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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 CLT and unique convergence to normal distribution? Since we can take $n=1000$ and use any other random variable. What am I missing in this interpretation?

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?

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 \,\,\textit{for n big enough}. \end{equation}

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 CLT and unique convergence to normal distribution? Since we can take $n=1000$ and use any other random variable. What am I missing in this interpretation?

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 CLT and unique convergence to normal distribution? Since we can take $n=1000$ and use any other random variable. What am I missing in this interpretation?