Let $X$ and $Y$ be two independent recurrent random walks with the same transition matrix $p$ on $\mathcal{Z}$.
Is $Z:=X-Y$ a recurrent random walk?
I would like to prove it by using this result (Theorem 17.41 of Probability Theory by A. Klenke, 3rd version):
A random walk on $\mathcal{Z}$ with $\sum_{-\infty}^{\infty}|x|p(0,x)$ is recurrent if and only if $\sum_{-\infty}^{\infty}xp(0,x)=0$.
My attempt
For starting points $x,y\in\mathcal{Z}$ of $X$ and $Y$, $z=x-y$ and $$ \bar p(z,z_1)=P_z[Z_1=z_1]=P_{x,y}[X_1-Y_1=z_1]=\sum_{y_1=-\infty}^{\infty}P_x[X_1=z_1+y_1|Y_1=y_1]P_y[Y_1=y_1]= \\\sum_{y_1=-\infty}^{\infty}p(x,z_1+y_1)p(y,y_1) $$
Then $$ \sum_{z_1=-\infty}^{\infty}z_1\bar p(z,z_1)=\sum_{x_1=-\infty}^{\infty}\sum_{y_1=-\infty}^{\infty}(x_1-y_1)p(x,z_1+y_1)p(y,y_1) \\ \sum_{y_1=-\infty}^{\infty}p(y,y_1)\sum_{x_1=-\infty}^{\infty}x_1p(x,x_1)-\sum_{x_1=-\infty}^{\infty}p(x,x_1)\sum_{y_1=-\infty}^{\infty}y_1p(y,y_1) \\ \sum_{y_1=-\infty}^{\infty}p(y,y_1)\sum_{s=-\infty}^{\infty}(x+s)p(0,s)-\sum_{x_1=-\infty}^{\infty}p(x,x_1)\sum_{t_1=-\infty}^{\infty}(y+t)p(0,t) \\=x-y=z $$
Question How can I conclude from here? Is it sufficient to take $z=0$ and apply the theorem above? How can I see that $Z$ is indeed a random walk?
Please, let me know if more details are needed. Thank you.
You may need to prove that: (i) $Z$ is a random walk on $\mathbb{Z}$ itself, (ii) that $\sum_z|z|p_Z(0,z)<\infty$ and, (iii) to conlude, that $\sum_zzp_Z(0,z)=0$.
We may write $Z_n=(x-y)+\sum_{k\leq n}(\xi_k^X-\xi_k^Y)$ where $x,y \in \mathbb{Z}$ are the initial states of $X,Y$ respectively and $(\xi^X_n)_n,(\xi^Y_n)_n$ are the IID increments of $X$ and $Y$ respectively. Write $z=x-y,\xi^Z_k=\xi^X_k-\xi^Y_k$ and note $(\xi^Z_n)_n$ are IID. So we have that $Z$ is a random walk on $\mathbb{Z}$ i.e. $Z_n=z+\sum_{k\leq n}\xi^Z_k$. We have $E[|Z_1-z|]\leq E[|\xi_1^X|]+E[|\xi_1^Y|]=\sum_x|x|p_X(0,x)+\sum_y|y|p_Y(0,y)<\infty$. Now, since $X,Y$ are recurrent, $\sum_zzp_Z(0,z)=E[Z_1-z]=E[\xi^X_1]-E[\xi_1^Y]=\sum_xx p_X(0,x)+\sum_yyp_Y(0,y)=0$.