Erlang distribution and Central Limit Theorem

Question

Let $X\sim\textrm{Erlang}(k,\lambda)$. Use the Central Limit Theorem to evaluate

$\underset{k\rightarrow\infty}{\lim}\mathbb{P}\left(\left|X-\mathbb{E}\left[X\right]\right|>\epsilon \cdot \mathbb{E}\left[X\right]\right)$

where $\epsilon>0$ but small.

Answer 1

According to Larsen and Marx, the central limit theorem states that:

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$$\underset{k\rightarrow\infty}{\lim}\mathbb{P}\left(\left|X-\mathbb{E}\left[X\right]\right|>\epsilon \cdot \mathbb{E}\left[X\right]\right)$$

$$= 1 - \underset{k\rightarrow\infty}{\lim}\mathbb{P}\left(\left|X-\mathbb{E}\left[X\right]\right|\le\epsilon \cdot \mathbb{E}\left[X\right]\right)$$

Let us try to use the central limit theorem on

$$\mathbb{P}\left(\left|X-\mathbb{E}\left[X\right]\right|\le\epsilon \cdot \mathbb{E}\left[X\right]\right)$$

$$= \mathbb{P}\left( -\epsilon \cdot \mathbb{E}\left[X\right] \le X-\mathbb{E}\left[X\right] \le \epsilon \cdot \mathbb{E}\left[X\right]\right)$$

$$= \mathbb{P}\left( -\epsilon \cdot \frac{k}{\lambda} \le (W_1 + ... + W_k)-\frac{k}{\lambda} \le \epsilon \cdot \frac{k}{\lambda} \right)$$

$$= \mathbb{P}\left( \frac{-\epsilon \cdot \frac{k}{\lambda}}{\frac{\sqrt{k}}{\lambda}} \le \frac{(W_1 + ... + W_k)-\frac{k}{\lambda}}{\frac{\sqrt{k}}{\lambda}} \le \frac{\epsilon \cdot \frac{k}{\lambda}}{\frac{\sqrt{k}}{\lambda}} \right)$$

$$= \mathbb{P}\left( -\epsilon \sqrt{k} \le \frac{(W_1 + ... + W_k)-\frac{k}{\lambda}}{\frac{\sqrt{k}}{\lambda}} \le \epsilon \sqrt{k} \right)$$

Unfortunately, our 'a' and 'b' here depend on k. Nevertheless, assuming I am interpreting my stochastic calculus class notes correctly, CLT is still applicable (if the random variable didn't depend on k, I think we could've used continuity of probability):

$$= \lim_{k \to \infty} \mathbb{P}\left( -\epsilon \sqrt{k} \le \frac{(W_1 + ... + W_k)-\frac{k}{\lambda}}{\frac{\sqrt{k}}{\lambda}} \le \epsilon \sqrt{k} \right)$$

$$= \mathbb{P}\left( \lim_{k \to \infty} -\epsilon \sqrt{k} \le \lim_{k \to \infty} \frac{(W_1 + ... + W_k)-\frac{k}{\lambda}}{\frac{\sqrt{k}}{\lambda}} \le \lim_{k \to \infty} \epsilon \sqrt{k} \right)$$

$$= \mathbb{P}\left( -\infty < Z < \infty \right) = 1$$

Hence

$$\underset{k\rightarrow\infty}{\lim}\mathbb{P}\left(\left|X-\mathbb{E}\left[X\right]\right|>\epsilon \cdot \mathbb{E}\left[X\right]\right)$$

$$= 1 - \underset{k\rightarrow\infty}{\lim}\mathbb{P}\left(\left|X-\mathbb{E}\left[X\right]\right|\le\epsilon \cdot \mathbb{E}\left[X\right]\right)$$

$$= 1 - 1 = 0$$