I have the followig sequences:

  1. $Pr(X_n=n)=Pr(X_n=-n)=0.5$

  2. $Pr(X_n=2^{n/2})=Pr(X_n=-2^{n/2})=0.5$

I have to show whether they satisfy Lindeberg's condition or not, but this condition is a bit unclear for me.

$\lim_{n\to\infty} \frac{1}{s_n^2} \sum^{n}_{k=1} \mathbb{E} [(X_k -\mu_k)^2 \cdot \mathbb{1}_{|X_k-\mu_k|>\epsilon s_n}]=0$

for all $\epsilon >0$.

The expected values of the r.v.-s are $0$.

The variance in the first case: $n^2$, in the second case $2^n$.

In both cases the sums of the variances go to infinity. Isn't that enough for the condition to go to zero?

And I don't really understand that expression in the indicator function.

I appreciate any help!

  • $\begingroup$ This looks like routine bookwork. Is this for some class? $\endgroup$
    – Glen_b
    Nov 23 '14 at 23:11
  • 1
    $\begingroup$ @Glen_b what routine bookwork for some may not be routine bookwork for someone else. $\endgroup$
    – lena242
    Nov 30 '14 at 18:15
  • $\begingroup$ The point is that as the kind of question one might find in a textbook and could be set for a class, it would then count as a self-study question. See the self-study tag wiki. Please add the tag and read the tag wiki info, modifying the question if needed. $\endgroup$
    – Glen_b
    Dec 1 '14 at 2:36

I will start with just some guidance, and perhaps return later to complete the answer.

Consider the first sequence of random variables, and note that $|X_n| = n$. In other words for given $n$, the absolute value of the random variable is a constant function (as is always the case for dichotomous random variables symmetric around zero).

Also, $$s^2_n = \sum_{i=1}^n\sigma^2_k = 1+2^2+3^2+...+n^2 =\frac {n(n+1)(2n+1)}6 = O(n^3)$$

Then the indicator function for some $k$ becomes

$$\mathbb{1}_{\left\{k>\epsilon \left(\frac {n(n+1)(2n+1)}6\right)^{1/2}\right\}}$$

Nothing random remains in here, so it can be taken out of the expected value, being a deterministic function.

Can you take it from here?

  • $\begingroup$ So, if n is very big even with a very small $\epsilon$ the right hand side will be bigger then k? $\endgroup$
    – lena242
    Nov 24 '14 at 10:18
  • $\begingroup$ Give the left-hand side its best advantage: examine the leading indicator function (where $k=n$). If it weren't for the indicator functions, I guess you see that in this case the limit under examination would be equal to unity. So the issue is whether the existence of the indicator functions makes the numerator to have a fixed value, or to be bounded, or to be of lower order than the denominator $\endgroup$ Nov 24 '14 at 12:02
  • $\begingroup$ After a while we are summing zeros, right? So the whole expression will go to infinity. But what about the 2.) case? $2^{k/2}>\epsilon (2^{n+1}-2)^{1/2}$ $\endgroup$
    – lena242
    Nov 24 '14 at 16:32
  • $\begingroup$ Square the expression $\endgroup$ Nov 24 '14 at 19:04
  • $\begingroup$ They have the same order. For The right hand side will be bigger for big epsilons, so again the condition doesn't hold, right? $\endgroup$
    – lena242
    Nov 24 '14 at 19:26

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