# Estimating the discrete random walk probability by error function

I am trying to work out the asymptotic large $$t$$ behavior of following function $$$$f(t ) = \sum_{x = 0}^{2t} { 2t \choose t + x} p^{ t+x } (1 - p)^{ t - x} = \sum_{x = 0}^{2t} { 2t \choose t + x} [p ( 1- p)]^{ t} \left( \frac{p }{1 -p } \right)^x$$$$ with $$p > \frac{1}{2}$$. Specifically, I want to find $$$$\alpha = \lim_{ t\rightarrow \infty} - \frac{\ln f(t)}{t}$$$$

I have two ways to approximate the sum, resulting in two different answers.

## 1

Since $$\left( \frac{p }{1 -p } \right)^x$$ is an exponentially decaying function about $$x$$ (because $$p > 1/2$$), we can take the saddle point approximation in the sum. The largest term in the sum is at $$x = 0$$. It will dominate the sum, hence $$$$f(t) = { 2t \choose t + x} [p ( 1- p)]^{ t} \sim [p ( 1- p)]^{ t} 2^{2t} \frac{1}{\sqrt{2t}}$$$$ which implies $$$$\alpha = - \ln [4p( 1- p) ] > 0$$$$

This should be the correct answer, because $$$$\frac{f(t)}{[p ( 1- p)]^{ t} 4^t} = {}_2 F_1 [1, -t, 1+t, \frac{p}{p - 1}]$$$$ We can plot the hypergeometric function and find that its large $$t$$ values seem to converge to a constant.

## 2

We consider a random walk $$X_t$$ with the probability $$p$$ of going right and $$1 - p$$ of going left, then $$$$f(t) = \text{Pr}( X_{2t} \ge 0 ) = 1 - \text{Pr}( X_{2t} < 0 )$$$$

In the large $$t$$ limit, the central limit theorem implies that the 2nd term will converge to the CDF of a Gaussian function with $$$$\mu = 2t [p - ( 1 - p )] \quad \sigma^2 = 2t ( 1 - (2p-1)^2 ) = 2t 4p ( 1- p)$$$$ Therefore $$$$f(t) \sim \int_0^{\infty} dx e^{-\frac{( x- \mu)^2}{2 \sigma^2}}\frac{1}{\sqrt{2\pi} \sigma} = \frac{1}{2}( 1 + \text{erf}( \frac{\mu}{\sqrt{2} \sigma} ) )$$$$ According to Wikipedia, we can approximate the error function by $$$$\text{erf}(x) \sim \text{sgn}(x) \sqrt{ 1 - e^{ -x^2 } }$$$$ when $$|x| \gg 0$$.

With this we have $$$$f(t) \sim \frac{1}{2} ( 1 + \sqrt{ 1 - e^{ - \frac{\mu^2}{2 \sigma^2}}} ) \sim \frac{1}{4} e^{ - \frac{\mu^2}{2 \sigma^2}}$$$$ Then $$$$\alpha \stackrel{?}{=} \frac{\mu^2}{2 \sigma^2 t} = \frac{( 1 - 2p)^2}{4p ( 1- p) }$$$$

## Questions

Why the 2nd approach does not give the correct asymptotic behavior?

My guess is that the central limit theorem only provides convergence to the scaled variable $$X_t / \sqrt{t}$$, not the distribution of $$X_t$$ itself.

Is it possible to fix the 2nd approach?