Does this converge in distribution in this case?

Question

Let $X_i$ be i.i.d. sequence of random variables with finite second moment.

Then, by the Central limit theorem, $\frac{\sqrt{n}(\bar{X}-E(X))}{\sigma_X} \to N(0,1)$ in distribution.

Let $\alpha_n$ be a sequence such that $\alpha_n\to E(X)$.

Then, $\frac{\sqrt{n}(\bar{X}-\alpha_n)}{\sigma_X} \to N(0,1)$ in distribution? If so, how do I prove this?

Answer 1

Let $\alpha_n = \mathbb{E}(X)- \frac 1 {\sqrt{n}}$.

Then :

$$\frac{\sqrt{n}(\bar{X}-\alpha_n)}{\sigma_X}=\frac{\sqrt{n}(\bar{X}-\mathbb{E}(X))}{\sigma_X} + \frac{1}{\sigma_X} \overset{d}\to \mathcal{N}\left( \frac{1}{\sigma_X},1 \right).$$

Even worse :

Let $(X_n)$ a sequence of bounded r.v. and $\alpha_n=\mathbb{E}(X)- \frac {X_n} {\sqrt{n}}$, hence :

$$\frac{\sqrt{n}(\bar{X}-\alpha_n)}{\sigma_X}=\frac{\sqrt{n}(\bar{X}-\mathbb{E}(X))}{\sigma_X}+\frac {X_n} {\sigma_X}.$$

Then everything can happen.

If $|\alpha_n - \mathbb{E}(X)| = o (\sqrt{n})$ then it is true. Indeed :

$$\frac{\sqrt{n}(\bar{X}-\alpha_n)}{\sigma_X}=\frac{\sqrt{n}(\bar{X}-\mathbb{E}(X))}{\sigma_X}+\frac{\sqrt{n}(\mathbb{E}(X)-\alpha_n)}{\sigma_X}.$$

You already know what happens to the first term. For the second one :

$$\left|\frac{\sqrt{n}(\mathbb{E}(X)-\alpha_n)}{\sigma_X}\right| = o(1)$$

almost surely.