3
$\begingroup$

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

  • 0
    Many thanks for your conclusion, Didier Piau2012-02-13

2 Answers 2

4

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.

2

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

  • 0
    Hi Duy Son, Thank alot for your answer2012-03-08