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I have a quick question. If we have a sequence of i.i.d random variables, $X_1,X_2,X_3,\ldots,X_n$. Then, we make sequence below: $$ S=X_1-X_2+X_3-X_4+\ldots+X_n $$ Does central limit theorem work in this situation? (can you refer me to a reference).

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    $\begingroup$ I edited your title to be more informative. If it is incorrect or you prefer other title feel free to edit it or revert my edit. $\endgroup$
    – Tim
    Mar 23, 2015 at 12:46

3 Answers 3

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  1. Consider $n$ even. Let $Y_i=X_{2i-1}-X_{2i}$, $i=1,2,...,n/2$. Then the $Y_i$ are iid and you can apply the CLT to the $Y$'s as long as they satisfy the required conditions, which they should do if the $X$'s do*.

  2. However, you don't even need to come up with that. There's more than one CLT, you just need to look at the right one. If some iid version of the CLT (Lindeberg-Lévy, presumably) applies to the $X$'s, the Lyapunov CLT should apply to that alternating sum.

* for example, if the $X$'s have finite variance, the $Y$'s will, and so on.

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  • $\begingroup$ Please consider mentioning the finite variance requirement in your answer for the sake of completeness. $\endgroup$
    – Sid
    Mar 24, 2015 at 20:58
  • $\begingroup$ @Sid isn't that already covered by "you can apply the CLT to the Y's as long as they satisfy the required conditions which they should if the X's do" in the first part and a similar comment in the second part? The question isn't about the conditions for the CLT for the $X$'s -- presumably someone dealing with this question can tell what conditions they needed on $X$ already. $\endgroup$
    – Glen_b
    Mar 24, 2015 at 21:05
  • $\begingroup$ @Sid I've added a little to my answer in any case. $\endgroup$
    – Glen_b
    Mar 24, 2015 at 21:14
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Yes, the central limit theorem works in these circumstances if you are willing to assume that $X$ has finite mean $\mu$ and finite variance $\sigma^2$ (something that you most likely had assumed already but just forgot to state explicitly).

Define $Y_i = \begin{cases}X_i, &i~~ \text{odd},\\ -X_i, &i~ ~\text{even.}\end{cases}$ Then, the $Y_i$'s are a sequence of independent random variables with the same variance $\sigma^2$ but not necessarily the same distribution, even in the case of $\mu =0$. Now, a paraphrased version of the Mathworld article on the central limit theorem says

Let $Y_1,Y_2,...,Y_N$ be a set of $N$ independent random variates and each $Y_i$ have an arbitrary probability distribution with mean $\mu_i$ and a finite variance $\sigma_i^2$. Then the normal form variate $$Y_{\text{norm}}= {\sum_{i=1}^N y_i- \sum_{i=1}^N \mu_i\over \sqrt{\sum_{i=1}^N \sigma_i^2}}$$
has a limiting cumulative distribution function which approaches a normal distribution.

This gives the desired result directly. As @Glen_b says, there are many CLTs and the above version seems the most directly applicable.

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The quantity whose convergence we are usually examining in a CLT-context is

$$\frac {S_n - E(S_n)}{\sqrt {{\rm Var}(S_n)}} \xrightarrow{d}\; ?\; N(0,1) \tag{1}$$

For the $S_n$ in question the fact that the sign alternates does not pose any issue as regards its variance, since variances are added irrespective of whether the RV's themselves are added or subtracted. Since the various sufficient conditions for the CLT to hold usually have to do with the variance(s), it appears that there is nothing special here: if such a condition holds, who cares if we add or subtract the RV's themselves?

But there is a naive question one can pose, looking at the left-hand side of $(1)$: What is the expected value of $S_n$?

Assume $n=\text {even}$. Then $E(S_n) = 0$. Assume $n = \text {odd}$. Then $E(S_n) = \mu$, where $\mu$ is the common mean. So if the common mean of the RV's is $\mu =0$, then $E(S_n) = 0\;\forall n$, and the CLT will hold for our $S_n$, if it also holds for the "add all" sum (and let's say it does, having finite common variance $\sigma^2$).

Assume now that the common mean is not zero, $\mu \neq 0$. In such a case the sequence of expected values for the $S_n$, $\{E(S_n)\}$, starting from $n=1$ is $\{\mu, 0, \mu, 0,...\}$, or $\mu\cdot \{1, 0, 1, 0,...\}$. Hmm, this is an oscillating sequence. Does this create a problem for the CLT?

The CLT determines the distribution of deviations around the mean. So now the question could be "Deviations around which mean? Zero or $\mu$?"

But, since, exactly, the CLT makes a statement about "deviations from the mean", it doesn't matter which mean is it, as long as we consider deviations from it.

So yes, the CLT holds.

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