MHB Weak Convergence to Normal Distribution

AI Thread Summary
The discussion revolves around proving that the expression (1/√n)(∑X_n - n/2) converges weakly to a normal distribution as n approaches infinity. Participants clarify the notation, suggesting that the summation index should be corrected for clarity. They agree that the problem likely stems from the Central Limit Theorem, which indicates that the normalized sum of independent random variables converges to a normal distribution. The expected mean of the random variables X_n approaches 1/2, supporting the connection to the Central Limit Theorem. Overall, the consensus is that the problem can be addressed using principles from probability theory.
joypav
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Problem:
Let $X_n$ be independent random variables such that $X_1 = 1$, and for $n \geq 2$,

$P(X_n=n)=n^{-2}$ and $P(X_n=1)=P(X_n=0)=\frac{1}{2}(1-n^{-2})$.

Show $(1/\sqrt{n})(\sum_{m=1}^{n}X_n-n/2)$ converges weakly to a normal distribution as $n \rightarrow \infty$.Thoughts:

My professor sent this problem over email and I am first off wondering about the notation. I think that the last line is meant to read $(\frac{1}{\sqrt{n}})(\frac{\sum_{m=1}^{n}X_m-n}{2})$? (I assume the summation index was simply a typo and that it is all over 2 not just the n?)

If that is the case, then I am thinking this is a consequence of the Central Limit Theorem.
Which has conclusion,

$\frac{S_n - \mu n}{\sigma \sqrt{n}} \implies \chi$

where $\chi$ is $N(0,1)$.
 
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joypav said:
Problem:
Let $X_n$ be independent random variables such that $X_1 = 1$, and for $n \geq 2$,

$P(X_n=n)=n^{-2}$ and $P(X_n=1)=P(X_n=0)=\frac{1}{2}(1-n^{-2})$.

Show $(1/\sqrt{n})(\sum_{m=1}^{n}X_n-n/2)$ converges weakly to a normal distribution as $n \rightarrow \infty$.Thoughts:

My professor sent this problem over email and I am first off wondering about the notation. I think that the last line is meant to read $(\frac{1}{\sqrt{n}})(\frac{\sum_{m=1}^{n}X_m-n}{2})$? (I assume the summation index was simply a typo and that it is all over 2 not just the n?)

If that is the case, then I am thinking this is a consequence of the Central Limit Theorem.
Which has conclusion,

$\frac{S_n - \mu n}{\sigma \sqrt{n}} \implies \chi$

where $\chi$ is $N(0,1)$.
I am not a probabilist, but I think that your formula should read $\frac{1}{\sqrt{n}} \Bigl(\left(\sum_{m=1}^{n} X_m\right)- \frac{n}{2}\Bigr)$.

If $m$ is large, then $X_m$ takes each of the values $0$ and $1$ with a probability close to $\frac12$ (and the large value $m$ with a very small probability $m^{-2}$). So the mean value of $X_m$ will be close to $\frac12$, and the mean value of $S_n = \sum_{m=1}^{n} X_m$ will be close to $\frac n2$. This seems to make it plausible that your formula should somehow relate to the central limit theorem.
 
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