Solution

Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 34 / 4 / d / Solution

{title2=$\mathbb E[M(\tau)^{-T_b}]=e^{-\tau b}$}

Let $S_n$ have integrable, mean-zero <independent and identically distributed random variables> as increments, taking integer values at most one, with $\mathbb P(X_1=1)>0$. For a positive integer $b$, put $T_b=\inf\{n:S_n=b\}$ and $M(\tau)=\mathbb E e^{\tau X_1}$, $\tau>0$. Then $T_b$ is finite <almost surely>. Indeed, $b-S_{n\wedge T_b}$ is a nonnegative <martingale>; finite convergence on $\{T_b=\infty\}$ would force its integer increments eventually to vanish, contrary to the <Borel-Cantelli lemmas> applied to $\{X_n=1\}$.

The <exponential martingale of a random walk> $e^{\tau S_n}/M(\tau)^n$ stopped at $T_b$ is bounded by $e^{\tau b}$ because $S_{n\wedge T_b}\leq b$ and $M(\tau)\geq1$. The <dominated convergence theorem> therefore preserves its mean one at the limit and proves the displayed <first-passage time> transform. This argument accommodates arbitrarily large negative jumps; the restriction concerns upward jumps. For a <simple symmetric random walk>, $M(\tau)=\cosh\tau$, giving $\mathbb E e^{-\alpha T_b}=(e^\alpha-\sqrt{e^{2\alpha}-1})^b$.

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 Incoming links: Exponential first-passage transform for an upward skip-free random walk