MHB Weak Convergence to Normal Distribution

joypav
Messages
149
Reaction score
0
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 $N(0,1)$.
 
Physics news on Phys.org
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.
 
Last edited:
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...
Thread 'Detail of Diagonalization Lemma'
The following is more or less taken from page 6 of C. Smorynski's "Self-Reference and Modal Logic". (Springer, 1985) (I couldn't get raised brackets to indicate codification (Gödel numbering), so I use a box. The overline is assigning a name. The detail I would like clarification on is in the second step in the last line, where we have an m-overlined, and we substitute the expression for m. Are we saying that the name of a coded term is the same as the coded term? Thanks in advance.
Back
Top