Results 1 to 3 of 3

Math Help - Calculating variance of u_n - u_2n

  1. #1
    Member
    Joined
    Jun 2008
    Posts
    148

    Calculating variance of u_n - u_2n

    Let u_n = (sum_(k=1,n) X_k)/n where X_i is an numerical observation of a random variable with density function f.

    Similarily let u_2n = (sum_(k=1,2n) X_k)/2n where again X_i is an observation of a random variable with same density function f

    Determine the value of E( (u_n - u_2n)^2 ) where 'E' represents the expected value.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Let's assume that X_i and X_j are independent \forall \ i \neq j. If it is not, tell me and throw my solution in the bin

    This implies that \text{cov}(X_i,X_j)=0 (it will be used later)


    u_n=\frac{\sum_{k=1}^n X_k}{n}

    u_{2n}=\frac{\sum_{k=1}^{2n} X_k}{2n}

    Therefore u_n-u_{2n}=\frac{\sum_{k=1}^n X_k}{n}-\frac{\sum_{k=1}^{2n} X_k}{2n}=\frac{\sum_{k=1}^n 2X_k-\sum_{k=1}^{2n} X_k}{2n}

    u_n-u_{2n}=\frac{1}{2n} \left(\sum_{k=1}^n X_k-\sum_{k=n+1}^{2n} X_k\right)


    -----------------------------
    Properties of the variance :

    \text{var}(aX)=a^2 \text{var}(X), where a is a constant.

    \text{var}(X+Y)=\text{var}(X)+\text{var}(Y)+2 \text{cov}(X,Y)
    -----------------------------

    Since the covariance between any X_i and X_j is 0, we can write :


    \text{var}(u_n-u_{2n})=\frac{1}{4n^2} \cdot \left(\sum_{k=1}^{n} \text{var}(X_k)+\sum_{k=n+1}^{2n} (-1)^2 \text{var}(X_k)\right)

    \text{var}(u_n-u_{2n})=\frac{1}{4n^2} \cdot \sum_{k=1}^{2n} \text{var}(X_k)=\frac{1}{4n^2} \cdot \left(2n \cdot \text{var}(X_i)\right)

    \boxed{\text{var}(u_n-u_{2n})=\frac{\text{var}(X_i)}{2n}}



    I hope I have not mistaken somewhere...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2008
    Posts
    148
    Moo

    Yes there is a mistake somewhere but its my fault for not clarifying in my initial thread.

    Every single one of the X_i in the second sum (from 1 to 2n ) is different from those in the first sum of X_i (ie its not just the second half from n + 1 to 2n which are different).

    I hope this helps.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Question about calculating Variance
    Posted in the Statistics Forum
    Replies: 3
    Last Post: August 13th 2011, 07:02 PM
  2. Calculating Variance - Poisson process
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: March 24th 2011, 02:13 PM
  3. Ways of calculating Mean and Variance
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: October 18th 2009, 04:25 AM
  4. Calculating Variance
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 6th 2008, 06:35 PM
  5. Calculating a Weighted Average and its Variance
    Posted in the Advanced Statistics Forum
    Replies: 7
    Last Post: August 29th 2007, 02:27 AM

Search Tags


/mathhelpforum @mathhelpforum