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

For the simple symmetric random walk, the moment-generating function isFor a given $\alpha >0$, choose the unique $\tau >0$ satisfying $\cosh \tau =e^{\alpha }$. The transform in part (d) becomes $\mathbb{E}{e}^{-\alpha T}=e^{-b\tau }$. To express this in terms of $\alpha$, put $r=e^{-\tau }\in (0,1)$. The equation $(r+r^{-1})/2=e^{\alpha }$ gives $r^2-2e^{\alpha }r+1=0$. Its smaller root is the one in $(0,1)$, soThe second expression follows by rationalizing the first root. As $\alpha \downarrow 0$ the transform tends to one, consistent with the almost-sure finiteness proved in part (c).