Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 2 / 12F / c / Solution

Because $u\mapsto (1+u/S{)}^{-n}$ decreases, $X_i<X_{i+1}$ exactly when $E_i>E_{i+1}$. Therefore $K\ge i$ exactly when $E_1\le \cdots \le E_i$, an event of probability $1/i!$ independent of $S$. The tail-sum rearrangement givesSince $\mathbb{E}[S^i]=(n+i-1)!/(n-1)!$, the $i$th term is $\left(\genfrac{}{}{0ex}{}{n+i-1}{i}\right)$, proving the formula.