# Same random sequence converge jointly in distribution then converge in probability

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?)

## 1 Answer

$$(X_n,X_n) \overset {p} {\to} (X,X)$$ 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.