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world:reflectional_coupling
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Reflection coupling
Let I be the identity matrix with d components.
We want to find a coupling of N(0,I) and N(a,I) where a \in \rset^d. For all b \in \rset^d, denote f_b the density of N(b,I). The reflection coupling is based on the following result: Set R_a=\Id -2 \frac{aa^T}{\|a\|^2} as the orthogonal reflection wrt to \{\rset a\}^\perp.
Lemma
- draw independently X \sim N(0,I) and U \sim \unif(0,1)
- set
- Y=X if U \leq \frac{f_0\wedge f_a(X)}{f_0(X)}
- Y=a+R_aX otherwise.
Then, Y \sim N(a,I).
Proof
Since R_a^2=\Id, we get R_a=R_a^{-1}. Then, y=a+R_a x is equivalent to R_a(y-a)=x. Moreover, (\det R_a)^2=\det R_a^2=1 so that |\det R_a|=1. Then, \begin{align*} \PE\lrb{h(Y)}&=\int h(x) \ f_0\wedge f_a(x)\ \rmd x+\int h(a+R_ax) (f_0- f_a)^+(x) \rmd x\\ &=\int h(x) \ f_0\wedge f_a(x)\ \rmd x+\int h(y) (f_0- f_a)^+(R_a(y-a)) \underbrace{|\det R_a|}_{1} \rmd y \end{align*} Now, noting that R_a is an isometry and that R_aa=-a, we get \begin{align*} \|R_a(y-a)\|^2&=\|y-a\|^2\\ \|R_a(y-a)-a\|^2&=\|R_ay\|^2=\|y\|^2 \end{align*} which implies that f_0(R_a(y-a))=f_a(y) and f_a(R_a(y-a))=f_0(y). Finally, \PE\lrb{h(Y)}=\int h(x) \ f_0\wedge f_a(x)\ \rmd x+\int h(y) (f_a- f_0)^+(y)\rmd y =\int h(x) f_a(x)\rmd x which concludes the proof.
Corollary
We now intend to construct a coupling of N(x,h I) and N(y,h I). We use the Lemma with N(0,I) and N(a,I) where a=(y-x)/\sqrt{h} to construct a coupling (Z,Z') and we set X=x+\sqrt{h}Z and Y=x+\sqrt{h}Z'. This is is equivalent to the following coupling. Write \varphi_{b,\Gamma} the density of N(b,\Gamma),
- Draw independently Z \sim N(0,I) and U \sim \unif(0,1)
- Set X=x+\sqrt{h}Z and set
- Y=X if U \leq \frac{f_0\wedge f_a(Z)}{f_0(Z)}=\frac{\varphi_{x,h I}\wedge \varphi_{y,h I}(X)}{\varphi_{x,h I}(X)}
- Y=y+R_{y-x} (\sqrt{h}Z) otherwise.
world/reflectional_coupling.txt · Last modified: 2022/04/11 13:42 by rdouc
