{{page>:defs}}

====== 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 $\{\rset a\}^\perp$. 

<WRAP center round box 80%> 
__**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, $Y \sim N(a,I)$.
</WRAP>
 

===== 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.