(t)] is constant and Cov(X(t), X(t+τ)) depends only on τ — which is exactly wide-sense stationarity. The converse is false: a Gaussian process that is wide-sense stationary is also strictly stationary (because Gaussian distributions are determined by their first two moments), but for non-Gaussian processes, wide-sense stationarity is strictly weaker.
Question 4 Short Answer
Explain why white noise (a process with R(τ) = σ²δ(τ)) cannot be a well-defined stochastic process with continuous sample paths.
Think about your answer, then reveal below.
Model answer: White noise has autocovariance R(τ) = σ²δ(τ), meaning the process is uncorrelated at every pair of distinct times — Cov(X(t), X(s)) = 0 for t ≠ s. For a continuous process, X(t) → X(s) as t → s, which would force Cov(X(t), X(s)) → Var(X(s)) = σ² as t → s. But R(τ) jumps from 0 to σ² at τ = 0, so the covariance is discontinuous — incompatible with path continuity. White noise exists as a generalized process (a distribution-valued process) or as the formal derivative of Brownian motion dW/dt, but not as a pointwise-defined continuous process.
This is why SDEs are written dX = σ dW rather than dX/dt = σ · (white noise). White noise ξ(t) = dW/dt doesn't exist as a function but does exist as a measure or distribution. The Itô integral ∫f dW is the rigorous substitute for ∫f(t)ξ(t)dt.