Results 1 to 7 of 7

Math Help - probability

  1. #1
    Senior Member Sampras's Avatar
    Joined
    May 2009
    Posts
    301

    probability

    How do you show  E(X) = \int_{0}^{\infty} P(X > \lambda) \ d \lambda ? We know that  E(X) = \int_{0}^{\infty} \lambda \ p(\lambda) \ d \lambda . In other words,  P(X > \lambda) controls all the moments.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Oct 2009
    Posts
    340
    Try integration by parts, noting that P(X > \lambda) = 1 - F(\lambda) where F is the CDF. Also, I think you are missing an assumption on the support of X.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2008
    Posts
    148
    Quote Originally Posted by Sampras View Post
    How do you show  E(X) = \int_{0}^{\infty} P(X > \lambda) \ d \lambda ? We know that  E(X) = \int_{0}^{\infty} \lambda \ p(\lambda) \ d \lambda . In other words,  P(X > \lambda) controls all the moments.
    That statement isn't true unless  \lambda is a nonnegative random variable.

    Did you really mean to prove the following:

     E(X) = \int_{0}^{\infty} P(X > \lambda) \ d \lambda - \int_{0}^{\infty} P(X \leq  -\lambda) \ d \lambda
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Sampras's Avatar
    Joined
    May 2009
    Posts
    301
     X is non-negative.

    So we know that  E(X) = \int_{0}^{\infty} \lambda p(\lambda) \ d \lambda . Let  u = \lambda and  dv = p(\lambda) \ d \lambda . Then  du = 1 and  v = P(X \leq \lambda) . So

     E(X) = \lambda P(X \leq \lambda)- P(X \leq \lambda)

     = \lambda P(X \leq \lambda)- E(X) = \int_{0}^{\infty} P(X \leq \lambda) \ d \lambda

     = \lambda P(X \leq \lambda)- E(X) = \int_{0}^{\infty} 1-P(X > \lambda) \ d \lambda

     = \lambda P(X \leq \lambda)- E(X) = \lambda- \int_{0}^{\infty} P(X > \lambda) \ d \lambda


    So we have:

     \int_{0}^{\infty} P(X > \lambda) \ d \lambda = E(X)-\lambda [\underbrace{P(X \leq \lambda)-1}_{0}]

    and so the result follows. Is this correct?

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Sampras's Avatar
    Joined
    May 2009
    Posts
    301
    Or more generally, one can show that  E(X^p) = p \int_{0}^{\infty} \lambda^{p-1} P(X > \lambda) \ d \lambda using induction on  p ? One could use Markov's inequality:  P(X \geq \lambda) \leq \frac{E(X)}{\lambda} ?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Jun 2008
    Posts
    148
    Quote Originally Posted by Sampras View Post
     X is non-negative.

    So we know that  E(X) = \int_{0}^{\infty} \lambda p(\lambda) \ d \lambda . Let  u = \lambda and  dv = p(\lambda) \ d \lambda . Then  du = 1 and  v = P(X \leq \lambda) . So

     E(X) = \lambda P(X \leq \lambda)- P(X \leq \lambda)

    Unfortunately integrating by parts won't help much here because were evaluating a definite integral. Your first line would actually be this:

     E(X) = \infty - \int_{0}^{\infty} P(X \leq \lambda)

    which is obviously no good.

    I don't know the answer off the top of my head. I'll post back if I have any useful hints.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1
    Quote Originally Posted by Sampras View Post
    How do you show  E(X) = \int_{0}^{\infty} P(X > \lambda) \ d \lambda ? We know that  E(X) = \int_{0}^{\infty} \lambda \ p(\lambda) \ d \lambda . In other words,  P(X > \lambda) controls all the moments.
    Hint:

    \int_0^\infty P(X > \lambda) \, d\lambda = \int_0^\infty \int_\lambda^\infty f(x) \,dx \, d\lambda

    Now interchange the order of integration in the double integral.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: January 21st 2011, 11:47 AM
  2. Replies: 3
    Last Post: May 29th 2010, 07:29 AM
  3. Replies: 1
    Last Post: February 18th 2010, 01:54 AM
  4. Replies: 3
    Last Post: December 15th 2009, 06:30 AM
  5. Continuous probability - conditional probability
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 1st 2009, 01:21 AM

Search Tags


/mathhelpforum @mathhelpforum