# A simple Central Limit Theorem question

Let $\{X_1,X_2,\ldots\}$ be random variables with mean $\mu$ and variance $\sigma^2$ both finite.

I know that the CLT gives

$$\frac{\sqrt{n}}{\sigma}\left(\frac{1}{n}\sum_{i=1}^nX_i - \mu\right)\overset{d}{\rightarrow}N(0,1)$$

But is it true that

$$\frac{\sqrt{n}}{\sigma}\left(\frac{1}{n}\sum_{i=1}^nX_i\right)\overset{d}{\rightarrow}N(\mu,1)$$

It seams right, but it cant prove it.

• 1. What makes you say "it seems right"? Can you explain the intuition? 2. Split the LHS of the first line into two terms (the first being the LHS of your second line) and look closely at the second term ... Commented Nov 18, 2016 at 6:19
• Got it, it's actually not true! Thx for your comment. Commented Nov 18, 2016 at 6:27
• If you'd care to put a less informal explanation as an answer, that might be good. Commented Nov 18, 2016 at 6:29

A simple rearrangement of terms proposed by @Glen_b shows that my intuition was wrong. Also a simple way to see why this is not true is to suppose that $X_i\sim N(\mu,\sigma^2)$. Then all we have to do is note that
$$\frac{\sqrt{n}}{\sigma}\left(\frac{1}{n}\sum_{i=1}^n{X_i}\right) \sim N\left(\frac{\sqrt{n}}{\sigma}\mu,1\right)$$
and that would never converge to $N(0,1)$.
• It can't really be said to converge to anything at all, because the distribution of $\sqrt n \bar{X}$ shifts further and further to the right as $n$ increases, beyond every finite bound. Commented Nov 18, 2016 at 10:53