= Solution
For the <waiting time for a new record after a fixed sample>, given $U=F(Y_n)$, no overflow in the first $k$ subsequent years has <probability> $U^k$. Averaging as in part (c) yields
$$
P(\tau>k)=n\int_0^1u^{n+k-1}\,du=\frac n{n+k},\qquad k\geq0.
$$
The <tail-sum formula for expectation> therefore gives
$$
E\tau=\sum_{k=0}^\infty P(\tau>k)
=n\sum_{k=0}^\infty\frac1{n+k}=\infty.
$$
\b[The answer is $\boxed{E\tau=\infty}$.] This does not contradict almost-sure finiteness: the <tail probability> tends to zero, but too slowly for its sum to converge.
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