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.


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.

Thanks for your time,

K. Kamal


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.


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