$X_1,…,X_{2n}$ i.i.d. standard Normal, find limiting distribution of $Z_n = U_n/V_n$
where $U_n = X_1/X_2 + X_3/X_4+ \dots + X_{2n-1}/X_{2n}$
$V_n = X_1^2+ \dots + X_n^2$
Attempt:
It can be shown that each $X_i/X_{i+1}$ follows a standard cauchy distribution. Further, it can easily be shown that $$(X_1^2+…+X_n^2)/n$$ converges in probability to 1.
Thus define we can find the solution by finding what $U_n/n$ converges to in distribution using Slutsky's theorem. Ths is where i am stuck. the solution that is provided is a standard cauchy distribution. By the provided answer I see that $U_n$/n must somehow converge in distribution to a standard cauchy distribution, but i don't see how this is.
Best Answer
Let $(\xi_i)_{i\in \Bbb N}$ i.i.d. with $\xi_1 \sim \text{Cauchy}(0,1)$. Then the characteristic function of $\xi_1$ is given by $$\varphi_{\xi_1} (t) = e^{-\vert t \vert}$$ and $$\varphi_{\frac{\xi_1}{n}} (t) = \Bbb E [e^{i\frac{\xi_1}{n} t}] = \varphi_{\xi_1} (\frac t n) = e^{-\frac 1 n\vert t \vert}$$ The characteristic function of $S_n := \frac 1 n \sum_{i=1}^n \xi_i$ is due to the independence $$\varphi_{S_n} (t) = \varphi_{\frac{\xi_1}{n}} (t) \cdot \ldots \cdot \varphi_{\frac{\xi_n}{n}} (t) = e^{-\frac 1 n\vert t \vert} \cdot \ldots \cdot e^{-\frac 1 n\vert t \vert} = e^{-\vert t \vert}$$ This shows (since the characteristic function is uniwue) that $S_n \sim \text{Cauchy}(0,1)$. Trivially it follows that $$S_n \to \text{Cauchy}(0,1)$$ in distribution.