# Thread: Is my professor wrong again?

1. ## Is my professor wrong again?

My professor has the annoying habit of consistently giving us problems which are stated wrong (after few hours of you eventually figure out it's not you that's making the mistake), but this time I think it happened on the midterm. Could someone please look at this:

For what s=? does $\displaystyle Var(X_n) ->_p0$ ?
$\displaystyle X_n$ is the sample mean of n independent $\displaystyle X_i$ (not identically distributed), where $\displaystyle X_i=i^s$ with $\displaystyle p=1/2$ and $\displaystyle X_i=-i^s$ with $\displaystyle p=1/2$.

Here's my reasoning. $\displaystyle Var(X_i)=1/2*(i^{2s}+i^{2s})=i^{2s}$ so then $\displaystyle Var(X_n)=Var(1/n*sum(X_i))=1/n^2*sum(Var(X_i))=$ (for i from 1 to n)
$\displaystyle =1/n^2*sum(i^{2s})=O_p(n^{2s}/n^2)=O_p(n^{2(s-1)})$ as n->oo
This means the highest order of n in the sequence is 2(s-1). The sequence will go to 0 as n goes to infinity (for all terms) if the highest order also goes to 0. So 2(s-1)<0, thus s<1.
However, the problem asked me to prove that s<1/2. I asked the professor to make sure my assumptions are correct; he confirmed, however the result still disagrees. To be diplomatic, I wrote the solution until the last line (with $\displaystyle O_p$), however I wrote that the conclusion is s<1/2. (just to make sure i'm wrong no matter what happens )) Opinions?

2. Originally Posted by svarog
My professor has the annoying habit of consistently giving us problems which are stated wrong (after few hours of you eventually figure out it's not you that's making the mistake), but this time I think it happened on the midterm. Could someone please look at this:

For what s=? does $\displaystyle Var(X_n) ->_p0$ ?
$\displaystyle X_n$ is the sample mean of n independent $\displaystyle X_i$ (not identically distributed), where $\displaystyle X_i=i^s$ with $\displaystyle p=1/2$ and $\displaystyle X_i=-i^s$ with $\displaystyle p=1/2$.

Here's my reasoning. $\displaystyle Var(X_i)=1/2*(i^{2s}+i^{2s})=i^{2s}$ so then $\displaystyle Var(X_n)=Var(1/n*sum(X_i))=1/n^2*sum(Var(X_i))=$ (for i from 1 to n)
$\displaystyle =1/n^2*sum(i^{2s})=O_p(n^{2s}/n^2)=O_p(n^{2(s-1)})$ as n->oo
This means the highest order of n in the sequence is 2(s-1). The sequence will go to 0 as n goes to infinity (for all terms) if the highest order also goes to 0. So 2(s-1)<0, thus s<1.
However, the problem asked me to prove that s<1/2. I asked the professor to make sure my assumptions are correct; he confirmed, however the result still disagrees. To be diplomatic, I wrote the solution until the last line (with $\displaystyle O_p$), however I wrote that the conclusion is s<1/2. (just to make sure i'm wrong no matter what happens )) Opinions?
Opinion:
1) I guess you mean $\displaystyle Var(S_n)$ where $\displaystyle S_n=X_1+\cdots+X_n$ (or $\displaystyle \bar{X}_n$)
2) $\displaystyle \sum_{i=1}^n i^{2s}\sim_{n\to\infty}\int_0^n x^{2s}dx=\frac{n^{2s+1}}{2s+1}$, hence $\displaystyle {\rm Var}(\frac{S_n}{n})\sim_n \frac{n^{2s+1}}{(2s+1)n^2}=C n^{2s-1}$...