# Weak Convergence to Normal Distribution

#### joypav

##### Active member
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 $N(0,1)$.

#### Opalg

##### MHB Oldtimer
Staff member
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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