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Thread: PDF of (nearly) collinear variables

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    PDF of (nearly) collinear variables




    My question is motivated by a problem with estimation by maximum likelihood. In simple terms, I want to estimate a parameter \beta and have three variables, X, Y and Z, such that Z = X+Y. Using all three variables is pointless and the joint pdf is singular. Now, suppose that instead of Z I have U such that U = Z + \epsilon, where \epsilon is independent from X and Y and its distribution does not depend on \beta. The joint pdf of X, Y and U is not singular. Seems to me obvious that U does not contain any additional information about \beta, given X and Y. So, I think it must be true that, in terms of the likelihood function, L(\beta;X,Y,U)=L(\beta;X,Y)*const. If I am correct, what would be the best way to prove it? Thanks.
    Last edited by Sam Vimes; 07-18-2014 at 01:29 PM. Reason: to textify

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    Re: PDF of (nearly) collinear variables


    The question is equivalent to show that the conditional pdf

    f_{U|X=x,Y=y}(u|x,y) = \frac {f_{U,X,Y}(u,x,y)} {f_{X,Y}(x,y)}

    is independent of the parameter \beta

    Consider the conditional CDF

    F_{U|X=x,Y=y}(u|x,y)

    = \Pr\{U \leq u|X = x, Y = y\}

    = \Pr\{X + Y + \epsilon \leq u |X = x, Y = y\}

    = \Pr\{\epsilon \leq u - x - y|X = x, Y = y\}

    = \Pr\{\epsilon \leq u - x - y\} (by independence)

    = F_\epsilon(u - x - y)

    By assumption, the distribution of \epsilon is independent of \beta, and therefore

    F_{U|X=x,Y=y}(u|x,y) = F_\epsilon(u - x - y)

    is independent of the beta, and similar for its derivative:

    f_{U|X=x,Y=y}(u|x,y) = \frac {\partial F_{U|X=x,Y=y}(u|x,y)} {\partial u}

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    Sam Vimes (07-18-2014)

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