theorem reproducing kernel from characteristic function let be an -valued random vector that is symmetric about the origin (that isx and - arer identically distributed)and let ps be its characteristic functionps(te eit eit (dxfor then (xx :ps( is valid reproducing kernel on example (gaussian kernelthe multivariate normal distribution with mean vector and covariance matrix is clearly symmetric around the origin its characteristic function is ps(texp ktk gaussian kernel taking / this gives the popular gaussian kernel on kx (xx exp bandwidth radial basis function (rbfkernel ( the parameter is sometimes called the bandwidth note that in the machine learning literaturethe gaussian kernel is sometimes referred to as "theradial basis function (rbfkernel from the proof of theorem we see that the rkhs determined by the gaussian kernel is the space of pointwise limits of functions of the form kx xi (xai exp = we can think of each point xi having feature xi that is scaled multivariate gaussian pdf centered at xi example (sinc kernelthe characteristic function of uniform[- random variable (which is symmetric around is ps(tsinc( :sin( )/tso (xx sinc( - is valid kernel inspired by kernel density estimation (section )we may be tempted to use the pdf of random variable that is symmetric about the origin to construct reproducing kernel howeverdoing so will not work in generalas the next example illustrates example (uniform pdf does not construct valid reproducing kerneltake the function ps( {| }which is the pdf of uniform[- unfortunatelythe function (xx ps( is not positive semidefiniteas can be seen for example by constructing the matrix [ (ti )ij for the points and as followsps( ps(- ps(- ps( ps(- ps( ps( ps( ps(