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Could you please help me to prove the inequality probability as follows: $\Pr\{X+Y where $X$ and $Y$ are non-negative independent random variables with common distribution. Many thanks for your helps

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    For $t=1$, $X$ 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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    Thanks 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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    I didn't see your edit, in this case $\{X+Y since $X$ and $Y$ are non-negative and take the probabilities.2012-02-13
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    To 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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    Many thanks for your conclusion, Didier Piau2012-02-13

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If $X(\omega)+Y(\omega), since $X(\omega)\geq 0$ we have $Y(\omega) and since $Y(\omega)\geq 0$ we have $X(\omega) so $\{X+Y and taking the probabilities, thanks to independence $$P(\{X+Y Note that the fact that $X$ and $Y$ have the same distribution wasn't used.

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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 $$\mathbb{P}(Y So we obtain $$\mathbb{P}(X+Y< t)\leq\mathbb{P}(Y< t)\int_{0}^{t}\mathbb{P}(X\in du)=\mathbb{P}(Y< t)\mathbb{P}(X< t).$$

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    Hi Duy Son, Thank alot for your answer2012-03-08