# 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