My question is; how does expansion of the conditional probability density $P_{1|1}(y_1,t_1|y_2,t_1+\tau)$ get put into the form of(6.23) of Reichl$^1$?
The term with the minus sign at the front, seems a bit mysterious, could some explanation of it be given, as part of an answer?
Reichl says, I quote
Let us expand the conditional probability density $P_{1|1}(y_1,t_1|y_2,t_1+\tau)$ in a Taylor series for small $\tau$, in such a way that to each order in $\tau$ the normalization of $P_{1|1}(y_1,t_1|y_2,t_1+\tau) $ (cf. Eq. 6.13) is preserved. If we note that $P_{1|1}(y_1,t_1|y_2,t_1)=\delta(y_1-y_2)$ (cf. Eq. 6.12) we obtain \begin{align} P_{1|1}(y_1,t_1|y_2,t_1+\tau)&=\delta(y_1-y_2)-\tau\int dy W_{t_1}(y_1,y)\delta(y_1-y_2) \\ &+\tau W_{t_1}(y_1,y_2)\tag{6.23}\end{align}
End of quote.
NB: (6.12) and (6.13) are as follows
$$P_1(y_2,t_2) =\int P_1(y_1,t_1) P_{1|1}(y_1,t_1|y_2,t_2) dy_1 \tag{6.12}$$
$$\int P_{1|1}(y_1,t_1|y_2,t_2) dy_2=1 \tag{6.13} $$
Reference:
1) Reichl, L.E., A Modern Course in Statistical Physics, Arnold, London, 1980.
Other Info
I thought I might encourage answers if I first provided some of the hard grind.
I must say typing up math equations is hard work , it's a pity I've no secretary.
First some $\delta$ function material. The defining property, the $f(0)$ property, is, for any continuous $f(x)$ $$ \int^\infty _{-\infty}f(x) \delta (x) dx = f(0)$$ Also,
\begin{align} \int^\infty _{-\infty}f(x) \delta (x-x_0) dx &= f(x_0) \nonumber \\ \int^\infty _{-\infty}f(x,y) \delta (x-x_0) dx &= f(x_0,y) \nonumber \end{align} With similar results for functions with more arguments.
A function with four arguments, two taking fixed values, like with $P_{1|1}(y_1,t_1|y_2,t_1)$ can be represented as a new function, with only two arguments, let us in this case, call this function $h(y_1,y_2)$ with rule $$h(y_1,y_2)=P_{1|1}(y_1,t_1|y_2,t_1)$$
So \begin{align} h(y_1,y_2)&=0 , y_1\neq y_2 \nonumber \\ \int^\infty _{-\infty} h(y_1,y_2) dy_2 &= 1 \nonumber \\ \int^\infty _{-\infty} h(y_1,y_2) f(y_2)dy_2 &= f(y_1) \int^\infty _{-\infty} h(y_1,y_2) dy_2 &= f(y_1) \nonumber \end{align} We have that, $h(y_1,y_2)$ behaves like $\delta (y_1-y_2)$ in an integral, so we put $$h(y_1,y_2)=\delta (y_1-y_2)=P_{1|1}(y_1,t_1|y_2,t_1) \tag{1}$$
The above material explains about $P_{1|1}(y_1,t_1|y_2,t_1)$ and part of it may also be useful latter.
Taylor expanding $P_{1|1}(y_1,t_1|y_2,t_1+\tau)$ to first order \begin{align} P_{1|1}(y_1,t_1|y_2,t_1+\tau)&= P_{1|1}(y_1,t_1|y_2,t_1) +\tau \left. \frac{\partial P_{1|1}(y_1,t_1|y_2,t)}{\partial t} \right|_{t=t_1} \nonumber\\ &+ f( y_1,t_1,y_2,\tau) \nonumber \end{align} where $ f( y_1,t_1,y_2,\tau)$ is a correction term to ensure normalisation in the sense of (6.13)
$$\int P_{1|1}(y_1,t_1|y_2,t_1 +\tau) dy_2=1 $$
Let $$ \left. \frac{\partial P_{1|1}(z_1,t_1|z_2,t)}{\partial t} \right|_{t=t_1}= W_{t_1}(z_1,z_2) $$ Then
\begin{align} \left. \frac{\partial P_{1|1}(y_1,t_1|y_2,t)}{\partial t} \right|_{t=t_1}= W_{t_1}(y_1,y_2) \nonumber\\ \left. \frac{\partial P_{1|1}(y_1,t_1|y,t)}{\partial t} \right|_{t=t_1}= W_{t_1}(y_1,y) \nonumber \end{align}
Hence our Taylor expansion may be written as
\begin{align} P_{1|1}(y_1,t_1|y_2,t_1+\tau)&=\delta(y_1-y_2)+\tau W_{t_1}(y_1,y_2) \nonumber\\ &+ f( y_1,t_1,y_2,\tau) \tag{2} \end{align} Somehow, we need to prove/justify that $$ f( y_1,t_1,y_2,\tau)= - \tau \int^\infty _{-\infty}dy W_{t_1}(y_1,y) \delta (y_1-y_2) \tag{3} $$
Please note that I do not regard the , preserving the normalization idea, as a fundamental way of deriving the 'Master Equation', getting to this equation is used as a stepping stone in Reichl$^1$ for getting to the Fokker-Planck equation, here is an "answer", that may be in need of correction. People who read the book could be interested in it.
In the 'Other Info' part of the question, it is explained how we may write the first order Taylor expansion of $P_{1|1}(y_1,t_1|y_2,t_1+\tau) $ as
\begin{align} P_{1|1}(y_1,t_1|y_2,t_1+\tau)&=\delta(y_1-y_2)+\tau W_{t_1}(y_1,y_2) \nonumber\\ &+ f( y_1,t_1,y_2,\tau) \nonumber \end{align}
for some $f$.
$f$ must preserve normalization in the sense
$$\int^\infty _{-\infty} \left[\delta(y_1-y_2)+\tau W_{t_1}(y_1,y_2) + f( y_1,t_1,y_2,\tau) \right] dy_2=1$$
Now $$\int^\infty _{-\infty} \delta(y_1-y_2) dy_2=1$$ so we must have an $f$ that satisfies $$\int^\infty _{-\infty} f( y_1,t_1,y_2,\tau) dy_2= -\int^\infty _{-\infty} \tau W_{t_1}(y_1,y_2) dy_2$$
With the choice
$$ f( y_1,t_1,y_2,\tau)= - \tau \int^\infty _{-\infty}dy W_{t_1}(y_1,y) \delta (y_1-y_2) $$
we have \begin{align} \int^\infty _{-\infty} f( y_1,t_1,y_2,\tau) dy_2&= - \int^\infty _{-\infty} \tau \int^\infty _{-\infty} W_{t_1}(y_1,y) \delta (y_1-y_2) dy dy_2 \nonumber \\ &= - \int^\infty _{-\infty} \tau W_{t_1}(y_2,y) dy \nonumber \end{align}
We see that we must have
$$\int^\infty _{-\infty} \tau W_{t_1}(y_1,y_2) dy_2 = \int^\infty _{-\infty} \tau W_{t_1}(y_2,y) dy$$
Or $$\int^\infty _{-\infty} \tau W_{t_1}(y_1,y) dy = \int^\infty _{-\infty} \tau W_{t_1}(y_2,y) dy$$
So actually, the book looks to be incorrect.