I am reading about the ANOVA kernel:
$$K_a(x,t)=\prod^n_{i=1}(1+x_it_i)$$
Where $x,t\in \mathbb{R^n}$
We have that $$\phi_a(x)=(1,x_1,...,x_n,x_1x_2,....,x_1x_2\cdot\cdot\cdot x_n)^T$$
$\phi_a(x)\in\mathbb{R}^{2^n}$
I am trying to show that $$K_a(x,t)=\langle\phi_a(x),\phi_a(t)\rangle,\forall x,t\in\mathbb{R}^n$$
My attempt to prove it:
Proof by induction:
Let $P(n)$ be the statement $$K_a(x,t)=\langle \phi_a(x),\phi_(t)\rangle$$
for $x,t\in\mathbb{R}^n$
Base case, $P(1)$:
$K_a(x,t)=1+x_1t_1=\phi_a^T(x)\phi_a(t)$
For the sake of induction, assume $$P(n):\prod^n_{i=1}(1+x_it_i)=\phi_a^T(x)\phi_a(t)$$
is true.
Let $\phi_a(x^{(n)})$ be the notation for when the basis function is applied to $x\in\mathbb{R}^n$
We know that $\phi_a(x^{(n+1)})$ is a vector which is concatenated by the vector $\phi_a(x^{(n)})$ and $x_{n+1}(\phi_a(x^{(n)}))$.
Hence, we have that $$ \langle \phi_a(x^{(n+1)}),\phi_a(t^{(n+1)})\rangle=\\=\phi_a(x^{(n+1)})^T\phi_a(t^{(n+1)})=\phi_a(x^{(n)})^T\phi_a(t^{(n)})+x_{n+1}t_{n+1}(\phi_a(x^{(n)})^T\phi_a(t^{(n)}))\\=(\phi_a(x^{(n)})^T\phi_a(t^{(n)}))(1+x_{n+1}t_{n+1})\\=\left(\prod ^n_{i=1}(1+x_it_i)\right)(1+x_{n+1}t_{n+1})\\=\prod^{n+1}_{i=1}(1+x_it_i)=K_a(x^{(n+1)},t^{(n+1)})$$
Is this okay? I'm not sure that I have shown rigorously enough that $$\phi_a(x^{(n+1)})^T\phi_a(t^{(n+1)})=\phi_a(x^{(n)})^T\phi_a(t^{(n)})+x_{n+1}t_{n+1}(\phi_a(x^{(n)})^T\phi_a(t^{(n)}))$$