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If we know $X_n \rightarrow_p Y$ and $X_n \rightarrow Z\space a.s.$ can we say that $P(Y=Z)=1$

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up vote 1 down vote accepted

Yes. Because $\forall \epsilon > 0$ $$P(|Y-Z| > \epsilon) = P(|Y-X_n + X_n - Z| > \epsilon) $$ $$\leq P(|X_n-Z|+|X_n - Y| > \epsilon)$$ $$\leq P(|X_n-Z| > \epsilon/2) + P(|X_n-Y|>\epsilon/2) \quad (*)$$ Take limits an $n\rightarrow \infty$ to get $$P(|Y-Z| > \epsilon)= 0 \quad \forall \epsilon > 0$$

This implies $P(|Y-Z| = 0) = 1$.

Proof of $(*)$ $$P(|X_n-Z|+|X_n - Y| > \epsilon) = P(|X_n-Z|+|X_n - Y| > \epsilon, |X_n -Y| > \epsilon/2) + P(|X_n-Z|+|X_n - Y| > \epsilon,|X_n -Y| \leq \epsilon/2)$$ $$\leq P(|X_n-Y|>\epsilon/2) + P(|X_n-Z| > \epsilon/2)$$ The last step follows as : $a+b >c, b< c/2 \Rightarrow a> c/2$.

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Thanks! SO, can we say Y=Z a.s.? – BigMike Dec 8 '12 at 7:44
Yes. Thats what $P(|Y-Z| = 0) = 1$ implies. – Gautam Shenoy Dec 8 '12 at 7:46
Additionally, if that epsilon logic is bothering you, you can replace it with $1/m$ and argue with that also. – Gautam Shenoy Dec 8 '12 at 7:48
Also, here we basically use the fact that $X\rightarrow Y\space a.s.$ implies $X \rightarrow_p Y$ correct? – BigMike Dec 8 '12 at 7:52
Exactly. Convergence in almost sure implies convergence in probability. – Gautam Shenoy Dec 8 '12 at 8:37

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