Past exam of the probability course of the University of Oxford / 2015 / 3 / c

Since $Y_n\to +\infty$ almost surely, only finitely many early values can attain the global minimum; it is a finite integer. Also $W\le Y_0=0$. For every nonnegative integer $m$, the unit-step property and translation invariance of the increments giveSubtracting consecutive tails yields the probability mass functionThus $-W$ has the geometric distribution counting failures before the first success, with success parameterThis support convention is essential: the minimum of an upward-biased random walk can equal its starting position, so $-W=0$ has positive probability.

Work on the single event of probability one where $Y_n/n\to \mu =2p-1>0$. On every sample path in that event, $Y_n\to +\infty$ and the unit-step property ensures that every positive integer $j$ is reached, so $T_j<\infty$. Starting from zero requires at least $j$ steps to reach $j$, hence $T_j\ge j$ and $T_j\to \infty$.

The convergent numerical sequence $Y_n/n$ on this fixed sample path also converges along its possibly random subsequence $n=T_j$. Since $Y_{T_j}=j$,Taking the reciprocal is valid because $\mu >0$. We obtainThis is a first-passage law of large numbers for an upward skip-free random walk. No independence between the hitting times and the path is needed: almost sure convergence already supplies convergence along every diverging subsequence on the same event.