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).
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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 \(\rho=\text{Cov}(Z_1,Z_2)/\sqrt{\text{Var}(Z_1)\text{Var}(Z_2)}\).
Step 1: Break the Brownian motion into independent increments.
Write $A=W(1)$, $B=W(2)-W(1)$, $C=W(3)-W(2)$. Since Brownian motion has independent increments, $A$, $B$, $C$ are independent, each with variance equal to its time length, so $\text{Var}(A)=\text{Var}(B)=\text{Var}(C)=1$.
Step 2: Rewrite Z1 and Z2 in terms of A, B, C.
Since $W(1)=A$, $W(2)=A+B$, $W(3)=A+B+C$,
\[ Z_1=W(1)+W(2)=2A+B \]
\[ Z_2=W(2)+W(3)=2A+2B+C \]
Step 3: Get Var(Z1) from the independent pieces.
Since $A,B,C$ are independent,
\[ \text{Var}(Z_1)=4\text{Var}(A)+\text{Var}(B)=4(1)+1=5 \]
Step 4: Get Var(Z2) the same way.
\[ \text{Var}(Z_2)=4\text{Var}(A)+4\text{Var}(B)+\text{Var}(C)=4+4+1=9 \]
Step 5: Get Cov(Z1,Z2).
Only matching independent terms survive:
\[ \text{Cov}(Z_1,Z_2)=\text{Cov}(2A+B,2A+2B+C)=4\text{Var}(A)+2\text{Var}(B)=4(1)+2(1)=6 \]
Step 6: Form the correlation and scale by 10.
\[ \rho=\frac{6}{\sqrt{5\times9}}=\frac{6}{3\sqrt5}=\frac{2}{\sqrt5}\approx0.8944 \]
\[ 10\rho\approx8.94 \]
\[ \boxed{10\rho\approx8.94} \]
