Suppose $\mathbb{P}(X=1)=\mathbb{P}(X=-1)=\frac{1}{6}$ and $\mathbb{P}(X=0) = \frac{2}{3}$. Is the random walk of $X$ transient or recurrent?
I wanted to use well known result that random walk is transient iff: $$\mathbb{E}N = \lim_{t \to 1^-} \int_{[-\pi , \pi]^d}\frac{ds}{(2 \pi)^d(1-t\varphi(s))} < \infty$$ where $N = \sum_{n \geq 0}\mathbb{1}_{\{S_n = S_0\}}$.
I need to find $\varphi_X(t)$. Using what I already know $$ \varphi_X(t)=\mathbb{E}e^{itX}=\frac{1}{6}e^{it}+\frac{1}{6}e^{-it}+\frac{2}{3}e^{it0} = \frac{\cos t + 2}{3} $$
I am stuck however, with the integral. $$ \mathbb{E}N = \lim_{t \to 1^-} \int_{[-\pi , \pi]}\frac{ds}{2 \pi(1-t[\frac{\cos s + 2}{3}])} = $$ function seems even so $$ = \lim_{t \to 1^-} \int_{[0, \pi]}\frac{ds}{\pi(1-t[\frac{\cos s + 2}{3}])} = \dots ? $$ Is this a good start of this exercise? What can I do next? Can I swap integral with the limit? If yes, why? I appreciate any help.
Recall that the random walk $Y$ on one dimension (i.e. $\mathbb{Z}$) is recurrent. Now, it is easy to convert your random walk to $Y$, by conditioning on $X\neq 0$. It should be much easier this way.
Edit: just saw that you wanted to use your method. I am having a look.
Ah okay, look at what happens as $s\rightarrow 0$ in your integral. The denominator goes to infinity as $t\rightarrow 1$. This behavior is proportional to a singularity like $1/u$, so your integral is not finite and thus the random walk is recurrent.
This seems more hand-wave-y than an argument based on conditioning.
Okay, it suffices to show that a lower bound of the integral still diverges. By drawing the graph of $\cos$ and connecting $(0,1)$ and $(\pi/2,0)$, we see that $\cos$ is bounded below by $1-(2/pi)*x$. Hence the denominator of the integral is bounded above by $⅓ (1-t)+(2t/\pi)x$.