I have a symmetric random walk that starts at the origin. With probability $1/6$ it goes right by one and with probability $1/6$ it goes left by one. With probability $4/6$ it stays put. After $n$ time steps, what is the probability that it is at the origin?
The answer should be the probability that you go left by the same amount you go right. I am having difficulty working this out however.
Best Answer
A moderate amount of (discrete) Fourier theory allows to provide an explicit formula for, and to estimate the asymptotic behaviour of $$P(X_n=0)$$ when $n\to\infty$, where $(X_n)_{n\geqslant0}$ is the random walk described in the question.
1. Namely, the steps $Z_n=X_n-X_{n-1}$ are i.i.d. and their distribution has Fourier transform $$u(t)=E(\mathrm e^{\mathrm i tZ})=\frac13(2+\cos t)$$ hence
2. This already yields an explicit formula for $p_n$, expanding the integrand, since, for every $k$, $$\int_{-\pi}^\pi\cos^{2k}t\,\mathrm dt=2\pi\cdot{2k\choose k}\frac1{2^{2k}}\qquad\int_{-\pi}^\pi\cos^{2k+1}t\,\mathrm dt=0 $$ hence the binomial identity $$u(t)^n=\frac1{3^n}\sum_k{n\choose k}\cdot\cos^kt\cdot2^{n-k} $$ yields $$P(X_n=0)=\frac1{3^n}\sum_k{n\choose2k}\cdot{2k\choose k}\frac1{2^{2k}}\cdot2^{n-2k}$$ or, equivalently,
3. But, perhaps more interestingly than the cumbersome sum above, this approach also yields precise asymptotics of $P(X_n=0)$ when $n\to\infty$, since one can rewrite the integral formula above as $$ 2\pi\sqrt{n}\cdot P(X_n=0)=\int_{-\pi\sqrt{n}}^{\pi\sqrt{n}} u(s/\sqrt{n})^n\mathrm ds. $$ When $t\to0$, $u(t)=1-\frac16t^2+o(t^2)$ hence $u(s/\sqrt{n})^n\to\mathrm e^{-s^2/6}$. Furthermore, $\pi\sqrt{n}\to+\infty$, hence a dominating condition such as $u(s/\sqrt{n})^n\leqslant\mathrm e^{-s^2/12}$ for every $|s|\leqslant\pi\sqrt{n}$ allows to apply Lebesgue dominated convergence theorem, showing that $$ 2\pi\sqrt{n}\cdot P(X_n=0)\to\int_{-\infty}^\infty\mathrm e^{-s^2/6}\mathrm ds=\sqrt{6\pi}, $$ that is,