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Sorry for very confusing title, I don't know how to describe the problem precisely.

I have two questions.

(1) Suppose $Xn\overset{p}{\to}X$ then $(X_n,X_n)\overset{d}{\to}(X,X)$.

Conversely,

(2) Suppose $(X_n,X_n)\overset{d}{\to}(X,X)$ for some X, then $X_n$ is cauchy in probability.

I am not very familiar with probability theorem but I do know the definition of convergence in probability and in distribution. However, I don't know what is the meaning of $(X_n,X_n)\overset{d}{\to}(X,X)$. Second, the lecturer quickly metioned that you need inifite dimensions rather than two to imply converge almost surely, do we have similar results for converge almost surely? (Ex: $Xn\overset{a.s.}{\to}X$ then $(X_n,X_n)\overset{p}{\to}(X,X)$. $Xn\overset{a.s.}{\to}X$ then $(X_n,X_n,...)\overset{d}{\to}(X,X,...)$. and conversely with some regularity conditions?)

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$(X_n,X_n) \overset {p} {\to} (X,X)$ means $P(|(X_n,X_n)-(X,X)| >\epsilon) \to 0$ as $ n \to \infty$ for every $\epsilon >0$. Here $|(a,b))|=\sqrt {a^{2}+b^{2}}$. If this property hods then $|X_n -X| > \epsilon$ implies $|(X_n,X_n)-(X,X)| >\epsilon$ so we get $X_n \overset {p} {\to} X$. Converse is also easy if you note that $|(a,a)-(b,b)|=\sqrt 2 |a-b|$.

If $X_n \to X$ almost surely then $(X_n,X_n) \to (X,X)$ almost surely which implies convergence in probability. However, for the infinite sequence $(X_n,X_n,\cdots)$ you have to specify what convergence of sequences means before an answer can be given.

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