I would like to check if my arguments are the correct ones in proving
$$ G_N(x,y):=\sum_{k=0}^N P_x[X_k=y]\leq G_N(y,y) $$
where $X$ is a random walk on $\mathcal{Z}$ with $E_0[|X_1|]<\infty$ and $E_0[X_1]=0$. This is part of the discussion that precedes Theorem 17.41 in Probability Theory, 3rd version, by A. Klenke.
My attempt, where $\tau_y^1=\inf\{n>0: X_n=x\}$, $N\in \mathcal{N}$:
$$\sum_{k=0}^N P_x[X_k=y]\stackrel{1}{=}\sum_{k=0}^N \sum_{l=0}^NP_x[X_k=y,\tau_y^1=l] \\\stackrel{2}{=}\sum_{l=0}^N\sum_{k=l}^{N}P_x[X_k=y,\tau_y^1=l]\\ \stackrel{3}{=}\sum_{l=0}^N\sum_{k=l}^{N}P_y[X_{k-l}=y]P_x[\tau_y^1=l] \\\stackrel{4}{=}\sum_{l=0}^N\sum_{k=0}^{N-l}P_y[X_k=y]P_x[\tau_y^1=l] \\\stackrel{5}{=}\sum_{l=0}^NP_x[\tau_y^1=l]G_{N-l}(y,y) \\ \leq \sum_{l=0}^NP_x[\tau_y^1=l]G_{N}(y,y) \\ \leq G_N(y,y). $$
I used:
- $1^{st}$ equality: Law of Total Probability
- $2^{nd}$ equality: $P_x[X_k=y,\tau_y^1=l]=0$ for $k<l$
- $3^{rd}$ equality: Markov property
- $1^{st}$ inequality: $G_{N-l}(y,y)\leq G_N(y,y)$ because the number of times the process visits $y$ in a "shorter period" is less than in a "greater period" starting from $y$
- $2^{nd}$ inequality: $\sum_{l=0}^NP_x[\tau_y^1=l] \leq 1$
Let me know if more context is needed. Thank you.
Almost correct, but there is a small error. On the right-hand side of the third equality, the correct summand is $P_{y}[X_{k-l} = y]P_{x}[\tau_{y}^{1} = l]$. This is because, due to the strong Markov property, the probability $P_{x}[X_{k} = y, \tau_{y}^{1} = l]$ is equal to the product of the probabilities "$X$ hits $y$ for the first time at $l$" and "$X$, starting from $y$, hits $y$ at $k-l$". The rest of the computation can be done in much the same way as you originally did.