6.2 Show that the following kernels K are PDS:

(a) K(x; y) = cos(x ???? y) over R  R.

(b) K(x; y) = cos(x2 ???? y2) over R  R.

(c) For all integers n > 0;K(x; y) =

PN i=1 cosn(x2i

???? y2 i ) over RN  RN.

(d) K(x; y) = (x + y)????1 over (0;+1)  (0;+1).

(e) K(x; x0) = cos\(x; x0) over Rn  Rn, where \(x; x0) is the angle between x and x0.

(f) 8 > 0; K(x; x0) = exp

????

???? [sin(x0 ???? x)]2

over R  R.

(Hint: rewrite [sin(x0 ???? x)]2 as the square of the norm of the dierence of two vectors.)

(g) 8 > 0;K(x; y) = e????kx????yk  over RN  RN.

(Hint: you could show that K is the normalized kernel of a kernel K0 and show that K0 is PDS using the following equality: kx ???? yk =

1 2????( 1 2 )

R +1 0

1????e????tkx????yk2 t

32 dt valid for all x; y.)

(h) K(x; y) = min(x; y) ???? xy over [0; 1]  [0; 1].

(Hint: you could consider the two integrals R 1 R 0 1t2[0;x]1t2[0;y]dt and 1 0 1t2[x;1]1t2[y;1]dt.)

(i) K(x; x0) = 1 p1????(x
x0)

over x; x0 2 X = fx 2 RN : kxk2 < 1g.

(Hint: one approach is to nd an explicit expression of a feature mapping 

by considering the Taylor expansion of the kernel function.)

(j) 8 > 0;K(x; y) = 1 1+kx????yk2

2 over RN  RN.

(Hint: the function x 7!

R +1 0 e????sxe????sds dened for all x  0 could be useful for the proof.)

(k) 8 > 0;K(x; y) = exp

PN i=1 min(jxij;jyij)

2



over RN  RN.

(Hint: the function (x0; y0) 7!

R +1 0 1t2[0;jx0j]1t2[0;jy0j]dt dened over R  R could be useful for the proof.)