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$P(X_n \leq X, |X_n - X|< \epsilon) \leq P(X \leq X + \epsilon)$ because $X_n - \epsilon < X < X_n + \epsilon < X + \epsilon$

For the right side of the equality...since we are not restricting X to be greater than $X_n - \epsilon$, can't the right side be less sometimes? Why is this statement true?

Thanks in advance

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$ P(X \leq X + \epsilon) = 1 $. – Haskell Curry Jan 31 '13 at 9:28
up vote 2 down vote accepted

If $\varepsilon$ is assumed to be non-negative then $$ \{X\leq X+\varepsilon\}=\Omega $$ and hence $P(X\leq X+\varepsilon)=1$ and so the inequality is trivial.

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Can we determine what $\Omega$ should be on our own? – user58289 Jan 31 '13 at 9:34
$\Omega$ is the set (sample space) of your probability space. $(\Omega,\mathcal{F},P)$ is pretty standard notation for a probability space: $\Omega$ is a set, $\mathcal{F}$ is a sigma-field on $\Omega$ and $P$ is a probability measure on $\mathcal{F}$. The point is that $\{X\leq X+\varepsilon\}$ is everything. There is not a sample point ($\omega\in\Omega$) such that it is false. – Stefan Hansen Jan 31 '13 at 9:37
Yes I know what $\Omega$ represents, but they didn't tell us what the space was. So can we just choose $\Omega$ to equal $\{ X \leq X + \epsilon \}$ just so the statement makes sense? – user58289 Jan 31 '13 at 9:39
No. The notation $\{X\leq X+\varepsilon\}$ really means the set $\{\omega\in\Omega: X(\omega)\leq X(\omega)+\varepsilon\}$. Now, the condition here is trivially satisfied (given that $\varepsilon\geq 0$) and hence the set is the whole of $\Omega$. – Stefan Hansen Jan 31 '13 at 9:41

Maybe you wanted $P(X_n≤X,|X_n−X|<\varepsilon)≤P(X≤X_n+\varepsilon)$? This is a little bit less trivial, since the right-hand-side might be smaller than $1$, and also true. Because any $\omega\in\Omega$ that satisfies $X_n(\omega)−\varepsilon<X(\omega)<X_n(\omega)+\varepsilon<X(\omega)+\varepsilon$ will also satisfy $X(\omega)≤X_n(\omega)+\varepsilon$, so that


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