How to derive the block differential equations for Wright–Fisher diffusion on the hierarchical model. I am studying the Lecture notes STOCHASTIC MODELS FOR GENETIC EVOLUTION by den Hollander.
The final chapter presents the Hierarchical model. The basic idea is that you have graph (like an $M$-regular tree) and that probability to move in it is a function of the distance between the points in the graph. The  distance is given by an ultrametric and so each point move can be seen as two independent decisions: First you decide how far you are going to jump, then you pick uniformly at random one of the points at that distance from you. 
After the basic definitions (to which we will come back) one reads

then consider:

$\ \ $ $\text{In}$ $\text{what}$ $\text{follows}$ $\text{we}$ $\text{will}$ $\text{make}$ $\text{a}$ $\textit{special}$ $\textit{choice}$ $\textit{for}$ $\textit{the}$ $\textit{migration}$ $\textit{kernel}$ $\text{(just}$ $\text{as}$ $\text{we}$ $\text{did}$ $\text{in}$ $(6.1.3)$ $\text{for}$ $\text{the}$ $\text{stepping}$ $\text{stone}$ $\text{model}$ $\text{on}$ $\mathbb{Z}^2),$ $\text{namely,}$ $$p(x,y)=\dfrac{1}{N_M}\sum_{k\geq\|x-y\|}\dfrac{1}{M^{2k-1}}\tag{7.1.7}$$

Now we are ready to define the blocks:

Finally one reads the following differential equation for the blocks:

Question: How do we prove 7.2.2?
Details and definitions:
The Hierarchical Lattice:

The distance on the Lattice:

The migration kernel:

$\ \ $ $\text{In}$ $\text{what}$ $\text{follows}$ $\text{we}$ $\text{will}$ $\text{make}$ $\text{a}$ $\textit{special}$ $\textit{choice}$ $\textit{for}$ $\textit{the}$ $\textit{migration}$ $\textit{kernel}$ $\text{(just}$ $\text{as}$ $\text{we}$ $\text{did}$ $\text{in}$ $(6.1.3)$ $\text{for}$ $\text{the}$ $\text{stepping}$ $\text{stone}$ $\text{model}$ $\text{on}$ $\mathbb{Z}^2),$ $\text{namely,}$ $$p(x,y)=\dfrac{1}{N_M}\sum_{k\geq\|x-y\|}\dfrac{1}{M^{2k-1}}\tag{7.1.7}$$

Relevant computations, discussion and attempt at solution:
Fix $x$ in $\Omega_M$. Let $y_i$ be such that $d(x,y_i) = \|x-y_i \|=i>0$
In this case $$p(x,y_i) = \frac{1}{N_M}\sum_{k \geq i} M^{1 - 2k} =\frac{1}{N_M} \frac{M^{1-2i}}{1 - M^{-2}}.$$
For the case $y =x$ 
$$p(x,x)= \frac{1}{N_M}\sum_{k \geq 0} M^{1 - 2k} =\frac{1}{N_M} \frac{M}{1 - M^{-2}}.$$
At exact distance $i$ we have $M^i - M^{i-1}$ therefore to compute normalizing constant on $p(x,y)$ we write:
$$1 = \sum_{y \in \Omega_M} p(x,y) = \frac{1}{N_M}\big[ \frac{M}{1 - M^{-2}} + \sum_{i = 1}^{\infty} \frac{M^{1-2i}}{1 - M^{-2}}(M^i - M^{i-1})\big]\\
= \frac{1}{N_M}\frac{M}{1 - M^{-2}}\big[ 1 + \sum_{i = 1}^{\infty} (M^{-i} - M^{-i-1})\big]\\
= \frac{1}{N_M}\frac{M^3}{M^2 - 1}\big[ 1 + \sum_{i = 1}^{\infty} (M^{-i} - M^{-i-1})\big]\\
= \frac{1}{N_M}\frac{M^3}{M^2 - 1}\big[ 1 + \frac{1}{M}\big]\\
= \frac{1}{N_M}\frac{M^3}{M^2 - 1}\big[ \frac{M + 1}{M}\big]\\
= \frac{1}{N_M}\frac{M^2}{M - 1}\\$$
So we conclude that $N_M = \frac{M^2}{M - 1}$
Remark: If the above computations are right then there is a typo on the notes, since one reads:

After computing $p(x,y)$ we move to the block differential equation, and let us ignore for the moment the noise term $dW_x$
$$ d Y^{[k]}_x(t) = \sum_{y : \|y-x \|\leq k} d Y_y(M^k t)\\
 = \sum_{y : \|y-x \|\leq k}\sum_{z \in \Omega_M} cp(y,z) [Y_z(M^k t) - Y_y(M^k t)]  $$
I can't find a way to arrive at the drift term of (7.2.2):
$$d Y^{[k]}_x(t) = c \sum_{l \in \Bbb{N}} \frac{1}{M^{l-1}}[Y_x^{[k+l]}(M^{-l t}) - Y^{[k]}_x(t)]$$

Vague ideas for the final computation:
For the deterministic part of the differential equation for the blocks, we have that:
 $$d Y^{[k]}_x(t) = \sum_{y : \|y-x \|\leq k} d Y_y(M^k t)\\
 = \sum_{y : \|y-x \|\leq k}\sum_{z \in \Omega_M} cp(y,z) [Y_z(M^k t) - Y_y(M^k t)]\\
\sum_{z \in \Omega_M} \sum_{y : \|y-x \|\leq k} cp(y,z) [Y_z(M^k t) - Y_y(M^k t)]\\
=\sum_{z \in \Omega_M} \sum_{y : \|y-x \|\leq k} cp(y,z) [Y_z(M^k t) - Y_y(M^k t)]\\
  $$
Consider now the distance $d(z,x) = l_z$ and rewrite the above sum with respect to this distance:
$$
=\sum_{l = 0}^\infty \sum_{z: d(x,z) = l} \sum_{y : \|y-x \|\leq k} cp(y,z) [Y_z(M^k t) - Y_y(M^k t)]\\
= \sum_{y : \|y-x \|\leq k} cp(y,x) [Y_x(M^k t) - Y_y(M^k t)] + 
 \sum_{l = 1}^\infty \sum_{z: d(x,z) = l} \sum_{y : \|y-x \|\leq k} cp(y,z) [Y_z(M^k t) - Y_y(M^k t)]\\
= \sum_{y : \|y-x \|\leq k} cp(y,x) [Y_x(M^k t) - Y_y(M^k t)] + 
 \sum_{l = 1}^k \sum_{z: d(x,z) = l} \sum_{y : \|y-x \|\leq k} cp(y,z) [Y_z(M^k t) - Y_y(M^k t)]\\
+ \sum_{l = k+1}^\infty \sum_{z: d(x,z) = l} \sum_{y : \|y-x \|\leq k} cp(x,z) [Y_z(M^k t) - Y_y(M^k t)]\\
  $$
where the last equality comes from the fact that we have the unltrametric property on the distance $d(y,z) = \max{d(x,y), d(y,z)}$.
It may be that at this point one should replace $p(y,z)$ by the explicit values we computed above and do some smart computations so as to derive the final result.  However I am stuck.
Do you have any ideas? 
 A: Here is a bit of computations that might be helpfull:
Remember that if $d(x,y_i) = i$ the probability of jumping from $x$ to $y_i$ is equal to $p(x,y_i) = \frac{M}{M+1} \frac{1}{M^{2i}}$. Therefore when we write down the expression for the drift term of $dY^{k}_x(t)$ we have:
$$d Y^{[k]}_x(t) = \sum_{y : \|y-x \|\leq k} d Y_y(M^k t)\\
=\sum_{y : \|y-x \|\leq k}\sum_{z \in \Omega_M}  cp(y,z) [Y_z(M^k t) - Y_y(M^k t)]\\
=\sum_{y : \|y-x \|\leq k}\sum_{i=0}^\infty\sum_{z : \|z-y\| = i}  cp(y,z) [Y_z(M^k t) - Y_y(M^k t)]\\
=\sum_{y : \|y-x \|\leq k}\sum_{i=0}^k\sum_{z : \|z-y\| = i}  cp(y,z) [Y_z(M^k t) - Y_y(M^k t)]\\
+ \sum_{y : \|y-x \|\leq k}\sum_{i=k+1}^\infty\sum_{z : \|z-y\| = i}  cp(y,z) [Y_z(M^k t) - Y_y(M^k t)]\\
 = \sum_{y : \|y-x \|\leq k}\sum_{z : \|z-x\| \leq k}  cp(y,z) [Y_z(M^k t) - Y_y(M^k t)]\\
 + \sum_{y : \|y-x \|\leq k}\sum_{i=k+1}^\infty\sum_{z : \|z-y\| = i}  cp(y,z) [Y_z(M^k t) - Y_y(M^k t)]\\
= \sum_{y : \|y-x \|\leq k}\sum_{i=k+1}^\infty\sum_{z : \|z-y\| = i}  cp(y,z) [Y_z(M^k t) - Y_y(M^k t)]\\
= \sum_{i=k+1}^\infty\sum_{z : \|z-y\| = i} \sum_{y : \|y-x \|\leq k} c \frac{M}{M+1} \frac{1}{M^{2i}} [Y_z(M^k t) - Y_y(M^k t)]\\ = \sum_{i=k+1}^\infty\sum_{z : \|z-y\| = i}  c \frac{M}{M+1} \frac{M^k}{M^{2i}} [Y_z(M^k t) - Y^{[k]}_x( t)]\\ $$
Let' us retain this result:
$$
\tag{1}
d Y^{[k]}_x(t) = \sum_{i=k+1}^\infty\sum_{z : \|z-y\| = i}  c \frac{M}{M+1} \frac{1}{M^{2i-k}} [Y_z(M^k t) - Y^{[k]}_x( t)] $$
Idealy we should be able to move from this point to the claim, that is, doing further computations we should find:
 $$\tag{2}
dY^{[k]}_x(t) = c\sum_{l =1}^\infty \frac{1}{M^{l-1}}[Y^{[k+l]}_x(M^{-l}t) - Y^{[k]}_x(t)]$$
In what follows, we will expand the above expressions so as to see if they coincide:
Starting with the term $Y^{k}_x(t)$ in $(1)$:
$$\sum_{i=k+1}^\infty \sum_{z : \|z-x\|=i}c \frac{M}{M+1}\frac{1}{M^{2i-k}}Y^{[k]}_x(t)\\
=\sum_{i=k+1}^\infty c \frac{M}{M+1}\frac{M^i - M^{i-1}}{M^i}\frac{1}{M^{2i-k}}Y^{[k]}_x(t)\\
=\sum_{i=k+1}^\infty c \frac{M}{M+1}\frac{M - 1}{M}\frac{1}{M^{i-k}}Y^{[k]}_x(t)\\
=c \frac{M-1}{M+1}\sum_{i=k+1}^\infty \frac{1}{M^{i-k}}Y^{[k]}_x(t)\\
=c \frac{M-1}{M+1}\frac{1}{M - 1}Y^{[k]}_x(t)\\
=c \frac{1}{M+1}Y^{[k]}_x(t)\\
$$
Now let us expand the values in $(2)$. First, the constant for $Y^{k}_x(t)$:
$$\sum_{l = 1}^\infty \frac{1}{M^{l-1}} Y^{[k]}_x(t) = c \frac{M}{M-1} Y^{[k]}_x(t) $$
Now, when it comes to the terms in $z$ we have to expand the terms in $$Y^{[k+l]}_x(M^{-l}t) = \sum_{i = 0}^{k+l} \sum_{z:\|z-x\| = i}\frac{1}{M^{l+k}} Y_z(M^k t).$$
So we have:
$$\sum_{l=1}^\infty \frac{1}{M^{l-1}}\sum_{z:\|z-x\| \leq l+k}\frac{1}{M^{l+k}} Y_z(M^k t)$$
We split in cases: $d(z,x)>k$ and $d(z,x)\leq k$. First, for $d(z,x)>k$. For a fixed  $z$ such that $d(z,x)=i>k$  we sum $Y_z(M^k t)$ many times, but with distinct weights:
$$ \sum_{l = {i-k}}^\infty \frac{1}{M^{l-1}} \frac{1}{M^{l+k}} Y_z(M^k t)\\
= c \frac{1}{M^{2i -k}}M \sum_j \frac{1}{M^2} Y_z(M^k t) \\
= c \frac{1}{M^{2i -k}}M \frac{M^2}{M^2-1} Y_z(M^k t)\\
= c \frac{1}{M^{2i -k}}\frac{M^3}{M^2-1} Y_z(M^k t)  $$
As for z $d(z,x)\leq k$ we have
$$sum_{l = {1}}^\infty \frac{1}{M^{l-1}} \frac{1}{M^{l+k}} Y_z(M^k t)\\
= c \frac{1}{M^{2 +k}} M \sum_j \frac{1}{M^2} Y_z(M^k t) \\
= c \frac{1}{M^{2 -k}} M \frac{M^2}{M^2-1} Y_z(M^k t)\\
= c \frac{1}{M^{2 -k}}\frac{M^3}{M^2-1} Y_z(M^k t) \\
= c \frac{M}{M^2-1} \frac{1}{M^{ -k}}Y_z(M^k t) \\ $$
summing on $z : \|z-y\|\leq k$ we arrive at 
$$\frac{M}{M^2-1} Y^{k}_x(t) $$
Finally  add this value with the one we had before (remeber they have opposite signs:)
$$c \big(\frac{M}{M^2-1}  - \frac{M}{M-1}\big) Y^{k}_x(t) = c \big(\frac{M - M(M+1)}{M^2-1} \big) Y^{k}_x(t)  = - c \big(\frac{M^2}{M^2-1} \big) Y^{k}_x(t)$$
Therefore if we put on a table, the values we found for $Y_z$ $\|z-x\|>k$ and $Y^{[k]}_x(t)$ we arrive at:
\begin{align}
&\text{terms} \quad& z: d(z,x)>k & \qquad Y^{[k]}_x(t)\\ \hline
&(1)&\quad c\frac{1}{M^{2i -k}}\frac{M}{M +1}&\quad-c\big(\frac{1}{M+1}\big)\\\hline
&(2)&\quad c\frac{1}{M^{2i -k}}\frac{M^3}{M^2-1}& \quad -c\big(\frac{M^2}{M^2-1}\big)
\end{align}
So to pass from $(1)$ to $(2)$ we need to multiply drift term by $\frac{M^2}{M-1}$ which means that we should have
$$ d Y^{[k]}_x(t)  = c\sum_{l \in \Bbb{N}} \frac{M+1}{M^2}\frac{1}{M^{l-1}} \big[Y_x^{[k+l]}(M^{-l}t) - Y_x^{[k]}(t)\big]\, dt\\
+ \sqrt{\frac{1}{M^k}\sum_{y : \|y-x\|\leq k} g(Y_y(M^k))} dW_x^{[k]}(t)$$
A: the factor you need to multiply is exactly $N_M$ the normalizing constant of the $M$-hiearchical lattice.
Note that if you redefine your transition probability to be transition rates, by dropping the normalizing constant, equation (1) beecomes 
$$\tag{1'}
d Y^{[k]}_x(t) = \frac{M^2}{M-1}\sum_{i=k+1}^\infty\sum_{z : \|z-y\| = i}  c \frac{M}{M+1} \frac{1}{M^{2i-k}} [Y_z(M^k t) - Y^{[k]}_x( t)] $$
This indicates that to get the desired block differential equations, one should not normalize the transition kernel, that is, instead of $p(x,y)$ as above, we should consider
$$\tag{3}r(x,y) = \sum_{k\geq \|x-y\|} \frac{1}{M^{2k-1}}. $$
Further more, the deduction above is not completely satisfying, one after reaching a certain point in your computations, you start using the desired equation to check if they indeed match. A better solution would be to go from $(1)$ to $(2)$ in a series of deductions.
To do this, it may be interesting to retain the original expression of $r(x,y)$. That is, don't sum the terms in $(3)$.
The computations will look like this:
$$d Y^{[k]}_x(t) = \sum_{y : \|y-x \|\leq k} d Y_y(M^k t)\\
=\sum_{y : \|y-x \|\leq k}\sum_{z \in \Omega_M}  cr(y,z) [Y_z(M^k t) - Y_y(M^k t)]\\
=\sum_{y : \|y-x \|\leq k}\sum_{l=0}^\infty\sum_{z: \|z-y\|=l}  cr(y,z) [Y_z(M^k t) - Y_y(M^k t)]\\
=\sum_{y : \|y-x \|\leq k}\sum_{l=0}^\infty\sum_{z: \|z-y\|=l}  c\sum_{j\geq l} \frac{1}{M^{2j-1}} [Y_z(M^k t) - Y_y(M^k t)]\\
$$
Now look at the term
$$\sum_{l=0}^\infty\sum_{z: \|z-y\|=l}  \sum_{j\geq l} \frac{1}{M^{2j-1}}\\
=\sum_{l=0}^\infty\sum_z 1_{\{z: \|z-y\|=l\}} \sum_{j = 0}^\infty  1_{\{j\geq l\}} \frac{1}{M^{2j-1}}\\
=\sum_{j = 0}^\infty \sum_{l=0}^\infty\sum_z 1_{\{z: \|z-y\|=l\}}   1_{\{l\leq j\}} \frac{1}{M^{2j-1}}\\
=\sum_{j = 0}^\infty \sum_{z\in B_j(y)}   \frac{1}{M^{2j-1}}\\$$
Where $B_j(x) = \{z: d(z,y)\leq j\} $ 
$$\sum_{l=0}^\infty\sum_z 1_{\{z: \|z-y\|=l\}}   1_{\{l\leq j\}}= 1_{B_j(y)} $$
Coming back to our equation, we may write:
$$d Y^{[k]}_x(t) =c\sum_{y : \|y-x \|\leq k}\sum_{l=0}^\infty\sum_{z: \|z-y\|=l}  \sum_{j\geq l} \frac{1}{M^{2j-1}} [Y_z(M^k t) - Y_y(M^k t)]
\\
=c\sum_{y : \|y-x \|\leq k}\sum_{j = 0}^\infty \sum_{z\in B_j(y)}   \frac{1}{M^{2j-1}} [Y_z(M^k t) - Y_y(M^k t)]\\
=c\big(\sum_{j = 0}^k + \sum_{j = k+1}^\infty\big) \sum_{y\in B_k(x)}  \sum_{z \in B_j(y)} \frac{1}{M^{2j-1}} [Y_z(M^k t) - Y_y(M^k t)]\\
$$
where in the last line we split the sum in $j$ in two parts. Let's consider the first one:
$$ \sum_{j = 0}^k  \sum_{y\in B_k(x)}  \sum_{z \in B_j(y)} \frac{1}{M^{2j-1}} [Y_z(M^k t) - Y_y(M^k t)]\\
=\sum_{j = 0}^k  \sum_{y\in B_k(x)}  \sum_{z \in B_k(x)} 1_{d(z,y)\leq j} \frac{1}{M^{2j-1}} [Y_z(M^k t) - Y_y(M^k t)] = 0$$
where the last equality follows from the symmetry in $z$ and $y$ of the expressions.
Now what we are left with is 
$$d Y^{[k]}_x(t) =c \sum_{j = k+1}^\infty \sum_{y\in B_k(x)}  \sum_{z \in B_j(y)} \frac{1}{M^{2j-1}} [Y_z(M^k t) - Y_y(M^k t)]\\
=c \sum_{j = k+1}^\infty \sum_{y\in B_k(x)}  \sum_{z \in B_j(x)} \frac{1}{M^{2j-1}} [Y_z(M^k t) - Y_y(M^k t)]\\
=c \sum_{j = k+1}^\infty  \sum_{z \in B_j(x)} \sum_{y\in B_k(x)}  \frac{1}{M^{2j-1}} [Y_z(M^k t) - Y_y(M^k t)]\\
=c \sum_{j = k+1}^\infty  \sum_{z \in B_j(x)} M^k  \frac{1}{M^{2j-1}} [Y_z(M^k t) - Y^{[k]}_x(t)]\\
=c \sum_{j = k+1}^\infty  M^k  \frac{1}{M^{2j-1}} \sum_{z \in B_j(x)} [Y_z(M^k t) - Y^{[k]}_x(t)]\\
=c \sum_{j = k+1}^\infty  M^k  \frac{1}{M^{2j-1}} M^j [Y^{[j]}_z(M^{k-j} t) - Y^{[k]}_x(t)]\\
=c \sum_{j = k+1}^\infty   \frac{1}{M^{j-k-1}}  [Y^{[j]}_z(M^{k-j} t) - Y^{[k]}_x(t)]\\
=c \sum_{j = 1}^\infty   \frac{1}{M^{j-1}}  [Y^{[j]}_z(M^{-j} t) - Y^{[k]}_x(t)]\\
$$
which is exactly what we wanted to prove.
