We know that $X\sim \exp(1),Y\sim \exp(2)$ and they are independent.
What is $P(Y>X)$?
exp=Exponential...
Thank you!
Asked
Active
Viewed 73 times
-1
CS1
- 2,047
-
Unfortunately this is ambiguous. Sometimes $\mathrm{Exp}(2)$ means an exponential distribution with expected value $2$, so that $\Pr(Y>y) = e^{-y/2}$ for $y\ge0$, and sometimes it means the rate is $2$, so that $\Pr(Y>y)=e^{-2y}$ for $y\ge0$ (and then the expected value would be $1/2$). ${}\qquad{}$ – Michael Hardy Jul 15 '14 at 17:55
-
I suspect that the reason someone down-voted this question is that you haven't included any thoughts on how to answer it. – Michael Hardy Jul 15 '14 at 17:56
-
I edit it, now it's clear? – CS1 Jul 15 '14 at 17:58
-
No, it's not. Either way, it's an exponential distribution. – Michael Hardy Jul 15 '14 at 17:58
-
So did you mean the expected value is $2$? – Michael Hardy Jul 15 '14 at 17:59
-
$f_X(X)=\lambda e^{-\lambda x}$ for $x>0$ @MichaelHardy – CS1 Jul 15 '14 at 18:01
-
This may help http://math.stackexchange.com/questions/115022/pdf-of-the-difference-of-two-exponentially-distributed-random-variables – Kamster Jul 15 '14 at 18:03
-
It don't :-( @user159813, I'm confused from this... – CS1 Jul 15 '14 at 18:04
-
1If you know pdf, of Y-X, than you can find answer of P(Y>X)=P(Y-X>0) and this may help – Kamster Jul 15 '14 at 18:07
-
@user159813, can you help me with the final answer please? I think that I need more hints. – CS1 Jul 15 '14 at 18:11
1 Answers
1
From this previous question(pdf of the difference of two exponentially distributed random variables) they derive for $Y\sim Exponential(\lambda)$ and $X\sim Exponential(\mu)$ letting $Z=X-Y$ a CDF of $Z$ $$P(X<Y)=P(X-Y<0)=P(Z<0)=F_{Z}(0)=1-\frac{\lambda}{\lambda+\mu}e^{-\mu (0)}=1-\frac{\lambda}{\lambda+\mu}=\frac{\mu}{\lambda+\mu}$$
in your example $\lambda=2$ and $\mu=1$ so we have $P(Y>X)=\dfrac{1}{3}$
If a theoretical method seems out of your grasp, You can always simulate large sample and get rough estimate of probability (guaranteed to be close by Law of Large Numbers) here is some R code that could estimate it
X=rexp(10000,1)
Y=rexp(10000,2)
Z=Y-X
positiveZ=Z[Z>0]
prob=length(positiveZ)/length(Z)
prob
PS Thanks to Robert Isarel and Andre Nicolas for their solution to the CDF
-
-
going off link looking at Andres argument, we look at joint distribution of X an Y (and since independent will just be product of distribution) then integrate at region below line $y-x=0$ – Kamster Jul 15 '14 at 18:41
-
-
1oh that rexp(n,1) just means simulate "n" amount of samples form distribution Exp(1), to get a good estimate you have to use a large sample size so I arbitrarily chose 10,000 – Kamster Jul 15 '14 at 19:32