# Exponetial Distribution Question.

• Sep 7th 2012, 05:01 AM
PhoenixFC
Exponential Distribution Question.
Have been given a challenge question to try and work out for this week, I have gotten part way through it but Im not really sure how to proceed from where I am...

Quote:

Questions reads: "Let X be an exponential random variable with rate parameter λ (lambda) > 0. Suppose it is known that X > k where k is a positive constant. Given (conditional on) this, what is the probability that X > k + x? Consequently, what is the distribution of X given X > k?"

So if I denote P() as a probability,

P(X > k+x) = 1 - P(X < k+x)
= 1 - F(k+x) where this is the CDF of X...

Not sure if I'm on the right track so any help would be much appreciated, Thanks!
• Sep 7th 2012, 06:30 PM
harish21
Re: Exponential Distribution Question.
Quote:

Originally Posted by PhoenixFC
Have been given a challenge question to try and work out for this week, I have gotten part way through it but Im not really sure how to proceed from where I am...

So if I denote P() as a probability,

P(X > k+x) = 1 - P(X < k+x)
= 1 - F(k+x) where this is the CDF of X...

Not sure if I'm on the right track so any help would be much appreciated, Thanks!

you are given that X>k, and you are told to find $P(X>k+x | X>k)$

now use the memoryless property of exponential distribution...
• Sep 7th 2012, 07:17 PM
PhoenixFC
Re: Exponential Distribution Question.
Quote:

Originally Posted by harish21
you are given that X>k, and you are told to find $P(X>k+x | X>k)$

now use the memoryless property of exponential distribution...

Okay thanks that helped a lot, didn't know about that memoryless property.

So in in the end I worked out
$P(X>k+x | X>k)$ = P(X>k+x)/P(X>K) = e^-(λx) = P(X > x)

So for the last part of the question how would I state the distribution?

"Consequently the distribution of X given X>k is exponential with rate λ and mean = 1/λ?"

Not sure if that's what my lecturer is meaning or not haha.
• Sep 7th 2012, 07:38 PM
harish21
Re: Exponential Distribution Question.
Quote:

Originally Posted by PhoenixFC
Okay thanks that helped a lot, didn't know about that memoryless property.

So in in the end I worked out
$P(X>k+x | X>k)$ = P(X>k+x)/P(X>K) = e^-(λx) = P(X > x)

So for the last part of the question how would I state the distribution?

"Consequently the distribution of X given X>k is exponential with rate λ and mean = 1/λ?" No. this is not exponential

Not sure if that's what my lecturer is meaning or not haha.

$P(X>k+x | X>k) = P(X>x) =1-P(X \leq x) = 1-F_X(x)=e^{-\lambda x}$

oh wait..the last part of your question says "what is the distribution of X given X > k?" which means find P(X=x|X>k)
• Sep 7th 2012, 08:01 PM
PhoenixFC
Re: Exponential Distribution Question.
Quote:

Originally Posted by harish21
$P(X>k+x | X>k) = P(X>x) =1-P(X \leq x) = 1-F_X(x)=e^{-\lambda x}$

oh wait..the last part of your question says "what is the distribution of X given X > k?" which means find P(X=x|X>k)

Are you sure...Similar questions in my textbook (without the complication of conditional probability) just state the distribution in words. Why would it not just be exponential.

Also how on earth would you find P(X=x|X>k)?