Results 1 to 3 of 3

Math Help - exponential distribution question

  1. #1
    Senior Member
    Joined
    Jan 2008
    From
    Montreal
    Posts
    311
    Awards
    1

    exponential distribution question

    Let Y be an exponentially distributed random variable with mean \beta. Define a random variable X in the following way: X=k if k-1 \leq Y \leq k \ \ \mbox{for} \ \ k =1,2,...,n

    a) Find P(X=k) \ \ \forall \ \ k=1,2,...,n

    b) Show that your answer to part a) can be written as:

    P(X=k) =(e^{-\frac{1}{\beta}})^{k-1}(1-e^{-\frac{1}{\beta}}) \ \ \forall \ \ k =1,2,...,n

    I would think for a) it would be \int_{k-1}^{k} \frac{1}{\beta} e^{-\frac{x}{\beta}} dy = -e^{-\frac{x}{\beta}} \bigg{|}^{k}_{k-1} = -e^{-\frac{k}{\beta}}+e^{-\frac{k-1}{\beta}}, but this doesn't seem right since k is not continuous.

    and for b) I would think that you have to manipulate the function you got in a) to get what is shown, but am clueless on how to do so. Any help would be greatly appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Your answer to a) is correct.

    Besides, this will be confirmed by b). You need to expand (e^{-\frac{1}{\beta}})^{k-1}(1-e^{-\frac{1}{\beta}}) (into a sum of two terms) and use the properties e^ae^b=e^{a+b} and \left(e^a\right)^b=e^{ab}, in order to write it like your result in a).
    (And if you can do that, then you can use the same computations in the reverse order to go from the result a) to the expected formula, if you prefer)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by lllll View Post
    Let Y be an exponentially distributed random variable with mean \beta. Define a random variable X in the following way: X=k if k-1 \leq Y \leq k \ \ \mbox{for} \ \ k =1,2,...,n

    a) Find P(X=k) \ \ \forall \ \ k=1,2,...,n

    b) Show that your answer to part a) can be written as:

    P(X=k) =(e^{-\frac{1}{\beta}})^{k-1}(1-e^{-\frac{1}{\beta}}) \ \ \forall \ \ k =1,2,...,n

    I would think for a) it would be \int_{k-1}^{k} \frac{1}{\beta} e^{-\frac{x}{\beta}} dy = -e^{-\frac{x}{\beta}} \bigg{|}^{k}_{k-1} = -e^{-\frac{k}{\beta}}+e^{-\frac{k-1}{\beta}}, but this doesn't seem right since k is not continuous.

    and for b) I would think that you have to manipulate the function you got in a) to get what is shown, but am clueless on how to do so. Any help would be greatly appreciated.
    Your answer to a) is correct. Why does the fact that k is discrete bother you?

    b) Note that e^{-(k-1)/\beta} - e^{-k/\beta} = e^{-(k-1)/\beta} - e^{-k/\beta} \, e^{1/\beta} \, e^{-1/\beta}

    = e^{-(k-1)/\beta} - e^{-(k-1)/\beta} \, e^{-1/\beta} = e^{-(k-1)/\beta} \left( 1 -  e^{-1/\beta}\right).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Exponential Distribution Question
    Posted in the Statistics Forum
    Replies: 5
    Last Post: November 5th 2011, 02:20 AM
  2. Exponential Distribution question
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: July 9th 2010, 10:22 PM
  3. Exponential Distribution Question
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: October 28th 2009, 12:04 PM
  4. Quick question on exponential distribution
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: December 15th 2008, 08:16 AM
  5. exponential distribution question
    Posted in the Advanced Statistics Forum
    Replies: 6
    Last Post: September 23rd 2008, 12:15 AM

Search Tags


/mathhelpforum @mathhelpforum