Results 1 to 5 of 5

Math Help - E(∑ |X_k|) < ∞ ? How can we prove it?

  1. #1
    Senior Member
    Joined
    Jan 2009
    Posts
    404

    E(∑ |X_k|) < ∞ ? How can we prove it?

    Let |X_k| ≥ 0 be a sequence of random variables.

    If we are given that

    ∑ |X_k| < ∞ and
    k=1

    +∞
    E(∑ |X_k|) < ∞ ,
    k=1

    does this imply that

    k=n
    E(∑ |X_k|) < ∞ ?
    k=1

    An infinite sum is by definition the limit of the sequence of partial sums and intuition seems to suggest that the above is true, but how can we prove it rigorously?

    Note: the approach of my book starts with the following axioms for expecation
    1. X≥0 =>E(X)|≥0
    2. E(cX+dY) = c E(X)+d E(Y)
    3. E(1)=1
    4. If X_1<X_2<...<X_n and lim Xn(ω)=X(ω), then lim E(X_n) = E(X) [same as monotone convergence theorem]

    I have been thinking about this for an hour now and I am pretty confused...can someone please help? I would really appreciate it!
    Last edited by kingwinner; October 16th 2009 at 03:24 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Aug 2009
    Posts
    80
    Quote Originally Posted by kingwinner View Post
    Let |X_k| ≥ 0 be a sequence of random variables.

    If we are given that

    ∑ |X_k| < ∞ and
    k=1

    +∞
    E(∑ |X_k|) < ∞ ,
    k=1

    does this imply that

    k=n
    E(∑ |X_k|) < ∞ ?
    k=1

    An infinite sum is by definition the limit of the sequence of partial sums and intuition seems to suggest that the above is true, but how can we prove it rigorously?
    Since |X_n| >= 0, the sequence of partial sums is increasing. The result follows by the monotone convergence theorem.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    No need of monotone convergence theorem here. Simply say E[\sum_{k=1}^n |X_k|]\leq E[\sum_{k=1}^\infty |X_k|]<\infty, where the first inequality is your first axiom (more explicitly, if X\leq Y then Y-X\geq 0 hence E[Y-X]\geq 0 by 1., i.e. E[X]\leq E[Y], called monotonicity of the expectation)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jan 2009
    Posts
    404
    Quote Originally Posted by Laurent View Post
    No need of monotone convergence theorem here. Simply say E[\sum_{k=1}^n |X_k|]\leq E[\sum_{k=1}^\infty |X_k|]<\infty, where the first inequality is your first axiom (more explicitly, if X\leq Y then Y-X\geq 0 hence E[Y-X]\geq 0 by 1., i.e. E[X]\leq E[Y], called monotonicity of the expectation)
    Hi,

    I follow your second point, but I don't understand the step before it.
    Why is it true that \sum_{k=1}^n |X_k|\leq \sum_{k=1}^\infty |X_k|?
    Here the right side is really defined as a LIMIT...(the result SEEMS obvious here, but how can we JUSTIFY it? i.e. how can we show that the left side is less than or equal to the LIMIT of the right side?)

    Thanks for explaining!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Aug 2009
    Posts
    80
    Quote Originally Posted by kingwinner View Post
    Hi,

    I follow your second point, but I don't understand the step before it.
    Why is it true that \sum_{k=1}^n |X_k|\leq \sum_{k=1}^\infty |X_k|?
    Here the right side is really defined as a LIMIT...(the result SEEMS obvious here, but how can we JUSTIFY it? i.e. how can we show that the left side is less than or equal to the LIMIT of the right side?)

    Thanks for explaining!
    Since each r.v. is nonnegative, the infinite sum |X_(n+1)| + |X_(n+2)| + ... is itself nonnegative. (This is because the partial sums of this series are increasing and nonnegative, and so their supremum cannot be negative.) Therefore |X_1| + |X_2| + ... + |X_n| <= (|X_1| + |X_2| + ... + |X_n|) + (|X_(n+1)| + |X_(n+2)| + ...) = |X_1| + |X_2| + ...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. answer of ∞ - ln∞
    Posted in the Calculus Forum
    Replies: 6
    Last Post: October 31st 2010, 08:31 AM
  2. Replies: 2
    Last Post: September 16th 2010, 06:34 AM
  3. Prove: Suppose |H|<∞. then
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 9th 2009, 12:00 AM
  4. Replies: 5
    Last Post: March 23rd 2009, 08:21 AM
  5. Replies: 2
    Last Post: November 8th 2008, 07:58 AM

Search Tags


/mathhelpforum @mathhelpforum