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The following theorem comes from the 7th edition of "Introduction to Mathematical Statistics" by Hogg, Craig and Mckean and it concerns the necessary and sufficient condition for the independence of two quadratic forms of normal variables.

This is a rather long extract but what I would appreciate some help with is only the transition from 9.9.6 to 9.9.7. I have included the previous steps just to give the overall picture in case a previous result is implicitly used. Could you please help me understand then why 9.9.6 and 9.9.7 are equivalent representations? I have tried deriving 9.9.7 on my own but all my attempts ended in frustration.

The proof goes on after that but I do not have any other problems. Thank you in advance.

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(9.9.6) states that$${\mathbf {AB}}=\{\mathbf{\Gamma_{11}'\Lambda_{11}}\}\mathbf{\Gamma_{11}\Gamma_{21}'}\{\mathbf{\Lambda_{22}\Gamma_{21}}\}={\mathbf U}\mathbf{\Gamma_{11}\Gamma_{21}'}{\mathbf V}$$Hence multiplying left and right by $\mathbf{U}'$ and $\mathbf{V}'$, we get $$\mathbf{U}'{\mathbf {AB}}\mathbf{V}'=\mathbf{U}'{\mathbf U}\mathbf{\Gamma_{11}\Gamma_{21}'}{\mathbf V}\mathbf{V}'$$and $$(\mathbf{U}'{\mathbf U})^{-1}\mathbf{U}'{\mathbf {AB}}\mathbf{V}'({\mathbf V}\mathbf{V}')^{-1}=\mathbf{\Gamma_{11}\Gamma_{21}'}$$I see no reason for the inner $\mathbf{U}'$ and $\mathbf{V}'$ to vanish so I would bet on a typo. However the conclusion remains the same, namely that $$\mathbf{\Gamma_{11}\Gamma_{21}'}=0$$ if and only if$$\mathbf{{AB}}=0$$

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  • $\begingroup$ Yes that was my train of thought precisely, thank you. I looked at the list of the errata for this book but this is not on it so I found it extremely puzzling. $\endgroup$ – JohnK May 30 '15 at 12:43
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I contacted the author Professor Joseph W. McKean who achknowledged the mistake and very kindly offered the correction. I am posting it here, in case anybody else studying on his or her own is in need of it.


After (9.9.6) write:

Let $\mathbf{U}$ denote the matrix in the first set of braces. Note that $\mathbf{U}$ has full column rank, so its kernel is null; i.e., its kernel consists of the vector $\mathbf{0}$. Let $\mathbf{V}$ denote the matrix in the second set of braces. Note that $\mathbf{V}$ has full row rank, hence the kernel of $\mathbf{V}^{\prime}$ is null.

For the proof then, suppose $\mathbf{AB}=\mathbf{0}$. Then

$$\mathbf{U}\left[\mathbf{\Gamma}_{11}\mathbf{\Gamma}_{21}^{\prime}\mathbf{V}\right]=\mathbf{0}$$

Because the kernel of $\mathbf{U}$ is null this implies that each column of the matrix in the brackets is $\mathbf{0}$. This implies that

$$\mathbf{V}^{\prime} \left[ \mathbf{\Gamma}_{21} \mathbf{\Gamma}_{11} \right]=\mathbf{0}$$

In the same way, because the kernel of $\mathbf{V}^{\prime}$ is null we have $\mathbf{\Gamma}_{11} \mathbf{\Gamma}_{21}^{\prime}=\mathbf{0}$. Hence by $\left(9.9.5 \right)$...

(and the proof continues for the other direction)


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