Sequence of vector of random variables

Question

I'm struggling a lot with below problem:

Let's take a sequence of independent two-dimmensional vectors of random variables $(A_n, B_n)_{n=1}^{\infty}$, where all vectors are uniformly distributed on square $[-2,2] \times [-2,2]$. Let $V_n=(S_n, T_n) = (\sum_{i=1}^n A_i, \sum_{i=1}^n B_i)$ and $|V_n| = \sqrt{(S_n)^2+(T_n)^2}$. Determine constant $c$ so that $lim_{n \to \infty} P(|V_n|<c\sqrt{n})=0,95$.

Any help is much appreciated.

Answer 1

Using the fact that $S_n\ and\ T_n$ are independent and approximately normal, you can write the joint density function approximately as a product of two normal densities. Convert to polar coordinates and the function of r (after integrating out the angle) is the approximate density function for $|V_n|$.

Answer 2

Note that $\frac{1}{n}S_n$ and $\frac{1}{n}T_n$ each converge to a normal distribution with mean 0 and variance $\frac{4/3}{n}$ by Central Limit Theorem. Thus, $$ \lim_{n\to\infty} {\left(\frac{S_n/n}{\sqrt{(4/3)/n}}\right)^2+\left(\frac{T_n/n}{\sqrt{(4/3)/n}}\right)^2} = \frac{3}{4n} (S_n^2+T_n^2) \quad\sim\quad \chi_2^2.$$

Let us write $Z=S_n^2+T_n^2$. Then, $$ \mathrm{Prob}(\sqrt{Z} \leq c\sqrt{n}) = \mathrm{Prob}(Z \leq c^2 n) = \mathrm{Prob}(\frac{3}{4n} Z \leq \frac{3}{4}c^2). $$

All that remains is to solve $F(3c^2/4)=0.95$ for $c$, where $F(x)$ is the CDF for the chi-squared distribution with two degrees of freedom.