Is $f(x)=e^{x^Tx'}$ a suitable kernel to be choosen?

Is $f(x)=e^{x^Tx'}$ a suitable kernel to be choosen? If so, to what dimension does it transform the data?

• Kernel for what? "Kernel" means many different things in mathematics and statistics. Even the data-transformation tag doesn't narrow the scope much!
– whuber
Dec 7, 2012 at 20:52
• @whuber: Right, sorry that I wasn't clear enough. I meant this one where $k(x,x')$ is referred to as a kernel or a kernel function. Dec 7, 2012 at 21:00
• What's the difference between $x^T$ and $x'$? Dec 7, 2012 at 21:08
• He means there's two different variables, $x$ and $x'$, and he's just taking their dot product. Dec 7, 2012 at 21:17
• I think he means $K(x,y)=e^{x^Ty}$. Is that correct? Dec 7, 2012 at 21:19

If you are referring to the kernel as a kernel in machine-learning literature, then yes, it is a kernel.

More generally, we can consider the family of Gaussian kernels, parametrized by $\sigma$:

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

Using the power series expansion of the function exponential, we can rewrite the expression of $K$ as:

$$K(x, x') = \sum_{n=0}^{\infty} \frac{(x^T \cdot x' )^n}{\sigma^{2n}n!}$$

Recall kernels are closed under summation, even infinite sums. $K$ then is sum of other (polynomial) kernels , thus is still kernel. The polynomial kernel, $(x\cdot x')^d$ can be shown to map $x$ to monomials of degree $d$. Thus the Gaussian kernel maps $x$ to all the polynomial kernels.

Example: In two dimensions, $x = (x_1,x_2), \; x' = (x_1', x_2')$, the secord order polynomial kernel $(x \cdot x')^2$ maps the data to look like a new inner product:

$$( x_1^2, x_2^2, x_1 x_2 ) \cdot (x_1'^2, x_2'^2, x_1' x_2')$$ The polynomial kernel is actually computing the above, which looks like it mapped $(x_1, x_2)$ to all monomials of degree 2. The Gaussian kernel does the same thing but for degree 1, degree 2, degree 3...

Side Note

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:

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

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

• Thank you for your answer. Could you please explain yourself a little more - what do you mean by "maps x to all powers of x"? Dec 7, 2012 at 21:28
• I edited my answer a bit to include a better explaination Dec 7, 2012 at 21:44