The statement is as follow:
Suppose $(S_n)$ is a 2d simple random walk, and $(S^1_n)_n, (S^2_n)_n$ are two independent random walks of the same type. Let $R_n = \sum_{j=1}^n\mathbf{1}\{S_j = 0\},$ and $ L_n = \sum_{j=1}^n \mathbf{1}\{ S^1_j = S^2_j\}$. Show that $$ L_n = R_{2n} \enspace \text{ in distribution}. $$
I was able to show that $\mathbf{P}(S^1_n = S^2_n ) = \mathbf{P}( S_{2n} = 0)$: $$ \mathbf{P}( S^1_n = S^2_n) = \sum_x \mathbf{P}(S^1_n = S^2_n = x) = \sum_x \mathbf{P}(S^1_n = x) \mathbf{P}(S^2_n = x) \enspace \text{ by independence of }S^1,S^2 $$ We can write $\mathbf{P}( S^2_n = x) = \mathbf{P}(S_{2n} -S_n =-x)$ and $\mathbf{P}(S^1_n = x) = \mathbf{P}(S_n = x)$, then $$ \mathbf{P}( S^1_n = S^2_n) = \sum_x \mathbf{P}(S_n = x) \mathbf{P}(S_{2n} -S_n =-x) = \mathbf{P}(S_{2n} = 0). $$ But I don't know how to go further. Can anyone help for the proof? Thanks.
I'm just going to give the proof myself. Write $S^i_n = \sum_{j=1}^n X^i_j$ for $i = 1,2$. Actually, it's not hard to establish that the following holds: $$ Z_1 := (X^1_1 - X^2_1, X^1_2- X^2_2 ,\dots , X^1_n - X^2_n) \overset{d}{=} (X_1+X_2 , X_3 +X_4, \dots, X_{2n-1}+X_{2n}) =:Z_2. $$ One possible way to show this is to use the properties of c.f. We then define a matrix $A = (A_{ij})_{ 1 \leq i ,j\leq n}$ by $$ A_{ij} = \begin{cases} 0 ,&\text{ if } j >i,\\ 1 &\text{ otherwise}. \end{cases}$$ The matrix $A$ is simply a lower triangular matrix with all nonzero entries equal to $1$. It's clear that $AZ_1 = (S^1_j -S^2_j)_{j=1}^n$ and $AZ_2 = (S_{2j})_{j=1}^n$, thus $AZ_1 \overset{d}{=} AZ_2$. Finally, observe that there exists a measurable function $f$ such that $L_n = f(AZ_1)$ and $R_{2n} = f(AZ_2)$. This in turn leads to the conclusion $L_n \overset{d}{=} R_{2n}$ and concludes the proof.