Edit: The probabilistic context below is probably not important. The question is simply why $(f_n)$ converges uniformly to $\pi^{-1}(x(1-x))^{-1/2}$ on compact subsets of $(0,1)$, where $f_n$ is the step function equal to $n{2k \choose k}{2\left(n-k\right) \choose n-k}2^{-2n}$ on $[\frac{k}{n},\frac{k+1}{n}]$. Pointwise convergence follows from Stirling's Approximation.
Originial formulation, including the context:
I am reading the proof of Theorem 4.9.5 in Probability: Theory and Examples by Durett, available online here (page 265 of the book / page 273 of the PDF file).
Let $X_1,X_2,\dotsc$ be i.i.d. with $P(X_i=1)=P(X_i=-1)=\frac{1}{2}$. Let $S_n=X_1+\cdots+X_n$ (a simple random walk). Let $L_{n}=\sup\{m\leq 2n\mid S_m=0\}$ (the last visit to $0$ among the first $2n$ steps).
What I do understand:
- $nP(L_{2n}=2k)=n{2k \choose k}{2\left(n-k\right) \choose n-k}2^{-2n}\to\pi^{-1}(x(1-x))^{-1/2}$ as $n\to\infty$ and $k/n\to x$.
- Defintion: $f_n$ is a step function equal to $nP(L_{2n}=2k)$ on $[\frac{k}{n},\frac{k+1}{n}]$.
What I do not understand:
A bit later in the proof, it is claimed that $f_n$ converges uniformly on compact sets (Edit: I mean on compact subsets of $(0,1)$).
Why is the convergence uniform on these compact sets?