Discriminant Rule Proof

**LackingArtistry** · 11-11-2010 12:10 PM

My HW question is:

Show that...

$-\frac{1}{2}(x-\mu_1)^{T}\Sigma^{-1}(x-\mu_1)+\frac{1}{2}(x-\mu_2)^{T}\Sigma^{-1}(x-\mu_2)$

is equal to...

$(\mu_1 - \mu_2)^{T}\Sigma^{-1}x-\frac{1}{2}(\mu_1-\mu_2)^{T}\Sigma^{-1}(\mu_1+\mu_2)$

My efforts:

Okay, so I started by expanding the first equation to this:

$-x^{2}2\Sigma^{-1}+x\mu_12\Sigma^{-1}+x\mu_12\Sigma^{-1}-\mu_1^{2}2\Sigma^{-1}+x^{2}2\Sigma^{-1}-x\mu_22\Sigma^{-1}-x\mu_22\Sigma^{-1}+\mu_2^{2}2\Sigma^{-1}$

After combining terms, I got this:

$2(x\mu_1)2\Sigma^{-1}-\mu_1^{2}2\Sigma^{-1}-2(x\mu_2)2\Sigma^{-1}+\mu_2^{2}2\Sigma^{-1}$

Then, I multiplied the entire equation by $\frac{1}{2}$ :

$\mu_1\Sigma^{-1}x-\mu_2\Sigma^{-1}x-\frac{1}{2}\mu_1^{2}\Sigma^{-1}+\frac{1}{2}\mu_2^{2}\Sigma^{-1}$

From there, I've been unable to make any more forward progress. I'm pretty sure the first two terms can be rewritten like...

$(\mu_1-\mu_2)^{T}\Sigma^{-1}x$

But I'm not sure about how the second two terms could be factorized. I can only assume that I've made an error in the expansion and simplification steps, but I've been unable the locate the error in my work (if one exists).

Please help!

**BGM** · 11-11-2010 12:36 PM

When you expand the quadratic form, you cannot write something like $x^2$ .

Here $x, \mu_1, \mu_2$ are vectors, not a square matrix.

So you should get something like

$\frac {1} {2} (-x^T\Sigma^{-1}x + ...)$

**LackingArtistry** · 11-12-2010 11:56 AM

I've been able to reduce the initial expression to this, after following BGM's advice:

$\frac{1}{2}(\mu_1-\mu_2)\Sigma^{-1}x^{T}+\frac{1}{2}(\mu_1-\mu_2)^{T}\Sigma^{-1}x+\frac{1}{2}(\mu_1+\mu_2)^{T}\Sigma^{-1}(\mu_1+\mu_2)$

I am unsure of how to eliminate or combine the first two terms such that they are equal to the equation in the original post, though. Any suggestions would be welcome.

**BGM** · 11-12-2010 01:04 PM

The first two terms are equal,
because this quadratic term is just a real number (or a 1 by 1 matrix if you like)

So you can always take transpose on it.

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