Let $(W_t)_{t\geq 0}$ be a standard one-dimensional Brownian motion and let
$$ Y_t := e^{-t}W(e^{2t}), \qquad X_t := \int^t_0 Y_s ds $$
Show that the joint process $(X_t,Y_t)$ is Markovian and find the generator of the process.
This is an exercise that I came across while reading a book. I can show the first part where the joint process is indeed Markovian. However, I don't know how to find the generator
$$ L[f](0,x) := \lim_{t \rightarrow 0}\frac{E_{(0,x)}[f(X_t,Y_t)] -f(0,x)}{t}. $$
for $f$ sufficiently regular.
My first thought is to use Ito, but I only know the form of $f(t,X_t)=...$ not the form $f(X_t,Y_t)=...$ (if there is any) and my second thought is to use the joint density but don't know how to proceed to find the joint density for now.
To find the generator I will look for the two-dimensional SDE that is solved by $(X_t,Y_t)\,.$
It is not hard to see that $$\tag{1} B_t:=\int_0^te^{-s}\,dW_{e^{2s}} $$ is a continuous martingale with quadratic variation $t\,.$ Therefore is is a Brownian motion that satisfies $$\tag{2} Y_t=e^{-t}\,W_{e^{2t}}=e^{-t}\int_0^te^u\,dB_u=\int_0^te^{u-t}\,dB_u\,. $$
It is easy to see that the SDE that is solved by $Y_t$ is the one of an Ornstein-Uhlenbeck process: $$\tag{3} dY_t=-Y_t\,dt+dB_t\,. $$ By definition, $$\tag{4} dX_t=Y_t\,dt\,. $$ The system we were looking for is (3) and (4).
This can be written in matrix/vector form as $$\tag{6} d\mathbf{Z}_t=\underbrace{\pmatrix{0&1\\0&-1}}_{\textstyle=:\mathbf{b}}\,\mathbf{Z}_t\;dt+\underbrace{\pmatrix{0&0\\1&0}}_{\textstyle=:\boldsymbol{\sigma}}\,d\mathbf{B}_t $$ where $$\tag{7} \mathbf{Z}_t=\pmatrix{X_t\\Y_t}\,,\quad \mathbf{B}_t=\pmatrix{B_t\\C_t} $$ and $C_t$ is a dummy Brownian motion independent of $B\,.$
Since $$\tag{8} \boldsymbol{\sigma}\boldsymbol{\sigma}^\top=\pmatrix{0&0\\0&1}=:\mathbf{a} $$ the generator of the Markov process $(X_t,Y_t)$ is $$\tag{9} A=\left(\mathbf{b}{x\choose y}\right)\cdot\nabla+\frac12\nabla^\top\mathbf{a}\nabla=y\,\partial_x-y\,\partial_y+\frac12\partial_y^2\,. $$