Combinatorial or probabilistic proof of $\sum_{k=0}^n C_{2k}C_{2n-2k}=2^{2n}C_n$

Question

I'm looking for a combinatorial or probabilistic proof for the following beautiful identity involving the Catalan numbers $C_n$:

$$ \sum_{k=0}^n C_{2k}C_{2n-2k}=2^{2n}C_n, \tag{1} $$

This has a simple algebraic proof: Note that

$$ f(z) := \sum_{n=0}^{\infty} C_n z^n = \frac{1-\sqrt{1-4z}}{2z}. $$

Using this explicit formula, it is easy to check:

$$ \left( \sum_{n=0}^{\infty} C_{2n} z^n \right)^2 = \left( \frac{f(\sqrt{z}) + f(-\sqrt{z})}{2} \right)^2 = f(4z) = \sum_{n=0}^{\infty} 2^{2n} C_n z^n. $$

However, the elegant form of the identity strongly suggests that there should be a more direct "proof by counting" or "probabilistic proof," which, if exists, likely utilizes the simple random walk on $\mathbb{Z}$ and its return times.

Does anyone have a combinatorial or probabilistic interpretation of this identity? I have no good lead on this, so I'd like to invite people to this problem.

Thought 1. A possible combinatorial interpretation is that the RHS of $\text{(1)}$ counts the number of Dyck paths of length $2n$ where each step is assigned one of two colors, and the LHS is likely the number of Dyck paths of length $4n$ with an "even diagonal point" $(2k, 2k)$ marked. However, I don't see a clear relationship between these two families of objects, so it might not be an appropriate choice.

Thought 2. Another idea I tried starts with the observation that, if $(X_n)_{n\geq0}$ is a simple random walk on $\mathbb{Z}$ started at $X_0=0$, then the probability generating function of the first time of return to $0$, $\tau = \inf\{n > 0 : X_0 = 0\}$, is given by

$$ \mathbb{E}[x^\tau] = 1 - \sqrt{1 - x^2} = \sum_{n=0}^{\infty} \frac{C_n}{2^{2n+1}} x^{2n+2}. $$

Hence,

$$ \sqrt{\mathbb{E}[x^\tau]} = \sqrt{1-\sqrt{1-x^2}} = \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{2}} = \sum_{n=0}^{\infty} \frac{C_{2n}}{2^{4n+1/2}} x^{2n+1} $$

and it is the PGF of some distribution. So, I considered the possbility of associating this distribution with some cleverly chosen process on $\mathbb{Z}$. To my disappointment, I found that $\sqrt{\mathbb{E}[x^\tau]}$ cannot be the PGF of the first return time of some symmetric random walk on $\mathbb{Z}$.

Answer 1

This is called Shapiro’s convolution formula and a bijective proof was given by Hajnal and Nagy [1]. The idea is to consider (instead of Dyck paths) a path defined as starting from $(0,0)$ and taking steps $\hat{\mathbf i} + \hat{\mathbf j}$ or $\hat{\mathbf i} - \hat{\mathbf j}$. A path is balanced if it ends on the $x$-axis, and it is non-negative if it never falls below the $x$-axis. So, in this notation, Dyck paths are non-negative balanced paths.

The authors then called a (balanced or non-balanced) path to be even-zeroed if its $x$-intercepts are all divisible by $4$. Then they proved that both the LHS and the RHS of the required identity counts the number of even-zeroed paths from the origin to $(4n + 1, 1)$.

[1] A bijective proof of Shapiro’s Catalan convolution, The Electronic Journal of Combinatorics, Volume 21 (2), 2014.