Probability Measures, and inequalities

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Let X and Y be random variables defined on the probability space $(Omega, mathcal{S}, mathcal{P})$.

I have:

(i) $mathcal{P}(|X+Y| >epsilon) leq mathcal{P}(X>epsilon/2) + mathcal{P}(Y>epsilon/2) $

(ii) $mathcal{P}(|X| >epsilon, |Y|> epsilon)= c$ implies $ mathcal{P}(|X+Y|>2epsilon) =c$

(The real problems are more complicated, I'm just trying to simplify the part I am confused. )

I'm getting caught up in combining and breaking a part these probability measures. I feel like some of these relationships have to do with the triangle inequality. Could use some help understanding the intuition.

edited Nov 18 at 4:11

asked Nov 18 at 4:05

kpr62

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up vote
0
down vote

favorite

Let X and Y be random variables defined on the probability space $(Omega, mathcal{S}, mathcal{P})$.

I have:

(i) $mathcal{P}(|X+Y| >epsilon) leq mathcal{P}(X>epsilon/2) + mathcal{P}(Y>epsilon/2) $

(ii) $mathcal{P}(|X| >epsilon, |Y|> epsilon)= c$ implies $ mathcal{P}(|X+Y|>2epsilon) =c$

(The real problems are more complicated, I'm just trying to simplify the part I am confused. )

edited Nov 18 at 4:11

asked Nov 18 at 4:05

kpr62

204

add a comment |

up vote
0
down vote

favorite

Let X and Y be random variables defined on the probability space $(Omega, mathcal{S}, mathcal{P})$.

I have:

(i) $mathcal{P}(|X+Y| >epsilon) leq mathcal{P}(X>epsilon/2) + mathcal{P}(Y>epsilon/2) $

(ii) $mathcal{P}(|X| >epsilon, |Y|> epsilon)= c$ implies $ mathcal{P}(|X+Y|>2epsilon) =c$

(The real problems are more complicated, I'm just trying to simplify the part I am confused. )

edited Nov 18 at 4:11

asked Nov 18 at 4:05

kpr62

204

Let X and Y be random variables defined on the probability space $(Omega, mathcal{S}, mathcal{P})$.

I have:

(i) $mathcal{P}(|X+Y| >epsilon) leq mathcal{P}(X>epsilon/2) + mathcal{P}(Y>epsilon/2) $

(ii) $mathcal{P}(|X| >epsilon, |Y|> epsilon)= c$ implies $ mathcal{P}(|X+Y|>2epsilon) =c$

(The real problems are more complicated, I'm just trying to simplify the part I am confused. )

probability-theory measure-theory probability-distributions

edited Nov 18 at 4:11

asked Nov 18 at 4:05

kpr62

204

edited Nov 18 at 4:11

asked Nov 18 at 4:05

kpr62

204

edited Nov 18 at 4:11

asked Nov 18 at 4:05

kpr62

204

asked Nov 18 at 4:05

kpr62

204

asked Nov 18 at 4:05

kpr62

204

add a comment |

1 Answer
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2) $X=-Y=2epsilon $ is a counterexample.

1) follows from ${|X+Y| >epsilon} subset {|X| >epsilon /2} cup {|Y| >epsilon /2} $

edited Nov 18 at 6:00

kpr62

204

answered Nov 18 at 5:44

Kavi Rama Murthy

43.6k31751

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1 Answer
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1 Answer
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up vote
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down vote

2) $X=-Y=2epsilon $ is a counterexample.

1) follows from ${|X+Y| >epsilon} subset {|X| >epsilon /2} cup {|Y| >epsilon /2} $

edited Nov 18 at 6:00

kpr62

204

answered Nov 18 at 5:44

Kavi Rama Murthy

43.6k31751

add a comment |

up vote
0
down vote

2) $X=-Y=2epsilon $ is a counterexample.

1) follows from ${|X+Y| >epsilon} subset {|X| >epsilon /2} cup {|Y| >epsilon /2} $

edited Nov 18 at 6:00

kpr62

204

answered Nov 18 at 5:44

Kavi Rama Murthy

43.6k31751

add a comment |

up vote
0
down vote

2) $X=-Y=2epsilon $ is a counterexample.

1) follows from ${|X+Y| >epsilon} subset {|X| >epsilon /2} cup {|Y| >epsilon /2} $

edited Nov 18 at 6:00

kpr62

204

answered Nov 18 at 5:44

Kavi Rama Murthy

43.6k31751

2) $X=-Y=2epsilon $ is a counterexample.

1) follows from ${|X+Y| >epsilon} subset {|X| >epsilon /2} cup {|Y| >epsilon /2} $

edited Nov 18 at 6:00

kpr62

204

answered Nov 18 at 5:44

Kavi Rama Murthy

43.6k31751

edited Nov 18 at 6:00

kpr62

204

edited Nov 18 at 6:00

kpr62

204

edited Nov 18 at 6:00

kpr62

204

answered Nov 18 at 5:44

Kavi Rama Murthy

43.6k31751

answered Nov 18 at 5:44

Kavi Rama Murthy

43.6k31751

answered Nov 18 at 5:44

Kavi Rama Murthy

43.6k31751

add a comment |

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