Results 1 to 9 of 9

Math Help - independently and identically distributed random variables

  1. #1
    Newbie
    Joined
    Nov 2008
    Posts
    14

    independently and identically distributed random variables

    Hi there,

    Assume that Y1, Y2, Y3 and Y4 are independently and identically distributed N(\mu, \sigma^2) random variables.


    how do I show that Y1 + Y2 - Y3 - Y4 and Y1 - Y2 + Y3 - Y4 are independent.


    Thanks
    Casper
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Nov 2008
    Posts
    58
    Thanks
    1
    Check if they are uncorrelated. Uncorrelated normal random variables are independent of each other.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2008
    Posts
    14
    Quote Originally Posted by cl85 View Post
    Check if they are uncorrelated. Uncorrelated normal random variables are independent of each other.
    could you explain a bit more?

    thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Nov 2008
    Posts
    58
    Thanks
    1
    Y_1 + Y_2 - Y_3 - Y_4 and Y_1 - Y_2 + Y_3 -Y_4 are both linear combination of independent normal random variables and are hence normal. To check if two normal random variables are independent, we just need to check if they are uncorrelated, ie, their covariance Cov(Y_1 + Y_2 - Y_3 - Y_4, Y_1 - Y_2 + Y_3 -Y_4) equal 0.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2008
    Posts
    14
    Quote Originally Posted by cl85 View Post
    Y_1 + Y_2 - Y_3 - Y_4 and Y_1 - Y_2 + Y_3 -Y_4 are both linear combination of independent normal random variables and are hence normal. To check if two normal random variables are independent, we just need to check if they are uncorrelated, ie, their covariance Cov(Y_1 + Y_2 - Y_3 - Y_4, Y_1 - Y_2 + Y_3 -Y_4) equal 0.
    Hi there,
    Before I continue, I got a stupid question to ask
    is Y_1 + Y_2 - Y_3 - Y_4 \sim N (0, 4 \sigma^2 ) ? and

    Y_1 - Y_2 + Y_3 -Y_4 \sim N (0, 4 \sigma^2 ) ,too?

    I know how to sum normal distribution but I forgot how to take the difference...

    Thanks
    Casper
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Nov 2008
    Posts
    14
    let

    <br />
W_1 = Y_1 + Y_2 - Y_3 -Y_4 \sim  N ( 0 , 4 \sigma^2 )<br />
    <br />
W_2 = Y_1 - Y_2 + Y_3 -Y_4 \sim  N ( 0 , 4 \sigma^2 )<br />


    when I calculate the

    <br />
Cov( W_1 , W_2 ) = E( W_1, W_2 ) - E( W_1 ) E ( W _ 2)<br />





    I cant find the E( W_1, W_2 )

    how should I proceed? I dont think it's a good way to proceed....

    anyone else got an idea?

    Thanks

    Casper
    Last edited by casperyc; December 3rd 2008 at 01:18 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Nov 2008
    Posts
    58
    Thanks
    1
    Quote Originally Posted by casperyc View Post
    Hi there,
    Before I continue, I got a stupid question to ask
    is Y_1 + Y_2 - Y_3 - Y_4 \sim N (0, 4 \sigma^2 ) ? and

    Y_1 - Y_2 + Y_3 -Y_4 \sim N (0, 4 \sigma^2 ) ,too?

    I know how to sum normal distribution but I forgot how to take the difference...

    Thanks
    Casper
    Yes and Yes, but you do not need to know it to solve for the question.

    I'll show through an example how to proceed
    Cov(Y_1 + Y_2, Y_1 - Y_2)
    = E[(Y_1 + Y_2 - E[Y_1 + Y_2])(Y_1 -Y_2 - E[Y_1 - Y_2])]
    = E[\{(Y_1-E[Y_1]) + (Y_2 -E[Y_2])\}\{(Y_1-E[Y_1])-(Y_2-E[Y_2])\}]
    =Var(Y_1) - Var(Y_2) - 2Cov(Y_1, Y_2)
    The trick here is to always rearrange terms so that the random variable is with its mean.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Nov 2008
    Posts
    14
    Quote Originally Posted by cl85 View Post
    Yes and Yes, but you do not need to know it to solve for the question.

    I'll show through an example how to proceed
    Cov(Y_1 + Y_2, Y_1 - Y_2)
    = E[(Y_1 + Y_2 - E[Y_1 + Y_2])(Y_1 -Y_2 - E[Y_1 - Y_2])]
    = E[\{(Y_1-E[Y_1]) + (Y_2 -E[Y_2])\}\{(Y_1-E[Y_1])-(Y_2-E[Y_2])\}]
    =Var(Y_1) - Var(Y_2) - 2Cov(Y_1, Y_2)
    The trick here is to always rearrange terms so that the random variable is with its mean.
    hmmmmmmmm
    Cov(Y_1 + Y_2, Y_1 - Y_2)=Var(Y_1) - Var(Y_2) - 2Cov(Y_1, Y_2)

    so in my case
    Cov(W_1 + W_2, W_1 - W_2)=Var(W_1) - Var(W_2) - 2Cov(W_1, W_2)
    to find Cov(W_1, W_2) = \dfrac{Var(W_1) - Var(W_2) -  Cov(W_1 + W_2, W_1 - W_2) }{2}

    and now i have to find  Cov(W_1 + W_2, W_1 - W_2) ??

    dose that make it more complicated or I have misunderstood it again?

    Casper
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Nov 2008
    Posts
    58
    Thanks
    1
    No! My example is not to give you a formula to use, it's to show you how to rearrange terms when doing covariance.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Normally distributed random variables
    Posted in the Statistics Forum
    Replies: 3
    Last Post: September 11th 2011, 04:19 AM
  2. Independent and identically distributed Random Variables
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: April 29th 2010, 08:35 PM
  3. Jointly distributed random variables
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: November 6th 2008, 11:06 PM
  4. Normally distributed random variables
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: June 14th 2008, 04:59 AM
  5. independent and identically distributed variables problem
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: May 24th 2008, 05:10 AM

Search Tags


/mathhelpforum @mathhelpforum