0
$\begingroup$

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!!

$\endgroup$
6
  • $\begingroup$ For the first, it is $e^{-a^2}$. $\endgroup$ Jul 16, 2013 at 7:08
  • $\begingroup$ @Did Is (a) correct now?? $\endgroup$
    – Natalie
    Jul 16, 2013 at 7:15
  • $\begingroup$ Is (a) changed? Not that I can see. $\endgroup$
    – Did
    Jul 16, 2013 at 7:16
  • $\begingroup$ yes @Did. It has changed and I have checked the definition. $\endgroup$
    – Natalie
    Jul 16, 2013 at 7:24
  • $\begingroup$ 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. $\endgroup$ Jul 16, 2013 at 7:47

2 Answers 2

1
$\begingroup$

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

$\endgroup$
-1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.