# Equality of two limits of r.v.

considering a sequence of real-valued r.v. $(X_n)$ convergent to $X$ in probability. Moreover we look at a sqeuence of r.v. $(Y_n)$, where $Y_n\in\operatorname{conv}(X_n,X_{n+1},\dots)$ and we suppose $Y_n\to Y$ $P$-a.s. $\operatorname{conv}$ denotes the convex hull, i.e. $\operatorname{conv}((X_k)_{k\ge n})=\{\sum_{k=n}^\infty \lambda_k X_k| \sum_{k=n}^\infty \lambda_k=1,\forall k\ge n:\lambda_k\ge 0,\lambda_k\not=0 \mbox{ for finitely many } \}$ I want conclude that $X=Y$ a.s. Since $P$-a.s. convergence implies convergence in probability, we have $Y_n\to Y$ in probability. Somehow I need that $Y_n$ converge also to $X$. Why is this true?

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what is conv$(X_n,X_{n+1},...)$? –  roger Apr 23 '13 at 12:32
@roger: $\mathrm{conv}$ is probably the convex hull. –  Stefan Hansen Apr 23 '13 at 12:36
Ah ok I see, thanks –  roger Apr 23 '13 at 12:38
@StefanHansen Indeed. I added this to my question –  user20869 Apr 23 '13 at 12:43
@roger $\lambda$ should be deterministic –  user20869 Apr 23 '13 at 13:13

Let $\Lambda$ denote the set of non-negative sequences $\lambda = ( \lambda_0,\lambda_1, \lambda_2,... )$ such that $\sum_{k=0}^\infty \lambda_k = 1$. I want to show that $$\Phi_n: \lambda \mapsto \sum_{k=0}^\infty \lambda_k X_{n + k}$$ satisfies $$\sup_{\lambda \in \Lambda} \vert \Phi_n(\lambda) - X \vert \overset{\mathbb P}{\longrightarrow} 0.$$ For all $\epsilon >0$, $$\mathbb P \left( \sup_{\lambda \in \Lambda} \vert \Phi_n(\lambda) - X \vert > \epsilon \right) = \mathbb P \left( \sup_{\lambda \in \Lambda} \vert \sum_{k=0}^\infty \lambda_k (X_{n + k} - X) \vert > \epsilon \right) \leq \mathbb P \left( \sup_{k \geq n} \vert X_{ k} - X \vert > \epsilon \right)$$ which converges to zero by assumption. So $$\sup_{\lambda \in \Lambda} \vert \Phi_n(\lambda) - X \vert \overset{\mathbb P}{\longrightarrow} 0,$$ and since $Y_n = \Phi_n(\lambda_n)$ for some sequence $\lambda_n \in \Lambda$, $Y_n$ converges in probability to $X$.