Weak Convergence and Random Index Suppose $X_n$ and $N_n$ are random variables ($N_n$ is integer-valued) such that   $(X_n, N_n/n)$ jointly weakly converges to $(X,1)$. There is NO assumption on independence between $X_n$ and $N_n$. Based on the paper:
Aldous, David J. "Weak convergence of randomly indexed sequences of random variables." Mathematical Proceedings of the Cambridge Philosophical Society. Vol. 83. No. 1. Cambridge University Press, 1978
one needs an additional assumption on $X_n$ to conclude the weak convergence of $X_{N_n}$  to the distribution of $X$. However, suppose one uses Skorokhod representation, so that on the same probability space, we have  $(X_n^*,N_n^*/n)$ converges a.s. to $(X^*, 1)$. Then would not this imply that $X_{N_n*}^*\rightarrow X^*$ a.s. and hence conclude the weak convergence of $X_{N_n}$ to $X$? Where is the mistake?
 A: If we know only that $(X_n, N_n) \overset{d}{=} (X_n^*, N_n^*)$ for every $n$, it does not follow that $X_{N_n^*}^* \overset{d} = X_{N_n}$. The distribution of $X_{N_n}$ depends on the entire joint distribution of $(N_n, X_1, X_2, X_3, \dots)$, which the Skorokhod construction does not necessarily preserve.
So even though $X_{N_n^*}^* \to X^*$ a.s., it does not follow that $X_{N_n} \to X$ in distribution.
For a trivial example, suppose $X_1 = N_2 = 0$, and that $X_2, N_1$ are each "coin flips" that are uniformly distributed on $\{1,2\}$, and that $X_2, N_1$ are independent. Let $X_1^* = N_2^* = 0$, let $X_2^*$ be uniformly distributed on $\{1,2\}$, and let $N_1^* = X_2^*$.  Then we have:

*

*$(X_1, N_1) \overset{d}{=} (X_1^*, N_1^*)$, as both pairs are uniformly distributed on $\{(0,1), (0,2)\}$


*$(X_2, N_2) \overset{d}{=} (X_2^*, N_2^*)$, as both pairs are uniformly distributed on $\{(1,0), (2,0)\}$


*$X_{N_1}$ takes the values $0,1,2$ with probabilities $1/2$, $1/4$, $1/4$ respectively


*$X^*_{N_1^*}$ takes the values $0$ and $2$ with probabilities $1/2$.
