# Is this inequality involving probabilities true?

I'm having trouble with the following inequality, I am not sure whether it is correct:

$$P((X_n-X)^2+(Y_n-Y)^2>\epsilon^2) \stackrel{?}{\le} P(|X_n-X|>\epsilon)+P(|Y_n-Y|>\epsilon)$$

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I think this doesn't always hold. It will be correct if we replace the left hand side by $P((X_n-X)^2+(Y_n-Y)^2 > 2\epsilon^2)$. To prove that inequality, note that $(X_n-X)^2+(Y_n-Y)^2 > 2\epsilon^2$ implies that either $|X-X_n| > \epsilon$ or $|Y-Y_n| > \epsilon$. –  Srivatsan Oct 26 '11 at 4:30

A simple counterexample: Let $X_n, Y_n$ be independent and take the value $1$ w.p. $\frac{1}{2}$ and $-1$ w.p. $\frac{1}{2}$. Let $X$ and $Y$ be identically $0$. Let $\epsilon=1$. Then the left hand side of your inequality takes the value $1$ and the right hand side takes the value $0$.

Tip: When you're trying to prove an inequality like this, it's usually good to change the variables to simplify things. In this case, your inequality holds iff the following holds for all random variables $X'$ and $Y'$ (it doesn't).

$$\mathbb{P}[X'^2+Y'^2>\epsilon^2]\leq \mathbb{P}[|X'|>\epsilon]+\mathbb{P}[|Y'|>\epsilon]$$

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It might help to visualize this. No, it is not universally correct. It depends on the distribution of the variables. The reason for this is because of the strict inequality

$$\{(x,y)\,|\, \|x\|>\epsilon\}\cup \{(x,y)\,|\, \|y\|>\epsilon\} \subsetneq\{(x,y)\, |\, x^2+y^2 > \epsilon^2\}.$$

The inequality will be true if the probability of the symmetric difference is 0.

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Let me reformulate my last comment: the inclusion in your answer is true but does not yield the inequality the OP is interested in, nor its corrected version. –  Did Oct 26 '11 at 6:21