If $(X_{n,k})_{n,k\geq 1}$ is a doubly-indexed sequence of random variables such that for every $n$, $X_{n,k}$ converges in law to a Gaussian random variable $X^*_n$ as $k \to \infty$, ($X_{n,k} \to^d X^*_n$) and the sequence $(X^*_n)_{n\geq 1}$ itself converges in law to a Gaussian random variable $X$, ($X^*_n\to^d X$), is it true that there is a convergence in law "along the diagonal" in the sense that $X_{n,n} \to^d X$ as $n\to \infty$ ?
Diagonal convergence in law
2
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sequences-and-series
probability-theory
convergence
random-variables
weak-convergence
1 Answers
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Unfortunately, the diagonal variables $X_{n,n}$ can have any distribution and not converge to anything. Ane example could be the following: $$ X_{n,k}=\begin{cases} Y,&\text{if $n\ne k$},\\ Y+n,&\text{if $n=k$}, \end{cases} $$ where $Y\sim\mathcal N(0,1)$. So, $Y_{n,k}$ converges in distribution to $Y_{n}\equiv Y$ (or even almost surely) if $k\to\infty$, and $Y_n$ converges as $n\to\infty$ as well. But the diagonal sequence doesn't converge at all.