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Thread: Describe the range of a random vector with multivariate normal distribution

  1. #1
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    Describe the range of a random vector with multivariate normal distribution

    Let $\displaystyle X = \begin{bmatrix} X_1 \\ X_2 \\ X_3 \end{bmatrix} $ have covariance matrix $\displaystyle \Sigma = \begin{bmatrix}3 & 1 & 1 \\ 1 & 4 & -7 \\ 1 & -7 & 15 \end{bmatrix}$.

    Note that $\displaystyle \Sigma \cdot \begin{bmatrix} 1 \\ -2 \\ -1 \end{bmatrix} = 0$, so $\displaystyle \Sigma $ have rank 2. If $\displaystyle \mu _X = \begin{bmatrix} 10 \\ 20 \\ 30 \end{bmatrix} $ and $\displaystyle X $ has a multivariate normal distribution, describe the range of X.

    Solution so far:

    Now, since $\displaystyle \Sigma \cdot \begin{bmatrix} 1 \\ -2 \\ -1 \end{bmatrix} = 0$, I have:

    $\displaystyle \begin{bmatrix} 1 & -2 & -1 \end{bmatrix} \Sigma \cdot \begin{bmatrix} 1 \\ -2 \\ -1 \end{bmatrix} = 0$

    $\displaystyle \begin{bmatrix} 1 & -2 & -1 \end{bmatrix} Cov(X,X) \begin{bmatrix} 1 \\ -2 \\ -1 \end{bmatrix} = 0 $

    $\displaystyle Cov( \begin{bmatrix} 1 & -2 & -1 \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \\ X_3 \end{bmatrix} ) = 0 $

    $\displaystyle Var( X_1-2X_2-X_3) = 0 $

    So I need to find out when does that happens, but I'm kind of stuck at the moment, any help please? Thanks!
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  2. #2
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    Re: Describe the range of a random vector with multivariate normal distribution

    Hey tttcomrader.

    What part of R^3 describes this linear equation involving X1, X2, and X3? (Hint: It will be on a plane in 3 dimensional space with two independent parameters).
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  3. #3
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    Re: Describe the range of a random vector with multivariate normal distribution

    So since $\displaystyle Var(X_1-2X_2-X_3)=0 $, then we must have $\displaystyle X_1-2X_2-X_3 = k $ for some constant k.

    That also means that $\displaystyle E[X_1-2X_2-X_3] = k $, which is $\displaystyle 10-40-30 = -60 $, so $\displaystyle X_1-2X_2-X_3 = -60 $?
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  4. #4
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    Re: Describe the range of a random vector with multivariate normal distribution

    Looks pretty good.
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