I know question on central limit theorem has been asked numerous times and I have searched around the website but still couldn't get what I wanted to find out (probably most people who asked the question is stuck at different parts of the theory).
Attached in my question is a lecture slide for the search for a transformation of x̅ that has a limiting distribution. I do not understand why x̅ and µ is multiply by √n instead of n. Also, why when the statistic √n(x̅ - µ) is used, µ becomes zero?
Thank you in advance!
A previous slide defines $$ \bar{X} = \frac{X_1+\cdots+X_n}{\sqrt{n}}, $$ where the random variables $X_1,\ldots,X_n$ are independent copies of the same distribution, which has mean $\mathbb{E}[X_i] = \mu$ and variance $\mathbb{V}[X_i] = \sigma^2$. One calculates that the mean of $\bar{X}$ is also $\mu$, and its variance is $\sigma^2/n$.
We want to normalize $\bar{X}$ so that it has mean $0$ and variance $1$. If we subtract $\mu$ from $\bar{X}$ we obtain a random variable $\bar{X} - \mu$ with mean $0$ and the same variance. If we multiply the result by $\sqrt{n}/\sigma$ then the mean remains $0$ and the variance becomes $1$. The new random variable is $$ \tilde{X} = \frac{\sqrt{n}(\bar{X} - \mu)}{\sigma}. $$ It so happens that as $n\to\infty$, $\tilde{X}$ tends to a normal distribution $N(0,1)$. This is the central limit theorem.
In order for a theorem of this sort to hold, the mean and variance of $\tilde{X}$ have to be fixed. This is why we normalized it to have mean $0$ and variance $1$. We could have chosen other values and would have obtained an equivalent central limit theorem, but the choice of zero mean and unit variance is the conventional standard.
The slide tries to explain why other normalizations won't work:
$\bar{X}$ has fixed mean but the variance tends to $0$. Therefore $\bar{X}$ tends to the constant $\mu$ distribution. This is the law of large numbers. The law tells us what $\bar{X}$ tends to, but not how the deviation from the mean behaves.
$\sqrt{n} \bar{X}$ has fixed variance $\sigma^2$ but the mean tends to infinity (or minus infinity). So $\sqrt{n} \bar{X}$ cannot converge to any fixed distribution, unless $\mu = 0$. This is because if $\sqrt{n} \bar{X}$ tends to a distribution with mean $m$, then the mean of $\sqrt{n} \bar{X}$ must tend to $m$.
$\sqrt{n}(\bar{X} - \mu)$ has mean $0$ and variance $\sigma^2$, and we can hope for a limit theorem. Indeed, the central limit theorem shows that $\sqrt{n}(\bar{X} - \mu)$ tends to $N(0,\sigma^2)$, a normal distribution with mean $0$ and variance $\sigma^2$. This is equivalent to the central limit theorem stated above.