Multiple-Try Metropolis question I read Multiple-Try Metropolis from Wikipedia and I do not understand some points. 

Suppose the current state is $\mathbf{x}$. The MTM algorithm is as
  follows:
  
  
*
  
*Draw ''k'' independent trial proposals
  $\mathbf{y}_1,\ldots,\mathbf{y}_k$ from $Q(\mathbf{x},.)$. Compute
  the weights
  $w(\mathbf{y}_j,\mathbf{x})=\pi(\mathbf{y}_j)Q(\mathbf{y}_j,\mathbf{x})\lambda(\mathbf{y}_j,\mathbf{x})$
  where $\lambda$ is non-negative and symmetric, but otherwise arbitrarily chosen.
  
*Select $\mathbf{y}$ from the $\mathbf{y}_i$ with probability
  proportional to the weights.
  
*Now produce a reference set by drawing
  $\mathbf{x}_1,\ldots,\mathbf{x}_{k-1}$ from the distribution
  $Q(\mathbf{y},.)$. Set $\mathbf{x}_k=\mathbf{x}$ (the current
  point).
  
*Accept $\mathbf{y}$ with probability :$$r=\text{min} \left(1,
     \frac{ w(\mathbf{y}_1,\mathbf{x} )+ \ldots+
     w(\mathbf{y}_k,\mathbf{x}) }{ w(\mathbf{x}_1,\mathbf{y})+ \ldots+
     w(\mathbf{x}_k,\mathbf{y}) } \right)$$
  



*

*why is probability proportional?

*Why do we sample a reference set at the step 3?

 A: [Warning: I used the same notations $Q(y,x)$ $-$and $w(y,x)$$-$ as in the question but it is counter-intuitive as we usually denote this proposal kernel $Q(y|x)$ or $Q(x,y)$.]
Multiple-Try Metropolis is a regular Metropolis move when including the other proposals $y_2,\ldots,y_k$ as auxiliary variables (assuming $y=y_1$ and when picking $Q(y_2,x)\cdots Q(y_k,x)$ as the target for those auxiliaries. Indeed, if one considers the Metropolis-Hastings ratio for the proposed move from the vector $(x,x_2,\ldots,x_k)$ to the vector $(y_1,y_2,\ldots,y_k)$, one gets
$$\begin{align*}\overbrace{\dfrac{\pi(y_1)Q(y_2|x)\cdots Q(y_k|x)}{\pi(x_1)Q(x_2|y_1)\cdots Q(x_k,y_1)}}^{\text{ratio of targets}} &\times
\overbrace{\dfrac{Q(x,y_1)Q(x_2,y_1)\cdots Q(x_k,y_1)}{Q(y_2,x)\cdots Q(y_k|x)}}^{\text{ratio of proposals}}\\
&\times \underbrace{\dfrac{\frac{w(x,y_1)}{w(x,y_1)+w(x_2,y_1)+\ldots+w(x_k,y_1)}}{\frac{w(y_1,x)}{w(y_1,x)+w(y_2,x)+\ldots+w(y_k,x)}}}_{\text{ratio of selection probabilities}}\\
\end{align*}
$$
Things then simplify to
$$\dfrac{w(y_1,x)+w(y_2,x)+\ldots+w(y_k,x)}{w(x,y_1)+w(x_2,y_1)+\ldots+w(x_k,y_1)}$$ when using $$w({y}_j,{x})=\pi({y}_j)Q({y}_j,{x})\lambda({y}_j,{x})$$ since
$$\dfrac{w(x,y_1)}{w(y_1,x)}=\dfrac{\pi(x)Q(x,y_1)\lambda(x,y_1)}{\pi(y_1)Q(y_1,x))\lambda(y_1,x)}$$
[An even more detailed validation is to establish that the move satisfies detailed balance, that is, that $$\pi(\mathbf{y})\mathfrak{h}(\mathbf{y},\mathbf{x})=\pi(\mathbf{x})\mathfrak{h}(\mathbf{x},\mathbf{y}),$$ where $\mathfrak{h}$ is the multiple-try kernel.]
A different validation is to use a pseudo-marginal approach since
$$\dfrac{w(y_1,x)}{w(y_1,x)+w(y_2,x)+\ldots+w(y_k,x)}\,Q(y_1,x)$$ is an unbiased estimator of the marginal distribution of $Y_1$ given $X=x$, when integrating out the auxiliary $y_j$'s.
