# Controlling the Lyapunov condition

I have been struggling with the following exercise for quite some time now.

Let $$(Z_n)_{n \geq 1}$$ be a sequence of independent random variables such that for $$j=1,2, \ldots$$ we have that for some constant $$a \in (1,1.5)$$.

$$P(Z_j=j^a)=P(Z_j=-j^a)=\frac{1}{6}j^{-2(a-1)},$$ and $$P(Z_j=0) = 1-\frac{1}{3}j^{-2(a-1)}.$$

Verify the Lyapunov condition.

I know that the Lyapunov condition is the following: Let $$E(Z_j) = \mu_j$$, $$V(Z_j) = \sigma_j^2$$ and $$s_n^2 = \sum_{j=1}^n\sigma^2_j$$.

Then, if for $$r > 2$$ the following function goes to zero as $$n\rightarrow \infty$$, then we have that the Lyapunov condition.

$$\beta(n,r)=\frac{\sum_{j=1}^nE(|Z_j-m_j|^r)}{s_n^r}.$$

## My Attempt:

We know that for any $$Z_j$$ we have $$E(Z_j) = j^a \cdot \frac{1}{6}j^{-2(a-1)} - j^a \cdot \frac{1}{6}j^{-2(a-1)} + 0 \cdot (1-\frac{1}{3}j^{-2(a-1)}) = 0.$$ Thus $$m_j = 0$$ for all $$j$$. Also, now we know that $$V(Z_j) = E(Z_j^2)$$.

$$\beta(n,r) = \frac{\sum_{j=1}^nE(|Z_j|^r)}{\sqrt(\sum_{j=1}^nE(Z_j^2))^r}.$$

We know that $$r = 2 + \delta$$ for $$\delta > 0$$, I feel as if perhaps we could use Hölders inequality to make sense out of this mess, but I'm not sure how.

Also, can I just choose some value for $$r$$ and if the condition holds for that value, is that enough for the Lyapunov condition?

I have also tried just calculating all of the expected values and by doing that I got the following:

$$\beta(n,r) = \frac{\sum_{j=1}^n\frac{1}{3}j^{a(r-2)+2}}{(\sqrt{\sum_{k=1}^n\frac{1}{3}j^2})^r}.$$

But I believe that there should be a smarter way to deal with this problem than what I have done thus far. Also maybe it is worth noting that I thought if we were to find an upper bound for $$\beta(n,r)$$ that goes to $$0$$ as $$n \rightarrow \infty$$, then we have that $$\beta(n,r)$$ will also go to $$0$$, since it is lower bounded by 0 in this case.

You have already done the hard part of the proof; note that $$\sum_{j=1}^n x^k$$ scales on the order of $$n^{k+1}$$ (herein denoted as $$\theta(n^{k+1})$$) for $$k \geq 0$$ (This can be checked via integral test). So, you know that in your fraction below, the top sum goes as $$\theta(n^{a \delta + 3})$$ and the bottom sum goes as $$\theta(n^\frac{6 + 3\delta}{2})$$. Now, your upper bound constraint on $$a$$ gives you the result you want.