Let be a Gaussian process. Define for . Prove that is also a Gaussian process, and find for .

Stochastic Processes Solution – Problem 26: Prove that is also a Gaussian process, and find for

Question

Let $\{B_{t}:t\geqslant0\}$ be a Gaussian process. Define $X_{t}=B_{t}-tB_{1}$ for $0<t<1$ . Prove that $X_{t}$ is also a Gaussian process, and find $\mathrm{cor}(X_{t},X_{s})$ for $0<t,s<1$ .

Step-by-step solution

Step 1. Prove $X_t$ is a Gaussian process. By definition, a stochastic process is Gaussian if every finite-dimensional distribution is multivariate normal. Since $\{B_t\}$ is Gaussian, for any times $t_1,\dots,t_n$ and $t=1$ , the vector $(B_{t_1},\dots,B_{t_n},B_1)$ is jointly normal. Each $X_{t_k} = B_{t_k} - t_k B_1$ is a linear combination of $B_{t_k}$ and $B_1$ . The vector $(X_{t_1},\dots,X_{t_n})$ is a linear transformation of the jointly normal vector $(B_{t_1},\dots,B_{t_n},B_1)$ . Since linear transformations preserve multivariate normality, $(X_{t_1},\dots,X_{t_n})$ is jointly normal. Hence $\{X_t\}$ is a Gaussian process.

Compute the mean: $E[X_t] = E[B_t] - tE[B_1] = 0$ .

Compute the covariance: $\mathrm{Cov}(X_t, X_s) = E[(B_t - tB_1)(B_s - sB_1)]$ $= E[B_tB_s] - sE[B_tB_1] - tE[B_1B_s] + stE[B_1^2]$ .

Using $E[B_uB_v] = \min(u,v)$ and $E[B_1^2]=1$ , with $s,t \in (0,1)$ : $\mathrm{Cov}(X_t, X_s) = \min(t,s) - st - ts + st = \min(t,s) - st$ .

Compute the correlation: $\mathrm{Var}(X_t) = t - t^2 = t(1-t)$ , $\mathrm{Var}(X_s) = s(1-s)$ . $\mathrm{cor}(X_t, X_s) = \frac{\min(t,s) - st}{\sqrt{t(1-t)s(1-s)}}$ .

Final answer

QED. $\frac{\min(t, s) - st}{\sqrt{t(1-t)s(1-s)}}$

Marking scheme

1. Checkpoints (max 7 pts total)

Proving $X_t$ is a Gaussian process (2 pts)

Identify that $(X_{t_1},\dots,X_{t_n})$ is a linear transformation of the multivariate normal vector $(B_{t_1},\dots,B_{t_n},B_1)$ . [additive, 1 pt]
Cite "linear transformations of multivariate normal vectors remain multivariate normal" (or equivalent characteristic function argument) to conclude $X_t$ is Gaussian. [additive, 1 pt]

Computing the covariance function (3 pts)

Set up $\mathrm{Cov}(X_t,X_s)$ or $E[X_tX_s]$ and correctly expand into terms involving $E[B_uB_v]$ . [additive, 1 pt]
Substitute $E[B_uB_v] = \min(u,v)$ and $E[B_1^2]=1$ (correctly handling $E[B_tB_1]=t$ and $E[B_sB_1]=s$ since $t,s<1$ ). [additive, 1 pt]
Simplify to obtain $\mathrm{Cov}(X_t,X_s) = \min(t,s) - st$ . [additive, 1 pt]

Computing the correlation coefficient (2 pts)

Correctly compute $\mathrm{Var}(X_t) = t(1-t)$ (either independently or by setting $s=t$ in the covariance). [additive, 1 pt]
Give the correct final expression $\frac{\min(t,s)-st}{\sqrt{t(1-t)s(1-s)}}$ (or the equivalent piecewise form). [additive, 1 pt]

Total (max 7)

2. Zero-credit items

Merely copying the definition of a Gaussian process without connecting it to the specific linear construction of $X_t$ .
Merely listing the generic correlation formula $\rho = \mathrm{Cov}/(\sigma_x\sigma_y)$ without computing expectations.
Assuming $B_t$ is a general Gaussian process without using standard Brownian motion properties ( $E[B_t]=0$ , $\mathrm{Cov}(B_s,B_t)=\min(s,t)$ ).

3. Deductions

Imprecise logic: Writing only $t - st$ in the final result without stating $t \le s$ or the $\min$ function: -1 pt.
Calculation error: Arithmetic mistakes in expectation expansion, variance computation, or algebraic simplification: -1 pt.
Property misuse: Incorrectly assuming $E[B_tB_1] = 1$ or other violations of the $t<1$ condition: -1 pt.
Maximum deduction rule: deductions for the same logical chain apply only once; total score cannot go below 0.

Stochastic Processes – Problem 26: Prove that is also a Gaussian process, and find for