Question

Let \(\{W(t):t\ge0\}\) be a standard Brownian motion, with \(W(0)=0\). Define
\[Z_1=W(1)+W(2)\quad\text{and}\quad Z_2=W(2)+W(3).\]
Let \(\rho\) be the correlation coefficient between \(Z_1\) and \(Z_2\). Then the value of \(10\rho\) is ______ (round off to two decimal places).

Hint:

Hint:
Use \(\text{Cov}(W(s),W(t))=\min(s,t)\) to expand \(\text{Var}(Z_1)\), \(\text{Var}(Z_2)\) and \(\text{Cov}(Z_1,Z_2)\), then form \(\rho=\text{Cov}(Z_1,Z_2)/\sqrt{\text{Var}(Z_1)\text{Var}(Z_2)}\).

Answer 1

Correct Answer: 8.94

Step 1: Recall the covariance structure of Brownian motion.
For a standard Brownian motion \(W(t)\) with \(W(0)=0\), the covariance between two time points is \(\text{Cov}(W(s),W(t))=\min(s,t)\), and \(\text{Var}(W(t))=t\). This single fact is enough to work out every variance and covariance we need.

Step 2: Write down what we need.
We are given \(Z_1=W(1)+W(2)\) and \(Z_2=W(2)+W(3)\). The correlation coefficient is
\[ \rho=\frac{\text{Cov}(Z_1,Z_2)}{\sqrt{\text{Var}(Z_1)\,\text{Var}(Z_2)}} \]
So we need \(\text{Var}(Z_1)\), \(\text{Var}(Z_2)\) and \(\text{Cov}(Z_1,Z_2)\).

Step 3: Find Var(Z1).
\[ \text{Var}(Z_1)=\text{Var}(W(1))+\text{Var}(W(2))+2\,\text{Cov}(W(1),W(2)) \]
Using \(\text{Var}(W(1))=1\), \(\text{Var}(W(2))=2\) and \(\text{Cov}(W(1),W(2))=\min(1,2)=1\),
\[ \text{Var}(Z_1)=1+2+2(1)=5 \]

Step 4: Find Var(Z2).
\[ \text{Var}(Z_2)=\text{Var}(W(2))+\text{Var}(W(3))+2\,\text{Cov}(W(2),W(3)) \]
Using \(\text{Var}(W(2))=2\), \(\text{Var}(W(3))=3\) and \(\text{Cov}(W(2),W(3))=\min(2,3)=2\),
\[ \text{Var}(Z_2)=2+3+2(2)=9 \]

Step 5: Find Cov(Z1,Z2).
Expand using bilinearity of covariance:
\[ \text{Cov}(Z_1,Z_2)=\text{Cov}(W(1),W(2))+\text{Cov}(W(1),W(3))+\text{Cov}(W(2),W(2))+\text{Cov}(W(2),W(3)) \]
Each term is a minimum: \(\min(1,2)=1\), \(\min(1,3)=1\), \(\min(2,2)=2\), \(\min(2,3)=2\). So
\[ \text{Cov}(Z_1,Z_2)=1+1+2+2=6 \]

Step 6: Compute the correlation coefficient.
\[ \rho=\frac{6}{\sqrt{5\times9}}=\frac{6}{\sqrt{45}}=\frac{6}{3\sqrt5}=\frac{2}{\sqrt5} \]
Numerically, \(\sqrt5\approx2.2361\), so \(\rho\approx0.8944\).

Final Answer:
Multiplying by 10 gives the required value. \[ \boxed{10\rho=\dfrac{20}{\sqrt5}\approx8.94} \]