Suppose I have the the sample mean from an iid sample $\bar{X}_n = \sum {X_i \over n}$ and define $\bar{E_n} = {2 \over n} \sum_i X_{2i}$.

Now I would like to show that given convergence of $\sqrt{n}(\bar{X}_n-\mu) \rightarrow_{d} F $, then $\sqrt{n \over 2}(\bar{E}_n-\mu) \rightarrow_{d} F $. Now I suppose I could argument somewhat heuristically saying that the even sample points compose an eqaully iid sample, that amounts to $n \over 2$ observations and the result would follow.

On the other hand, I was thinking of applying the delta method, by making a smart transformation. That is, doing the taylor expansion of the transformations and etc, as if deriving the delta method. The one I tried was $\bar{E_n} = 2\bar{X}_n - \bar{O_n}$, where $\bar{O_n}$ is the odd counterpart. In the end, I reached: $\sqrt{n}(\bar{E}_n-\mu) + \sqrt{n}(\bar{O}_n-\mu) \rightarrow_{d} 2F$. I suppose my transformation is not appropriate since it is also a function of $\bar{O_n}$ is also a function of $\bar{X_n}$.

Is such an approach valid or possible? Also, I cant see why the Delta Theorem is restricted to normal distribution. I suppose I could do equally same taylor expansions for any transformations on any convergent distributions, couldn't I?

Thanks in advance.


For simplicity, I assume that $\mu$ is the expected value of each $X_i$, and I will denote the variance by $\sigma^2$.

Another way to express $\bar E_n$ is to write it using indicator functions

$$\bar E_n = \frac 2n\sum_{i=1}^n X_i\cdot I_{\{i=even\}}$$

In this way it becomes clear that as $n$ will start to increase, the number of random variables in the collection does not increase equally. Now consider that the expected value of $\bar E_{n}$ is not $\mu$, rather, it fluctuates in one side: if, say, $n=10$ the expected value is $\mu$. When $n=11$ the expected value is $\frac {10}{11}\mu$. When $n=12$ the expected value becomes again $\mu$. When $n=13$ the expected value is $\frac {12}{13}\mu$, etc. In general

$$\mathbb E(\bar E_n) = I_{\{n=even\}}\cdot \mu + \left[1-I_{\{n=even\}}\right]\cdot \frac {n-1}{n}\mu$$

$$=\frac {n-1}{n}\mu + I_{\{n=even\}}\cdot \frac 1n \mu = \left(1-\frac {I_{\{n=odd\}}}{n}\right)\mu$$

Analogously one can find that

$$\operatorname{Var}(\bar E_n) = 2\left(1-\frac {I_{\{n=odd\}}}{n}\right)\frac {\sigma^2}{n}$$

Therefore setting $Z_n\equiv \sqrt{n \over 2}(\bar{E}_n-\mu)$ we have that

$$\mathbb E(Z_n) = -\sqrt{n \over 2}\frac {I_{\{n=odd\}}}{n}\mu,\;\; \operatorname{Var}(Z_n) = \left(1-\frac {I_{\{n=odd\}}}{n}\right)\sigma^2$$

and so

$$\lim_{n\rightarrow \infty}\mathbb E(Z_n) = 0,\;\; \lim_{n\rightarrow \infty}\operatorname{Var}(Z_n) = \sigma^2$$

These are finite, so they will also be the moments of the limiting distribution of $Z_n$ (if it exists), and equal to the moments of $F$ (where $F$ is the normal distribution since it is the limiting distribution of a scaled and centered sample mean from an i.i.d sample). Is it possible that $Z_n$ will not converge? No (why?). Is it possible that $Z_n$ will converge to another distribution than the normal? No because it is an affine transformation of a sample mean of an i.i.d sample. Since moreover the normal is completely characterized by its mean and variance, we arrive at $Z_n \rightarrow_d F$ (the same $F$).

  • $\begingroup$ The reason $Z_n$ must converge is its finite variance? $\endgroup$ – user191919 Feb 2 '14 at 10:56
  • $\begingroup$ Yes, finite, and non-zero, to converge to a non-degenerate r.v. $\endgroup$ – Alecos Papadopoulos Feb 2 '14 at 11:18

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