The strength of the linear combination is the same linear combination of the unit strengths:
$$
X_r(\omega)=\sum_{\gamma:o\to z_r}
\frac{\mu(\gamma)\mathbf1_{\{\gamma\text{ open}\}}}
{\mathbb P(\gamma\text{ open})}.
$$
Taking expectations cancels each denominator. Since the truncated paths partition the path space,
$$
\mathbb EX_r
=\sum_{\gamma:o\to z_r}\mu(\gamma)=1.
$$

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 212 / 2 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-212/2/b/solution