Generating function for Random Walk Hitting Time, taking the wrong root - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T16:02:46Z http://mathoverflow.net/feeds/question/107981 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107981/generating-function-for-random-walk-hitting-time-taking-the-wrong-root Generating function for Random Walk Hitting Time, taking the wrong root David 2012-09-24T14:38:58Z 2013-05-07T00:22:00Z <p>In a calculation of the hitting time for a Bernoulli random walk we have to calculate the hitting time $\tau(1)=\inf\{n\ge 0:S_n=1\}$ to reach $+1$ and the generating function has the recursion relationship $E[s^\tau]=G_\tau(s)=ps+qsE(s^{\tau'+\tau''})$ where $\tau'$ and $\tau''$ have the same distribution as $\tau$. The generating function is then one of the two solutions to $G(s)=\frac{1\pm\sqrt{1-4pqs^2}}{2qs}$, and it is easily seen to be the negative $\frac{1-\sqrt{1-4pqs^2}}{2qs}$. </p> <p>My question is: What is the other solution $\left(\frac{1+\sqrt{1-4pqs^2}}{2qs}\right)$ a solution to? Does it have any meaningful interpretation?</p> http://mathoverflow.net/questions/107981/generating-function-for-random-walk-hitting-time-taking-the-wrong-root/107985#107985 Answer by Robert Israel for Generating function for Random Walk Hitting Time, taking the wrong root Robert Israel 2012-09-24T15:19:59Z 2012-09-24T15:19:59Z <p>It's $\dfrac{1}{qs} - G(s)$, so you could call it $E\left[ \dfrac{1}{qs} - s^\tau \right]$. Of course it's not a probability generating function, because it has negative coefficients except for the $1/(qs)$ and is not $1$ at $s=1$. </p>