Reproducing Kernel Hilbert Space

Since writing about the simulation of fractional Brownian motion, I’ll spend an hour to write RKHS.

Definition: A Hilbert space \mathcal{H} is a RKHS if |\mathcal{F}_t[f]|=|f(t)|\leq M\|f\|_{\mathcal{H}}, \forall f\in \mathcal{H}

Theorem: If \mathcal{H} is a RKHS, then for each t \in X there exists a function K_t \in \mathcal{H} (called the representer of t) with the reproducing property

\mathcal{F}_t[f]=<K_t,f>_{\mathcal{H}}=f(t), \forall f\in\mathcal{H}

Therefore, K_t(t^{'})=<K_t,K_{t^{'}}>_{\mathcal{H}}.

Definition: K: X\times X\rightarrow\mathbb{R} is a reproducing kernel if it’s symmetric and positive definite.

Theorem: A RKHS defines a corresponding reproducing kernel. Conversely, a reproducing kernel defines a unique RKHS.

Once we have the kernels, if f(\cdot)=\sum \alpha_i K(t_i,\cdot), g(\cdot)=\sum \beta_i K(t^{'}_i,\cdot),

then <f,g>_{\mathcal{H}}=\sum\sum \alpha_i \beta_j K(t_i, t^{'}_j)

When it comes to the fractional Brownian motion

Theorem: For fBm RKHS K(x,x^{'})=R(x,x^{'}) is symmetric and positive definite, , there exists K^H(x,\cdot) s.t K(x,x^{'})=\int K^H(x,y)K^H(x^{'},y)dy

Take Wiener integral as an example:

\begin{pmatrix}    \textit{dual space}&\textit{inner product}&E&\leftrightarrow&<\cdot,\cdot>_{\mathcal{H}}&\leftrightarrow&<\cdot,\cdot>_{L^{2}}\\    \delta_{t}(\cdot)& &B^H_t&\leftrightarrow&R(\cdot,t)&\leftrightarrow&K^H(t,\cdot)\\    f(\cdot)& &\textit{"}\int_0^T f(t)B^H_tdt\textit{"}&\leftrightarrow&K^H\circ(K^H)^{*}f&\leftrightarrow&(K^H)^{*}f    \end{pmatrix}

where K^H\circ f(t)=\int_0^T K^H(t,s)f(s)ds,

(K^H)^{*}\circ f(t)=\int_0^T K^H(s,t)f(s)ds and

K^H\circ(K^H)^{*}f(t)=\int_0^T R(t,s)f(s)ds

For example, when H=\frac{1}{2}, K^H(t,s)=1_{[0,t]}(s), , we can check

E(B_tB_s)=<t\wedge\cdot, s\wedge\cdot>_{\mathcal{H}}=<1_{[0,t]}(\cdot),1_{[0,s]}(\cdot)>_{L^2}

Claim: By previous theorem, the reproducing kernel R(t,s) uniquely defines a RKHS L(R(t,\cdot)) with the inner product \mathcal{H}

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s