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
\begin{equation}
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  \,\,
\end{equation}
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?
 A: The convergence to Normal in the CLT is a stronger statement than that.

*

*The (weak) Law of Large Numbers says $\bar Y_n\stackrel{p}{\to} \mu$

*Using Chebyshev's inequality you can strengthen this to get the rate of convergence $\bar Y_n-\mu=O_p(n^{-1/2})$

*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
A: 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.
