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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1For $t=1$, $X$ and $Y$ which follow the exponential distribution of parameter $1$, the LHS is $1-2e^{-1}$ whereas the RHS is $1-2e^{-1}+e^{-2}$. But if you substitute $t$ by $t/2$ in the RHS it's correct. – 2012-02-13
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0Thanks for your answer. But is it possible for a general distribution function. Here, I mean that the $X$ and $Y$ have common distribution. – 2012-02-13
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1I didn't see your edit, in this case $\{X+Y
since $X$ and $Y$ are non-negative and take the probabilities. – 2012-02-13 -
1To conclude on @Davide's solution, the result uses the fact that $X$ and $Y$ are both almost surely nonnegative and that they are independent, but not that they have the same distribution. – 2012-02-13
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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