# Expected number of steps before leaving a ball

Consider an infinite undirected graph $$G$$, like for example $$\mathbb{Z}^d$$ with edges connecting nearest neighbours sites. Let $$X(t)$$ be a simple random walk starting from the origin, $$o$$, define $$B_L := \{ x \in \mathbb{Z}^d \, \, : \, \, d(x,o) \leq L\}$$ namely the set of sites whose graph distance from the origin is at most $$L$$. Let $$E_1$$ be the expected number of steps performed by the simple random walk before leaving the set $$B_L$$ or returning to the origin, $$E_1 = E \big ( \sum\limits_{t=0}^{\infty} \mathbb{1}\{ t < \tau_{B_L^c \cup \{o\}} \} \big ),$$ where $$\mathbb{1}$$ is the indicator function, $$E$$ is the expectation of the simple random walk, and $$\tau_{B_L^c \cup \{o\}}$$ is the hitting time of the set $$B_L^c \cup \{o\}$$, where $$B_L^c:= \mathbb{Z}^d \setminus B_L$$. Let $$E_2$$ be the expected number of distinct vertices visited by the simple random walk before leaving $$B_L$$ or returning to the origin, $$E_2 = \sum\limits_{x\in B_L}^{\infty} \mathbb{1}\{ \exists t < \tau_{B_L^c \cup \{o\}} \, \, \, s.t. \, \, X_t = x\}$$

Are $$E_1$$ and $$E_2$$ of the same order of magnitude in $$L$$? The answer is easy and positive if the graph is transient, but what about recurrent graphs?

It would be enough for me to receive some hint!

First, note that $$E_1 = E(\tau_{B_L^x \cup \{o\}})$$, ie is just the expected hitting time.
I expect this can depend a lot on the graph. Take, for example, just $$\mathbb Z$$. Recall that the expected return time to the origin (for a standard symmetric SRW) is infinite. On the other hand, the expected exit time of $$[-L,L]$$ is order $$L^2$$. Given that the walk exits $$[-L,L]$$ before returning to $$0$$, it hits precisely $$L$$ distinct vertices.
Now, a slight caveat: I've said "expected return time to the origin is $$\infty$$" and "expected exit time of $$[-L,L]$$ is order $$L^2$$" -- these are both correct statements, but they don't, a priori, imply that "expected exit time of $$[-L,L]$$, given that the origin is not returned to, is order $$L^2$$". However, this should be pretty easy to prove: first walk from $$0$$ to $$L/2$$ directly (this is the worst that conditioning on not returning to $$0$$ can do for the first $$L/2$$ steps); now hitting $$0$$ or $$L$$ is the same as exiting the interval $$[L/2 - L/2, L/2 + L/2]$$, which is order $$(L/2)^2$$, ie order $$L^2$$. I'm sure one can make this rigorous.
Using the fact that a SRW on $$\mathbb Z^2$$ is just a pair of independent SRWs on $$\mathbb Z$$ (in continuous time), one can likely extend this result to $$\mathbb Z^2$$ without too much change or additional ideas.