# Cental limit theorem, Chebyshev's inequality, and convergence of distributions through rescaling

I've been thinking about this issue for a few days and although read some of relevant questions on this site, still couldn't get it off my mind.

Suppose we have $$n$$ i.i.d random variables $$X_i$$ with mean $$\mu$$ and variance $$\sigma^2$$. Suppose also that $$Y$$ is a random variable with mean zero and variance $$1$$ that is independent of $$X_i$$s, then using Chebyshev's inequality

$$$$P\Big( |\frac{\sum_{i=1}^{n}X_i-n\mu}{n\sigma}-\frac{1}{\sqrt{n}}Y |\geq \epsilon\Big)\leq \frac{V[\frac{\sum_{i=1}^{n}X_i-n\mu}{n\sigma}-\frac{1}{\sqrt{n}}Y ]}{\epsilon^2}=\frac{2}{n\epsilon^2}\rightarrow 0\quad \,\,$$$$

for big enough $$n$$. Therefore for any random variable $$Y(0,1)$$, and scaling it with $$\frac{1}{\sqrt{n}}$$, meaning $$\frac{Y}{\sqrt{n}}$$, we can approximate the distribution of $$\frac{\sum_{i=1}^{n}X_i-n\mu}{n\sigma}$$.

Doesn't it contradict the CLT and unique convergence to a normal distribution? Since we can take $$n=1000$$ and use any other random variable. What am I missing in this interpretation?

• What do you mean by "Since we can take n=1000 and use any other random variable."?? Commented Apr 16, 2021 at 17:38
• @Fiodor1234 I mean it s not needed for "n" to go to infinity, since the RHS of the limit will still me a small number, close enough to zero, and we can take the probability almost zero. Like when we say in central limit theorem, that n=>30 will do the job. Commented Apr 16, 2021 at 17:55

The convergence to Normal in the CLT is a stronger statement than that.

1. The (weak) Law of Large Numbers says $$\bar Y_n\stackrel{p}{\to} \mu$$
2. Using Chebyshev's inequality you can strengthen this to get the rate of convergence $$\bar Y_n-\mu=O_p(n^{-1/2})$$
3. The CLT strengthens this even further to say what happens when you rescale the $$O_p(n^{-1/2})$$ remainder so it doesn't collapse to zero $$\sqrt{n}(\bar Y_n-\mu)\stackrel{d}{\to} N(0,\sigma^2)$$

3 implies 2 which implies 1

To begin with, I'm going to use some referent parts to simplify your statement. Using the sample mean $$\bar{X}_n$$ you can write the difference and scaled difference of interest as:

$$D_n \equiv \frac{\bar{X}_n - \mu}{\sigma / \sqrt{n}} - Y \quad \quad \quad S_n \equiv \frac{1}{\sqrt{n}} \bigg[ \frac{\bar{X}_n - \mu}{\sigma / \sqrt{n}} - Y \bigg].$$

It is simple to show that:

$$\begin{matrix} \mathbb{E}(D_n) = 0 & & & \mathbb{V}(D_n) = 2, \\[6pt] \mathbb{E}(S_n) = 0 & & & \mathbb{V}(S_n) = 2/n. \\[6pt] \end{matrix}$$

With this machinery in place, your statement uses Chebychev's rule to say that:

$$\lim_{n \rightarrow \infty} \mathbb{P}(S_n \geqslant \epsilon) \leqslant \lim_{n \rightarrow \infty} \frac{2}{n \epsilon^2} = 0.$$

Now, the limiting result you observe is merely showing that the scaled difference $$S_n$$ converges in probability to zero. It does not say anything about the difference quantity $$D_n = \sqrt{n} S_n$$ (which has a higher order than the scaled difference). Your conclusion that you can adequately approximate the distribution of $$(\bar{X}_n-\mu)/\sigma$$ with the distribution of $$Y$$ is mistaken. Your confusion here seems to arise from conflating the difference with the scaled difference. If we look at the asymptotic properties of the difference quantity $$D_n$$ instead of the scaled difference, we can see that this converges to the difference between a standard normal random variable and another (arbitrary) standardised random variable. That is what the CLT tells us in this case.