Symmetry of last hitting time of a simple random walk before a fixed time.

Question

It is known that for the simple random walk on $\mathbb Z$ started at $0$, say $(S_m, m \in \mathbb Z^+)$, the last hitting time of $0$ before $2n$ $$T_{2n} = \max\{m \le 2n : S_m =0\}$$ has distribution :

$$\mathbb P(T_{2n}=2k)= \frac{1}{2^{2n}} {2k \choose k} {2(n-k) \choose n-k}$$

see Durrett's book for instance. This relation is usually proven using the ballot theorem. Taking the $n \to \infty$ limit, one find (after rescaling) the Arcsine distribution.

A striking feature of the previous formula is the symmetry when exchanging the role of $k$ and $n-k$.

My question : is there a (simple?) pathwise/geometric transformation that would help to understand this symmetry ?

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Since $\mathbb P(T_{2n}=2k) = \mathbb P(S_{2k}=0) \; \mathbb P(S_1 \neq 0, S_2 \neq 0, \ldots, S_{2(n-k)} \neq 0 )$ and $\mathbb P(S_{2k}=0)= 2^{-2k} {2k \choose k}$, the question is equivalent to finding a bijective proof of the relation : $$\mathbb P(S_{2k}=0) = \mathbb P(S_1 \neq 0, S_2 \neq 0, \ldots, S_{2k} \neq 0)$$

Answer 1

Well, it appears Mark van Leeuwen answers my question (in its second formulation) in his neat answer to this post :

Identity for convolution of central binomial coefficients: $\sum\limits_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$

A very nice piece of maths to my mind. Hence I shall perhaps delete my question.

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EDIT : actually the aforementioned bijection is between the set of balanced walks

$$\Big\{(x_1, \ldots, x_{2n}) \in \{-1,1\} : \sum_{i=1}^{2n} x_i=0\Big\} $$

and non-negative walks

$$\Big\{(x_1, \ldots, x_{2n}) \in \{-1,1\} : \sum_{i=1}^{2n} x_i \ge 0\Big\} $$

whereas what is needed here is a bijection between the set of balanced walks and non-zero walks:

$$\Big\{(x_1, \ldots, x_{2n}) \in \{-1,1\} : \sum_{i=1}^{2n} x_i \ne 0\Big\} $$

Although the step from one bijection to the other is not that big, this makes the link by Mike more appropriate :

people.bath.ac.uk/masadk/papers/catalan.pdf