Question on generators in the proof of Kolmogorov's Backward Equation 
Here is a part of the proof of the Kolmogorov's Backward Equation. I cannot see why $Y_t$ has been picked as it has. In particular, I cannot see why you would want to subtract t in the first bit of the bracket. 
Furthermore I cannot see how $Y_t$ has generator $\tilde{A}$. The definition does not seem like it could be easily applied. I can see that $\tilde{A} \omega := -1 + A \omega$, but cannot see how this would be further evaluated.
 A: Let $(Z_t)_{t \geq 0}$ be a Markov process with generator $B$. Then we know from Dynkin's formula that
$$u(Z_t)-u(Z_0)- \int_0^t Bu(Z_s) \, ds$$
is a martingale if $u$ is "nice" (e.g. bounded and in the domain of the generator). In particular, if $u$ satisfies $Bu=0$, then we see that
$$\mathbb{E}^xu(Z_t) = \mathbb{E}^x u(Z_0) = u(x).$$
This means the following: Given any function $u$ satisfying $Bu=0$ for some operator $B$. Now if we succeed in finding an stochastic process $(Z_t)_t$ with generator $B$, then $u(x) = \mathbb{E}^x u(Z_t)$. In particular, the solution to $Bu=0$ is unique and we have an explicit representation formula.

That's exactly the idea of the proof: By assumption, $w$ satisfies $\tilde{A}w=0$ where
$$\tilde{A}w := - \frac{\partial}{\partial t} w + Aw$$
and $A$ is the generator of a Markov process $(X_t^{0,x})_{t \geq 0}$. Applying the above idea, we would like to find a (Markov) process $(Y_t)_{t \geq 0}$ such that $Y$ has generator $\tilde{A}$. If we define
$$Y_t := (s-t,X_t^{0,x})$$
it is not difficult to see that this is the case. Indeed: By definition, the generator $B$ of $Y$ is given by
$$Bu(s,x) = \lim_{t \to 0} \frac{\mathbb{E}^{s,x} u(Y_t)-\mathbb{E}^{s,x} u(Y_0)}{t} = \lim_{t \to 0} \frac{\mathbb{E}^x u(s-t,X_t^{0,x})-u(s,x)}{t}.$$
Hence,
$$\begin{align*} Bu(s,x) &= \lim_{t \to 0} \frac{\mathbb{E}^{x} u(s-t,X_t^{0,x})-\mathbb{E}^{x} u(s,X_t^{0,x})}{t}+ \lim_{t \to 0} \frac{\mathbb{E}^x u(s,X_t^{0,x}) - u(s,x)}{t}. \end{align*}$$
It follows from the very definition of the generator that the second term converges to $Au(s,x)$. For the term it is not difficult to see that
$$ \lim_{t \to 0} \frac{\mathbb{E}^x u(s-t,X_t^{0,x})-\mathbb{E}^x u(s,X_t^{0,x})}{t} = - \frac{\partial}{\partial s} u(s,x)$$
if the limit exists (i.e. if $u$ is differentiable with respect to $s$). Consequently, $B = \tilde{A}$.
