Results 1 to 15 of 15

Math Help - Cummulative distribution function

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    195
    Thanks
    1

    Cummulative distribution function

    Let X be an exponentially distributed stocastic variable with probability density:

    f(x;b) = (1/b)e^(-1/b), for x larger than or equal to 0, else 0)

    How can I find the cummulative distribution function F(x)?

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by gralla55 View Post
    Let X be an exponentially distributed stocastic variable with probability density:

    f(x;b) = (1/b)e^(-1/b), for x larger than or equal to 0, else 0) Mr F says: Where is the x-dependency in your pdf!?

    How can I find the cummulative distribution function F(x)?

    Thanks!
    You integrate the pdf. Where are you stuck?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2009
    Posts
    195
    Thanks
    1
    Sorry, correct pdf is of course:

    f(x;b) = (1/b)e^(-x/b)

    So if I intergrate with regards to x it should give:

    -1/x * e^(-x/b)

    I think, and then I intergrate with regards to b?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by gralla55 View Post
    Sorry, correct pdf is of course:

    f(x;b) = (1/b)e^(-x/b)

    So if I intergrate with regards to x it should give:

    -1/x * e^(-x/b)

    I think, and then I intergrate with regards to b?
    F(x) = 0 when x < 0 and \int_0^x f(t;b) \, dt for x > 0.
    Last edited by mr fantastic; March 18th 2010 at 02:14 AM. Reason: Fixed a latex typo
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2009
    Posts
    195
    Thanks
    1
    I don't get how "t" is suddently in the picture, but if I continue to intergrate with regards to "b", I get (b/x^2)e^(-x/b)...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by gralla55 View Post
    I don't get how "t" is suddently in the picture, but if I continue to intergrate with regards to "b", I get (b/x^2)e^(-x/b)...
    t is a dummy variable since you have to get a function of x after integrating. Replace x with t in f(x;b) and then integrate wrt t like I said. b is a parameter, not a variable. You do not integrate wrt b.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2009
    Posts
    195
    Thanks
    1
    So, the integral should be -1/t * e^(-t/b)?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    no, you are integrating wrt t
    and t is just a dummy variable
    you cannot integrate x wrt x
    x is a limit in the integration
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Oct 2009
    Posts
    195
    Thanks
    1
    I'm still confused... =(

    f(x;b) = (1/b)e^(-x/b) , repacing x with t

    f(t;b) = (1/b)e^(-t/b), intergrating wrt t

    F(t) = -1/t * e^(-t/b), adding limits?

    F(x) = -1/x * e^(-x/b) - lim x->0 ( -1/x * e^(-0/b) ) ?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by gralla55 View Post
    I'm still confused... =(

    f(x;b) = (1/b)e^(-x/b) , repacing x with t

    f(t;b) = (1/b)e^(-t/b), intergrating wrt t

    F(t) = -1/t * e^(-t/b), adding limits?

    F(x) = -1/x * e^(-x/b) - lim x->0 ( -1/x * e^(-0/b) ) ?
    I don't see what the trouble can possibly be.

    F(x) = \int_0^{x} \frac{1}{b} e^{-t/b} \, dt for x > 0 and you should surely know how to evaluate this integral!
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    Quote Originally Posted by mr fantastic View Post
    I don't see what the trouble can possibly be.

    F(x) = \int_0^{x} \frac{1}{b} e^{-t/b} \, dt for x > 0 and you should surely know how to evaluate this integral!
    I wouldn't be that confident.
    I have a calculus three student who didn't know what the square root of 36 was.
    And I'm confident she couldn't integrate this either.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Oct 2009
    Posts
    195
    Thanks
    1
    Haha, well I do know the square root of 36.

    The anti-derivative of e^kx is (1/k)e^kx. So the antiderivative when k = -1/b should be -b*e^(-x/b). Multiplying with 1/b negates the b and I'm left with just:

    -e^(-x/b)

    All right, for some reason I got the antiderivative wrong the last time, which made evaluating the intergral alot harder than it should have been... so proceeding to evaluate:

    (-e^(-x/b)) - (-e^(0/b) = (-e^(-x/b)) + 1 = 1 - e^(-x/b)

    Is this correct?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by gralla55 View Post
    Haha, well I do know the square root of 36.

    The anti-derivative of e^kx is (1/k)e^kx. So the antiderivative when k = -1/b should be -b*e^(-x/b). Multiplying with 1/b negates the b and I'm left with just:

    -e^(-x/b)

    All right, for some reason I got the antiderivative wrong the last time, which made evaluating the intergral alot harder than it should have been... so proceeding to evaluate:

    (-e^(-x/b)) - (-e^(0/b) = (-e^(-x/b)) + 1 = 1 - e^(-x/b)

    Is this correct?
    (And about time too).
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member
    Joined
    Oct 2009
    Posts
    195
    Thanks
    1
    Finally!

    Now to prove that the exponential distribution is memoryless, that is:

    P(X > x0 + x|X > x0) = P(X > x)

    Which should be simple enough, I'll just have to do something to the left side so that it becomes the right?
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by gralla55 View Post
    Finally!

    Now to prove that the exponential distribution is memoryless, that is:

    P(X > x0 + x|X > x0) = P(X > x)

    Which should be simple enough, I'll just have to do something to the left side so that it becomes the right?
    \Pr(X > x_0 + x | X > x_0) = \frac{\Pr(X > x_0 + x)}{\Pr(X > x_0)}.

    Review Bayes Theorem. Get the required probabilities by integrating the pdf.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding a Cumulative Distribution Function and Density Function
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: November 17th 2011, 09:41 AM
  2. cumulative distribution function using a gamma distribution
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: September 11th 2010, 10:05 AM
  3. Moment generating function and distribution function?
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: January 25th 2009, 02:02 PM
  4. Cumulative distribution function of binomial distribution
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: October 31st 2008, 03:34 PM
  5. single varibale cummulative dist. func.
    Posted in the Advanced Statistics Forum
    Replies: 6
    Last Post: April 13th 2008, 04:13 AM

Search Tags


/mathhelpforum @mathhelpforum