Could you please help me to prove the inequality probability as follows: $\Pr\{X+Y
How can I prove this inequation \Pr\{X+Y
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statistics
probability-theory
inequality
probability-distributions
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0Many thanks for your conclusion, Didier Piau – 2012-02-13
2 Answers
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If $X(\omega)+Y(\omega)
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This is my answer for your question.
Since X and Y are independent, so we have $\mathbb{P}(X+Y< t)=\int_{0}^{t}\mathbb{P}(Y< t-u)\mathbb{P}(X\in du).$
On the other hand, $\{Y< t-u\}\subset\{Y< t\}$ this implies $\mathbb{P}(Y
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0Hi Duy Son, Thank alot for your answer – 2012-03-08