[Soln] Central Limit Theorem: Strange Result!

For n \in \mathbb{N}, define the random variable

X_n = \begin{cases} \pm 1 &\text{each with probability } \frac{1}{2}\left( 1 - \frac{1}{n^2} \right), \\ \pm n^2 &\text{with probability } \frac{1}{2n^2}. \end{cases}

Let S_n = \displaystyle\sum_{k = 1}^n X_k. Prove that as n \rightarrow \infty,

a) the distribution of \displaystyle\frac{S_n}{\sqrt{n}} converges to \mathcal{N}(0, a) for some real number a \neq 2,

b) but \text{Var} \displaystyle\frac{S_n}{\sqrt{n}} converges to 2.

(Credits: I learnt of this problem from Persi Diaconis in my probability class.)

For each n \in \mathbb{N}, let Y_n be the simple truncation of X_n, i.e. Y_n = X_n 1_{\{ X_n < n \}}, or equivalently

Y_n = \begin{cases} \pm 1 &\text{each with probability} \frac{1}{2}\left(1 - \frac{1}{n^2} \right) \\ 0 &\text{otherwise.} \end{cases}

It is easy to calculate the following:

\begin{aligned} \mathbb{E} Y_n &= 0, \\  \text{Var} Y_n &= 1 - \frac{1}{n^2}, \\  \mathbb{E} |Y_n|^3 &= 1 - \frac{1}{n^2}. \end{aligned}

If we let s_n^2 = \displaystyle\sum_{k=1}^n \text{Var} Y_k, then

s_n^2 = n - \displaystyle\sum_{k=1}^n \frac{1}{k^2}.

Let us check that Lyapounov’s condition holds for \delta = 1:

\begin{aligned} \frac{1}{s_n^3} \sum_{k=1}^n \mathbb{E} |Y_k|^3 &= \frac{1}{\left(n - \sum_{k=1}^n \frac{1}{k^2} \right)^{3/2}} \sum_{k=1}^n 1 - \frac{1}{k^2} \\  &= \frac{n - \sum_{k=1}^n \frac{1}{k^2}}{\left(n - \sum_{k=1}^n \frac{1}{k^2} \right)^{3/2}} \\  &= \frac{1}{\sqrt{n - \sum_{k=1}^n \frac{1}{k^2}}}. \end{aligned}

Since \displaystyle\lim_{n \rightarrow \infty}\sum_{k=1}^n \frac{1}{k^2} = \frac{\pi^2}{6} < \infty, as n \rightarrow \infty, the denominator of the above goes to infinity, which means that the entire RHS goes to zero, i.e. Lyapounov’s condition holds. Hence, we can use Lyapounov’s Central Limit Theorem to conclude that

\displaystyle\frac{\sum_{k=1}^n Y_k}{s_n} \Rightarrow \mathcal{N}(0,1).

(Here, \Rightarrow means “converges in distribution”.) Next, note that

\displaystyle\sum_{n=1}^\infty \mathbb{P} \{ X_n \neq Y_n \} = \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} < \infty,

By the first Borel-Cantelli lemma, we conclude that \mathbb{P} \{ X_n \neq Y_n \text{i.o.} \} = 0. From this, we can further conclude that

\begin{aligned} \displaystyle\frac{\sum_{k=1}^n X_k}{s_n} =\frac{\sum_{k=1}^n X_k}{\sqrt{n - \sum_{k = 1}^n \frac{1}{k^2}}}  &\Rightarrow \mathcal{N}(0,1), \\  \frac{S_n}{\sqrt{n}} \frac{\sqrt{n}}{\sqrt{n - \sum_{k = 1}^n \frac{1}{k^2}}} &\Rightarrow \mathcal{N}(0,1), \\  \frac{S_n}{\sqrt{n}}&\Rightarrow \mathcal{N}(0,a),  \end{aligned}

where a = \displaystyle\lim \frac{n - \sum_{k=1}^n \frac{1}{k^2}}{n} = 1.

It is straightforward to show that \text{Var} \displaystyle\frac{S_n}{\sqrt{n}} converges to 2:

\begin{aligned}\text{Var} \displaystyle\frac{S_n}{\sqrt{n}} &= \frac{1}{n} \sum_{k=1}^n \text{Var} X_k \\  &= \frac{1}{n} \sum_{k=1}^n \left( 2 - \frac{1}{k^2} \right) \\  &= 2 - \frac{1}{n}\sum_{k=1}^n \frac{1}{k^2} \\  &\rightarrow 2. \end{aligned}

Done!

Advertisements
This entry was posted in Undergraduate and tagged . Bookmark the permalink.

Leave a Reply

Please log in using one of these methods to post your comment:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s