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

Any hints/answers would help!!

  • For the first, it is $e^{-a^2}$. – André Nicolas Jul 16 '13 at 7:08
  • @Did Is (a) correct now?? – Natalie Jul 16 '13 at 7:15
  • Is (a) changed? Not that I can see. – Did Jul 16 '13 at 7:16
  • yes @Did. It has changed and I have checked the definition. – Natalie Jul 16 '13 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. – markovchain Jul 16 '13 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 $\{X>a\}$.

ad b) and c): The link below discusses the same problem. pdf of the difference of two exponentially distributed random variables

$$ \mathbb P(X>Y) = \frac b{a+b}ae^{-ax} $$

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