Let $X_n$ be a sequence of almost surely positive real-valued random variables s.t. $$\sqrt{n} \, \left( X_n -a \right) \to_D \mathcal{N} ( 0, 1)$$ where $\to_D$ denotes convergence in distribution and $a>0$. Now I'm interested in what happens to$$\sqrt{n} \, \log \left( \frac{X_n}{a} \right)$$as $n \to \infty$.
My thoughts were the following: By expanding $\log(x)$ to a series at x=1 we get $$\sqrt{n} \, \log \left( \frac{X_n}{a} \right)= \sqrt{n} \, \frac{X_n-a}{a} + \sqrt{n} \,\mathcal{O} \left( \left( X_n-a \right)^2 \right),$$ and because the higher order terms tend to zero in probability $$\sqrt{n} \, \log \left( \frac{X_n}{a} \right) \to_D \mathcal{N} \left( 0, \frac{1}{a^2} \right).$$
THIS WAS DUE TO A SIMPLE ERROR IN MY CODE
However, numerical simulations seem to suggest that actually $$\sqrt{n} \, \log \left( \frac{X_n}{a} \right) \to_P 0,$$where $\to_P$ denotes convergence in probability. I'd appreciate any comments on why my reasoning may not be valid. Even better if someone has an idea of how to do this correctly.
Not a homework.
Indeed $\sqrt{n}\log(X_n/a)$ converges in distribution to a Gaussian $\mathcal{N}(0,1/a^2)$. One way to prove is to use the identity: $$ \frac{x}{1+x} \leq \log(1+x) \leq x $$ which holds for all $x>-1$ (i.e., whenever $\log(1+x)$ is nicely defined).
So now define $G_n = \sqrt{n}(X_n-a)$. Then $\log(X_n/a) = \log(1 + \frac{G_n}{a\sqrt{n}})$ and so the above identity gives: $$ \frac{\frac{G_n}{a\sqrt{n}}}{1+\frac{G_n}{a\sqrt{n}}} \leq \log(X_n/a) \leq \frac{G_n}{a\sqrt{n}} $$ Multiplying by $\sqrt{n}$ gives: $$ \frac{G_n}{a}\left(\frac{1}{1+ \frac{G_n}{a\sqrt{n}}}\right) \leq \sqrt{n}\log(X_n/a) \leq \frac{G_n}{a} $$
Now define: \begin{align} M_n &=\sqrt{n}\log(X_n/a) \\ Z_n &= \frac{1}{1+\frac{G_n}{a\sqrt{n}}} \end{align} Thus, $$ \frac{G_nZ_n}{a} \leq M_n \leq \frac{G_n}{a} $$ Define $N$ as a Gaussian $\mathcal{N}(0,1/a^2)$. Note that $G_n/a$ converges to $N$ in distribution, and $Z_n$ converges to 1 in distribution.
Upper bound: For all $x \in \mathbb{R}$ we have: $$ Pr[M_n\leq x] \geq Pr[G_n/a \leq x]$$ and so $$ \liminf_{n\rightarrow\infty} Pr[M_n \leq x] \geq Pr[N \leq x] $$
Lower bound: For simplicity, fix $x>0$ (similar techniques can be used for $x \leq 0$). We have: $$ Pr[M_n \leq x] \leq Pr[G_nZ_n/a \leq x] $$ For all $\epsilon>0$: $$ \{G_nZ_n/a \leq x\} \subseteq \{G_n/a \leq x + \epsilon\} \cup \{Z_n < x/(x+\epsilon)\} $$ So: $$Pr[M_n \leq x] \leq Pr[G_n/a \leq x+ \epsilon] + Pr[Z_n < x/(x+\epsilon)]\rightarrow Pr[N\leq x+\epsilon] $$ This holds for all $\epsilon>0$, and so (assuming $x>0$): $$ \limsup_{n\rightarrow\infty}Pr[M_n \leq x] \leq Pr[N\leq x] $$
The upper and lower bounds together imply $\lim_{n\rightarrow\infty} Pr[M_n\leq x] = Pr[N\leq x]$.