3. (10 pts) (Constructing Kernels) One convenient property of kernels is that they can be combined...
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3. (10 pts) (Constructing Kernels) One convenient property of kernels is that they can be combined together to build new kernels. In this questions, we consider building a kernel K(u, v), u, vE R, using existing kernels G(u, v) and H(u, v). Remember that a function K(u, v) is a (Mercer) kernel if and only if it satisfies the following two conditions: • Symmetry K(u, v) = K (v, u) • Positive semi-definiteness: E,E-1 a;a;K(x;, L;) > 0, for any data points r1,..., Ln and any values a1, .. ., an E R. And, under such conditions, there exists a mapping o such that K(u, v) = ¢(u)T¢(v). 2 Conversely, if one can show such a mapping o exists such that K(u, v) = ¢(u)T¢(v), then this also shows it is a valid kernel. Show that the following are valid kernels: (a) (2 pts) (Scaling) K(u, v) = c. G(u, v), with c> 0 = C: (b) (3 pts) (Sum) K(u, v) = G(u, v) + H(u, v) (c) (3 pts) (Product) K(u, v) = G(u, v) · H(u, v) (d) (2 pts) (Polynomial) K(u, v) = E, a;G(u, v)', with a; 20 3. (10 pts) (Constructing Kernels) One convenient property of kernels is that they can be combined together to build new kernels. In this questions, we consider building a kernel K(u, v), u, vE R, using existing kernels G(u, v) and H(u, v). Remember that a function K(u, v) is a (Mercer) kernel if and only if it satisfies the following two conditions: • Symmetry K(u, v) = K (v, u) • Positive semi-definiteness: E,E-1 a;a;K(x;, L;) > 0, for any data points r1,..., Ln and any values a1, .. ., an E R. And, under such conditions, there exists a mapping o such that K(u, v) = ¢(u)T¢(v). 2 Conversely, if one can show such a mapping o exists such that K(u, v) = ¢(u)T¢(v), then this also shows it is a valid kernel. Show that the following are valid kernels: (a) (2 pts) (Scaling) K(u, v) = c. G(u, v), with c> 0 = C: (b) (3 pts) (Sum) K(u, v) = G(u, v) + H(u, v) (c) (3 pts) (Product) K(u, v) = G(u, v) · H(u, v) (d) (2 pts) (Polynomial) K(u, v) = E, a;G(u, v)', with a; 20
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