# How can I prove this inequation $\Pr\{X+Y<t\} \le \Pr\{X<t\} \Pr\{Y<t\}$

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

-
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. – Davide Giraudo Feb 13 '12 at 19:23
Thanks for your answer. But is it possible for a general distribution function. Here, I mean that the $X$ and $Y$ have common distribution. – Tran Tam Feb 13 '12 at 19:31
I didn't see your edit, in this case $\{X+Y<t\}\subset \{X<t\}\cap \{Y<t\}$ since $X$ and $Y$ are non-negative and take the probabilities. – Davide Giraudo Feb 13 '12 at 19:39
Thanks a lot for your answer. – Tran Tam Feb 13 '12 at 19:41
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. – Did Feb 13 '12 at 19:42

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

-
Perfect answer, Ngo Bao Chau is a Vietnamese. He has got Field medal because he studied the mathematic in French. Thanks a lot your help. – Tran Tam Feb 13 '12 at 19:51
You're welcome! – Davide Giraudo Feb 13 '12 at 19:54

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

-
Hi Duy Son, Thank alot for your answer – Tran Tam Mar 8 '12 at 12:45