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By central limit theorem, we know that for independent and identically distributed random variables $X_1,X_2,...,X_n$, as $n\rightarrow\infty$, that:

$$\frac{\sqrt{n}(\bar X-\mu)}{\sigma}\rightarrow N(0,1)$$

Where, in this context, $\rightarrow$ represents convergence in distribution, $\bar X=\frac{\sum_{i=1}^n X_i}{n}$, $\mu$ is the mean of the $X_i$'s, and $\sigma$ is the standard deviation of the $X_i$'s.

However, what is the the limiting distribution of the following quantity, for $\alpha\in\mathbb{R}_{>0}$ and $\beta\in\mathbb{R}$:

$$\frac{\alpha\sqrt{n}(\bar X-\beta\mu)}{\sigma}$$

Similarly, what is the limiting distribution of:

$$\frac{\alpha(\bar X-\beta\mu)}{\sigma}$$

Is it possible to determine a limiting distribution for either of these quantities for all values $\alpha$ and $\beta$? Remember, the underlying $X_i$'s are not necessarily normally distributed; all we know is that they are i.i.d. If not, what additional assumptions need to be made to determine these limiting distributions?

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Let us start from the second expression. Note that $\bar{X}_n$ converges in probability to $\mu$ (WLLN), and all the rest are constants, hence by the continuous mapping theorem you have that $$ \sigma^{-1}(\bar{X}_n - \mu \beta) = g(\bar{X}_n) \xrightarrow{p} g(\mu) = \sigma^{-1}(\mu - \mu \beta). $$ For the first expression is slightly more subtle. $\alpha$ doesn't bother us, however let us look at $\beta$. If $\beta = 1$ then you limit is $\alpha N(0,1)$ (by the continuous mapping, regular calculations - whatever). Now, let $\beta \neq 1$ (and $\mu \neq 0$), thus $\pm \mu$ in the brackets you get $$ \frac{\alpha \sqrt{n}}{\sigma}( \bar{X}_n - \mu + \mu(1-\beta) )= \frac{\alpha \sqrt{n}}{\sigma}( \bar{X}_n - \mu) + \frac{\alpha \sqrt{n}}{\sigma}\mu(1-\beta) \xrightarrow{D}\alpha N(0,1) \pm \infty = \pm \infty. $$ Namely, if $\beta < 1$ then it goes to $\infty$, if $\beta >1$ then to $-\infty$.

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  • $\begingroup$ In your first statment, you say that $\sigma^{-1}(\bar{X}_n - \mu \beta) = g(\bar{X}_n) \xrightarrow{p} g(\mu)$. However, $g(\mu)$ is just a constant, is it not? Should it not converge to a random variable? I'm specifically interested in the distribution of what it's converging to. $\endgroup$ Commented Feb 16, 2018 at 3:33
  • $\begingroup$ @jippyjoe4 There is no reason for $g(\bar{X}_n)$ to converge to a non degenerate r.v. $\bar{X}_n$ converges a.s. to $\mu$ and $g()$ is just some continuous transformations, hence $g(\bar{X}_n)$ converges a.s. to $g(\mu)$. Intuitively, In the second expression the $\sqrt{n}$ is the factor that doesn't let $\bar{X}_n - \mu$ to go to $0$ as it "pools" it further as $\bar{X}_n$ tends to $\mu$, while in the first expression there is no such factor. $\endgroup$ Commented Feb 16, 2018 at 10:44
  • $\begingroup$ Okay, I think that I understand now. Thank you. $\endgroup$ Commented Feb 16, 2018 at 17:56

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