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Let $\{X_n\}$ is a sequence of independent random variables with $P\{X_n=\pm n^\alpha\}=\frac{1}{2}$, $n=1,2,\cdots$.

The sequence $\{X_n\}$ obeys weak law of large numbers if

1) $\alpha <1/2$

2) $\alpha =1/2$

3) $1/2<\alpha \leq 1$

4) $\alpha >1$

I find $E(S_n)=0, Var(X_n)=n^{2 \alpha}$ and I have to show $\frac{E(S_n^2)}{n^2}$ tend to zero as $n \rightarrow \infty$ for satisfying WLLN. So

$\frac{E(S_n^2)}{n^2}=\frac{Var(S_n)}{n^2}=\frac{1}{n^2}\sum_{i=1} ^n Var(X_i)$

after that how to handle to find the condition on $\alpha$, Please help.

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  • $\begingroup$ Well... you computed $Var(X_i),$ so just plug it in. So do you need help evaluating that limit and seeing which $\alpha$ make it zero? (And from a multiple choice perspective this is pretty easy... you know $\alpha = 0$ obeys it.) $\endgroup$ – spaceisdarkgreen Jan 21 '18 at 7:03
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I assume is the summation what's troubling you... since you're almost there. Following you, $$\frac{Var(S_n)}{n^2}=\frac1{n^2}\sum_{k=1}^n Var(X_k)=\frac1{n^2}\sum_{k=1}^n k^{2\alpha}.$$

Well when this sequence converges is not that obvious, but a trick that works sometimes in these situations of mixed "n's" and "k's" is to try to form a Riemann sum, most of the times taking $n$ equally spaced intervals and a gauge formed by points like $c_k=\frac kn$ or so. This will give something of the form $\frac1n \sum_{k=1}^n f\left(\frac kn \right)$ for some $f$, although there might be variations.

Here we have $$\frac1{n^2}\sum_{k=1}^n k^{2\alpha}=\frac{n^{2\alpha}}{n}\cdot \frac1n \sum_{k=1}^n \left(\frac kn\right)^{2\alpha}=n^{2\alpha-1}\cdot \frac1n \sum_{k=1}^n f\left(\frac kn\right),$$ where $f(x)=x^{2\alpha}$. So we have $$\frac1n \sum_{k=1}^n f\left(\frac kn\right)\to \int_0^1x^{2\alpha}\,dx,$$ which is a positive real number for $\alpha\ge 0$ (and even for some negative values, as improper integral).

On the other hand, the limit of $n^{2\alpha -1}$ as $n\to \infty$ might be infinity, $1$ or $0$ depending on the value of $\alpha$. In the first case the limit will diverge, but in the last two will converge, although only in the last case it will go to zero.

Can you find possible $\alpha$ values from here?

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    $\begingroup$ Yes, I got it,$\alpha <1/2$, many thanks $\endgroup$ – 1256 Jan 22 '18 at 5:43

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