# Exponential Random Variables

QUESTION: Let $X$ and $Y$ be exponentially distributed random variables with parameters $a$ and $b$ respectively. Calculate the following probabilities: (a) $P(X>a)$ (b) $P(X>Y)$ (c) $P(X>Y+1)$ ANSWER: (a) since $X$ is exponentially distributed, we have $P(X>a) = 1 - P(X \leq a) = 1 - (1 - e^{-a^2}) = e^{-a^2}$ . (b) Not sure (c) Not sure

• For the first, it is $e^{-a^2}$. Jul 16, 2013 at 7:08
• @Did Is (a) correct now?? Jul 16, 2013 at 7:15
• Is (a) changed? Not that I can see.
– Did
Jul 16, 2013 at 7:16
• yes @Did. It has changed and I have checked the definition. Jul 16, 2013 at 7:24
• I havent brushed up on this, but for (b) and (c), you are supposed to look at their parameters and compare them. For example, (b) is just asking the question, what is the chance that the exponential RV $X$ is going to be greater than $Y$? This is similar to the question, if the RVs refer to time, what is the chance that $X$ takes longer to complete than $Y$? Hint: All you need is division and addition. Jul 16, 2013 at 7:47

ad a): As noted by others,$P(X>a)=1-P(X\leq a)$, if you have troubles with this use the expected value of the indicator functions on sets like $\{X>a\}$.
$$\mathbb P(X>Y) = \frac b{a+b}ae^{-ax}$$