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The main problem is about computer science, trying to show that $f(x)=e^{x^Tx'}$ is of the form $\exp{\Big( \frac{||x - x'||^2}{2\sigma^2} \Big) }$, so it could be a kernel function. (see here for more details). In the following paragraph, I don't get how the last line is equal to the previous one.

Often in ML literature, the Gaussian kernel is defined as

$$K'(x,x') = \exp{\Big( \frac{||x - x'||^2}{2\sigma^2} \Big) }$$

But this is actually the normalized Gaussian kernel. A normalized kernel, $K'$, is defined:

$$ K'(x,x') = \frac{K(x,x')}{\sqrt{ K(x,x) K(x',x') } } $$

If we use $K(x,x') = \exp{\Big( \frac{x^Tx'}{\sigma^2 } \Big) }$, we get:

$$ K'(x,x') = \frac{e^{x^Tx'/\sigma^2 }}{ \exp{ \Big(\frac{||x||^2}{2\sigma^2} \Big)} \exp{\Big(\frac{||x'||^2}{2\sigma^2}\Big) } } $$

$$ = \exp{ \Big( -\frac{||x' - x||^2}{2\sigma^2} \Big) }$$

And $\exp{\text{(something)}}$ and $e^{\text{something}}$ equal, aren't they?

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Yes, $\exp a$ is another name for $e^a$. You can get the last equality if you take $\dfrac{\exp a}{\exp b\exp c}=\dfrac{\exp a}{\exp(b+c)}=\exp(a-(b+c))$ and simplify. –  Rahul Jan 25 '13 at 12:21
    
@RahulNarain: Right, thank you. But I still don't understand how to get that Euclidean norm. –  Gigili Jan 25 '13 at 12:26

1 Answer 1

It's easier to go the other way.

$$\begin{align} \lVert x'-x\rVert^2&=(x'-x)^T(x'-x)\\ &=x'^Tx'-2x^Tx'+x^Tx\\ &=\lVert x'\rVert^2-2x^Tx'+\lVert x\rVert^2\\ \end{align}$$ and you can hopefully see how to get there from here.

$$\exp\left(-\frac{\lVert x'-x\rVert^2}{2\sigma^2}\right)=\exp\left(\frac{-\lVert x'\rVert^2+2x^Tx'-\lVert x\rVert^2}{2\sigma^2}\right)=\cdots$$

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