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I am on a project and I had to study about signal detection, Bayes decision theory etc. The basic paper I am reading, gives an equation for LLR, referring to these equations:

$H_0 : Y\sim N(\mu_0,\Sigma_0)$

$H_1 : Y\sim N(\mu_1,\Sigma_1)$

The log-likelihood ratio is given then by:

$$\log(L(y))=\frac{1}{2}\left[\log\frac{| \Sigma_0 |}{| \Sigma_1 |} + y^T(\Sigma_0^{-1} - \Sigma_1^{-1})y+2 (\mu_1^T\Sigma_1^{-1} -\mu_0^T \Sigma_0^{-1} )y+\mu_0^T \Sigma_0^{-1}\mu_0-\mu_1^T\Sigma_1^{-1}\mu_1\right]$$

But as I wanted to reproduce the final equation, I ended in this: $$\log(L(y))=\frac{1}{2}\left[\log\frac{| \Sigma_0 |}{| \Sigma_1 |}+y^{T}(\Sigma _0^{-1}-\Sigma_1^{-1})y+(\mu_1^T\Sigma_1^{-1}-\mu_0^T \Sigma_0^{-1}) y + \mu_0^T \Sigma_0^{-1} \mu_0-\mu_1^T\Sigma_1^{-1}\mu_1+y^T(\Sigma_1^{-1}\mu_1-\Sigma_0^{-1} \mu_0)\right]$$

The above are equal if only:

$$y^T(\Sigma_1^{-1}\mu_1-\Sigma_0^{-1}\mu_0) = (\mu_1^T\Sigma_1^{-1}-\mu_0^T\Sigma_0^{-1})y$$

which I find obviously wrong.

Can someone help me with this?

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  • $\begingroup$ The equality holds if the covariance matrices are symmetric, which is true. $\endgroup$
    – Chester
    Commented Jul 13, 2015 at 23:49
  • $\begingroup$ Ok, I really missed that! Thank you! $\endgroup$
    – user7127
    Commented Jul 14, 2015 at 0:11

1 Answer 1

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$$ y^T(\Sigma_1^{-1}\mu_1-\Sigma_0^{-1}\mu_0) = (\mu_1^T\Sigma_1^{-1}-\mu_0^T\Sigma_0^{-1})y $$ Either of these is the transpose of the other. But these are $1\times1$ matrices. Every $1\times1$ matrix of real numbers is its own transpose.

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