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Math Help - X,Y Random Variables, X+Y , 1 are independent?

  1. #1
    Newbie noname's Avatar
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    X,Y Random Variables, X+Y , 1 are independent?

    Hi!

    I take my statistics exam two days ago and I'm thankful for your help

    Only one exercise gave me trouble. It was:

    X and Y are random variables, check if random variables X+Y , 1 are independent.

    The professor tell us that we should have used an appropriate example to check the independence of the variables.

    Does anyone know a way to solve this exercise?

    Thanks

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  2. #2
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    Quote Originally Posted by noname View Post
    Hi!

    I take my statistics exam two days ago and I'm thankful for your help

    Only one exercise gave me trouble. It was:

    X and Y are random variables, check if random variables X+Y , 1 are independent.

    The professor tell us that we should have used an appropriate example to check the independence of the variables.

    Does anyone know a way to solve this exercise?

    Thanks
    To prove X and Y being independent random variables:

    If h(xi,yj)= sigma f(xi)*g(yj), then X and Y are independent;
    otherwise, X and Y are not independent random variables.

    where h(xi,yj) is the joint distribution and i and j being the subscripts
    where f(xi) is the probability distribution of X, and
    g(yj) the probability distribution of Y.

    The joint distribution of X+Y is the sum of h(xi,yj), where X=xi and Y=yj

    I do not quite get your question about 1 being independednt of X+Y as I have not heard about it.
    If 1 being a member in the probability distribution, then there must be only one random variable in the sample space, since 0< or = P and < or = 1.
    If this is the case then P(X) = 1 and P(Y) =1, and P(X)*P(Y) = h(xi,yi) = 1.
    Last edited by novice; September 30th 2009 at 11:14 AM.
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  3. #3
    MHF Contributor matheagle's Avatar
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    by a constant, I interprete that as P(W=1)=1, and that is independent of any other random variable
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  4. #4
    Newbie noname's Avatar
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    Thank you guys for your answer

    Quote Originally Posted by novice View Post
    If 1 being a member in the probability distribution, then there must be only one random variable in the sample space, since 0< or = P and < or = 1.
    If this is the case then P(X) = 1 and P(Y) =1, and P(X)*P(Y) = h(xi,yi) = 1.
    Sorry but I can't understand this. Could you please explain it?

    Quote Originally Posted by matheagle
    by a constant, I interprete that as P(W=1)=1, and that is independent of any other random variable
    Then i should have this, shouldn't I?

    P(Z=z,W=1)=P(Z=z)P(W=1)=P(Z=z)*1

    How can I prove that P(Z=z,W=1)=P(Z=z)?
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  5. #5
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    Quote Originally Posted by noname View Post
    Thank you guys for your answer



    Sorry but I can't understand this. Could you please explain it?



    Then i should have this, shouldn't I?

    P(Z=z,W=1)=P(Z=z)P(W=1)=P(Z=z)*1

    How can I prove that P(Z=z,W=1)=P(Z=z)?
    The hardest part of answering this question is that we do not know whether "1" is a random variable or a probability in the distribution in the sample space, but I will try to simplify my thinking here:

    Suppose "1" is in the distribution, then the sample space S must look like this: S={a}, let say P(X=1)= P({a}) = 1.
    Then P(Y=1)=P({a})=1.

    P(X=1,Y=1)=1 , which is the P(X=1)P(Y=1)=1, which also proved X and Y being independent.
    In this case, P(X+Y) = P({ \emptyset})

    I have never heard of anyone looking for the probability of non-existence, i.e. P({ \emptyset}), which render this nonsense.

    Suppose "1" is one of the random variables, and
    if P(X+Y)P(1)=P(X+Y,1), then they are independent, but You must know what P(X+Y) and P(1).
    Say P(X+Y)=0.5 and P(1) =0.3, if P(X+Y)P(1)=0.15, they are independent.
    Last edited by novice; September 30th 2009 at 03:25 PM.
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  6. #6
    Newbie noname's Avatar
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    Now it's more clear. Thank you

    The last thing you said resembles the hint of the professor.
    He said that we should use an example like:

    cosx^{2}+senx^{2}=1

    and calculate the derivatives (maybe are there some properties of indipendent random variables involving derivatives?)
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    Quote Originally Posted by noname View Post
    Now it's more clear. Thank you

    The last thing you said resembles the hint of the professor.
    He said that we should use an example like:

    cosx^{2}+sinx^{2}=1

    and calculate the derivatives (maybe are there some properties of indipendent random variables involving derivatives?)
    I hope you don't mind that I am guessing. I am thinking as follows, and hope that I am not thinking wishfully. If not, you might be able to use same idea to approach the correct solution:

    Since cosx^{2}+senx^{2}=1 ---(eq1)is given,we can use triangle as representation
    where the base is X(x), height is Y(x)=1-X(x), \sqrt{X(x)^2+Y(x)^2} is hypotenuse, and 0\leq x \leq 1

    let

    P(Y(x))=1-P(X(x)) where  P(X(x)) = \cos x and
     P(Y(x)) = 1-P(X(x)) = \sin x


    The derivative of eq1 is

     2 \sin \theta \cos \theta -\sin \theta \cos \theta =0
    Since 0\leq P \leq 1, \theta\leq \pi x/2

    Let

    P(X=x) = \cos \theta, P(Y=y) = \sin \theta


    P(X,Y) = P(Y) P(Y)= \\sin \theta \cos \theta
    which shows X and Y are independent.

    X+Y= 1 always.

     P(X+Y,1) = P(X,Y)\cap P(1) = P(X=0,Y=1)\cap P(Y=1) P(X,Y) is a proper subset of P(X+Y+1)

    or

     P(X=1,Y=0)\cap P(X=1)= P(X=1,Y=0) or P(X=0,Y=1) ; Both P(Y=1) and P(X=1) are proper subsets of P(X+Y,1).

    P(X=1)=0, P(X=0)=1
    P(Y=1)=1, P(Y=0)=0

    P(X+Y,1) =P(X=1,Y=0)=P(X=1)*P(Y=0) = 0
    P(X+Y,1) =P(X=0,Y=1)=P(X=0)*P(Y=1) = 0

    Consequently, X, Y, X+Y, and 1 are independent.
    Last edited by novice; October 1st 2009 at 04:11 PM. Reason: final fix
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