If the sample mean converges in distribution to F, the mean of the odd and even observations converge as well

Question

could somebody please help with this question:

Consider a random sample of $X_1, X_2, \ldots ,X_n$, such that $E[X_i] = \mu$ and $n $ is even. Define $P_n = \frac{2}{n}\sum_{i}X_{2i}$ and $I_n = \frac{2}{n}\sum_{i}X_{2i-1}$, the average of the even and odd terms in the sample.

Suppose we know that $\sqrt{n}(\tilde{X}_n - \mu) \xrightarrow[]{D} F$ (but we don't know that $F$ is normal). Argue that $\sqrt{\frac{n}{2}}({P}_n - \mu) \xrightarrow[]{D} F$ and $\sqrt{\frac{n}{2}}({I}_n - \mu) \xrightarrow[]{D} F$.

I don't know how to do this. I thought that maybe doing by contradiction could work, by arguing that if $P_n$ does not converge to $F$ and $I_n$ does, the sum of both terms would be $\tilde{X}_n$, and it would not be necessarily true that the some of them should have $F$ as its assymptotic distribution.

Another argument could be through moment generating functions, although I do not quite know how to show that $P_n$, $I_n$ and $\tilde{X}_n$ would have the same ones.

Answer 1

$P_n$ has same distribution as $\frac 2 n \sum\limits_{k=1}^{n/2} X_i$ because $\{X_i\}$ is i.i.d. Hence $\sqrt {\frac 2 n} (P_n-\mu) $ converges to $F$ in distribution, (If a whole sequence converges, so does any subsequence). Similar argument for the second sequence.

Answer 2

This follows trivially from the fact that you are using "random sampling". In the context of an infinite series of random variables, the requirement that the first $n$ values constitutes a "random sample" of the whole means that this sequence is exchangeable, which means that they are IID conditional on the underlying distribution (via the representation theorem). Thus, your starting point is that you have an exchangeable sequence of random variables $X_1, X_2, X_3, ... \sim \text{IID Dist}$ for some distribution. If this is the case then it follows trivially that for any subsequence of indices $0 < t_1 < t_2 < t_3 < ...$ we also have:

$$X_{t_1}, X_{t_2}, X_{t_3}, ... \sim \text{IID Dist}.$$

In particular, for the subsequences of odd and even values we have:

$$X_1, X_3, X_5, ... \sim \text{IID Dist} \quad \quad \quad \text{and} \quad \quad \quad X_2, X_4, X_6, ... \sim \text{IID Dist}.$$

Thus, if the conditions for convergence in distribution hold with respect to the overall sequence of random variables then those same conditional also hold for every subsequence, including the sequences of odd and even values.