My Math Forum Exponential Random Variable

 October 26th, 2013, 03:30 PM #1 Newbie   Joined: Oct 2013 Posts: 13 Thanks: 0 Exponential Random Variable Hi, I have a quick question. Let R and S be two independent exponentially distributed random variables with rates ? and ?. How would I compute P{S < t < S + R}? I am a little bit confused because of the variables on either side of the inequalities. I have tried conditioning on both S and R but I am not sure if I'm doing it right here. I can compute something like P{S < R} but the t is throwing me off! Any help is appreciated!
 October 26th, 2013, 06:28 PM #2 Senior Member   Joined: Feb 2013 Posts: 281 Thanks: 0 Re: Exponential Random Variable $P(t)= P\{S
 October 27th, 2013, 10:53 AM #3 Newbie   Joined: Oct 2013 Posts: 13 Thanks: 0 Re: Exponential Random Variable Thank you, I think I understand where that came from. What exactly are the function? Is it as simple as f(R) = L*exp(-LR) (L=lambda) and f(S)* = M*exp(-MS) (M=mu)? With the outer integral, should it be from 0 to infinity? As the exponential random variable isn't defined for below zero. Because doing this gives me M/(L-M)[exp(-Mt)-exp(-Lt)] which I'm not sure is correct because I want to show that M*P{S < t < S + R} = L*P{R < t < R + S}.
 October 28th, 2013, 02:47 PM #4 Newbie   Joined: Oct 2013 Posts: 13 Thanks: 0 Re: Exponential Random Variable Bump, I still can't progress.
 October 28th, 2013, 05:29 PM #5 Senior Member   Joined: Feb 2013 Posts: 281 Thanks: 0 Re: Exponential Random Variable I edited my post a bit. f(S) was not correct, it should be f(S,M). I think your result is okey. You can show this: L*P{S < t < S + R} = M*P{R < t < R + S}. (compare it with yours)
 October 28th, 2013, 07:04 PM #6 Newbie   Joined: Oct 2013 Posts: 13 Thanks: 0 Re: Exponential Random Variable Okay, I guess it could be a typo in the question. Should I get P{S < t < S + R} = M/(L-M)(exp(-Mt)-exp(-Lt))?
 October 29th, 2013, 03:23 AM #7 Senior Member   Joined: Feb 2013 Posts: 281 Thanks: 0 Re: Exponential Random Variable yes
 October 30th, 2013, 11:26 AM #8 Newbie   Joined: Oct 2013 Posts: 13 Thanks: 0 Re: Exponential Random Variable Thank you so much!

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