In analyzing the thermodynamic limit of a certain system in statistical mechanics, I've encountered the following situation: $f_n$ is a sequence of probability density functions on the real line with the property that if we define\begin{align} \mu_n = \frac{2}{3}n, \qquad \sigma_n = \frac{2}{2\sqrt{3}}\sqrt{n}, \qquad g(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}\end{align}then for each $x\in\mathbf R$,\begin{align} \lim_{n\to\infty} \sigma_n f_n(\sigma_n x + \mu_n) = g(x) \tag{1}\end{align}I'd like to now show that for each $x\in\mathbb R$\begin{align} f_n(x) \sim \frac{1}{\sigma_n}g\left(\frac{x-\mu_n}{\sigma_n}\right)\, \quad \text{for $n\to\infty$}\end{align}So let $x\in\mathbf R$, and notice that\begin{align} \lim_{n\to\infty} \frac{f_n(x)}{\frac{1}{\sigma_n}g\left(\frac{x-\mu_n}{\sigma_n}\right)}&= \lim_{n\to\infty} \frac{\sigma_nf_n(x)}{g\left(\frac{x-\mu_n}{\sigma_n}\right)}= \lim_{n\to\infty} \frac{\sigma_nf_n\left(\sigma_n\cdot \frac{x-\mu_n}{\sigma_n} +\mu_n\right)}{g\left(\frac{x-\mu_n}{\sigma_n}\right)}\end{align}I'm tempted to now invoke (1) and set the limit on the very right to $1$, but it's not clear to me that I can do this since for given $x$, the expression $(x-\mu_n)/\sigma_n$ scales like $\sqrt{n}$ when $n$ is large. Are there conditions on $f_n$ that enable this final conclusion about the limit? For completeness, here is $f_n$ for the particular case I'm interested in:\begin{align} f_n(x) = \frac{1}{n}\frac{\Gamma\left(\frac{3}{2}n + \frac{3}{2}\right)}{\Gamma\left(\frac{1}{2}n + \frac{1}{2}\right)\Gamma\left(n + 1\right)}\theta\left(\frac{x}{n}\right)\theta\left(1-\frac{x}{n}\right)\left(\frac{x}{n}\right)^n\left(1-\frac{x}{n}\right)^{(n-1)/2}\end{align}where $\theta$ is the Heaviside step function.
↧
