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In my probability class we learned the Central Limit Theorem in the following form.

Theorem: Let $\{X_i\}_{i=1}^\infty$ be a sequence of independent identically distributed random variables and suppose that $E(X_i)=\mu$, $\operatorname{Var}(X_i)=\sigma^2<\infty$. Then,

$$\sqrt{n}\frac{\frac{1}{n}\sum_{i=1}^nX_i-\mu}{\sigma}\stackrel{P}{\longrightarrow}N(0,1),\quad\text{as }n\to\infty$$

in probability, where the notation $N(\mu,\sigma^2)$ means normal distribution of mean $\mu$ and variance $\sigma^2$.

Now, if we remove $\sigma$ and $\mu$, do we get $$\sqrt{n}\frac{1}{n}\sum_{i=1}^nX_i\stackrel{P}{\longrightarrow}N(\mu,\sigma^2)?$$ This seems very natural, but I cannot prove it. Is it true? I think I can prove it by removeing only $\sigma$, but I am not sure for $\mu$.

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    $\begingroup$ Hint: $\displaystyle E\left[\sqrt{n}\frac 1n\sum_{i=1}^nX_i\right] = \sqrt{n}\frac 1n E\left[\sum_{i=1}^nX_i\right] = \sqrt{n}\frac 1n n\mu = \sqrt{n}\mu$. $\endgroup$ – Dilip Sarwate Dec 6 '14 at 23:48
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    $\begingroup$ I'm not sure what you mean by the superscript $P$ over the arrow, but the usual central limit theorem says that that random variable converges "in distribution" or "in law" to the normal distribution. I'd be likely to understand the letter $P$ as indicating convergence in probability rather than convergence in distribution. $\endgroup$ – Michael Hardy Dec 7 '14 at 0:31
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$$\sqrt{n}\frac{\frac{1}{n}\sum_{i=1}^nX_i-\mu}{\sigma}\stackrel{d}{\longrightarrow}N(0,1),\quad\text{as }n\to\infty$$ $$\implies \sqrt{n}\left(\frac{1}{n}\sum_{i=1}^nX_i-\mu\right)\stackrel{d}{\longrightarrow}N(0,\sigma^2),\quad\text{as }n\to\infty$$ $$\implies \frac{1}{\sqrt{n}}\sum_{i=1}^nX_i-(\sqrt{n}-1)\mu \stackrel{d}{\longrightarrow}N(\mu,\sigma^2),\quad\text{as }n\to\infty$$ but I am not sure that is more useful

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    $\begingroup$ @Michael Hardy: I would have thought it was obvious you can add constants to both sides or multiply both sides by constants without affecting the convergence. What you cannot do is put variables (random or systematic) into the limit, e.g. you cannot properly say that $\frac1n \sum_{i=1}^nX_i$ converges in distribution to $N(\mu,\sigma^2/n)$ even if it commonly done. $\endgroup$ – Henry Dec 7 '14 at 0:47
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    $\begingroup$ @Michael Hardy: Really? If $|x_n-x|\lt \epsilon$ then $|(ax_n+b)-(ax+b)|\lt |a|\epsilon$ $\endgroup$ – Henry Dec 7 '14 at 10:19
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    $\begingroup$ @Michael Hardy: Or perhaps you want for $a \gt 0$ the result $\Pr(aX+b\le ak+b)=\Pr(X\le k)$ $\endgroup$ – Henry Dec 7 '14 at 10:30
  • $\begingroup$ No, that's not what I had in mind. $\endgroup$ – Michael Hardy Dec 8 '14 at 20:49
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Be glad that you have not been able to prove that $\displaystyle\sqrt{n}\frac{1}{n}\sum_{i=1}^nX_i\stackrel{P}{\longrightarrow}N(\mu,\sigma^2)$ because the result is not true, not only for the desired mode of convergence but also for any other mode of convergence. As my comment on your question pointed out, $$\displaystyle E\left[\sqrt{n}\frac 1n\sum_{i=1}^nX_i\right] = \sqrt{n}\frac 1n E\left[\sum_{i=1}^nX_i\right] = \sqrt{n}\frac 1n n\mu = \sqrt{n}\mu \to \infty ~\text{as}~ n \to \infty$$ and so the random variable $\displaystyle\sqrt{n}\frac{1}{n}\sum_{i=1}^nX_i$ does not converge to a fixed random variable in any sense.

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