Why does propagator for SDE involve response variable? Being trained as a mathematician I am trying to understand stochastic field theory (i.e. field theory applied for dynamics of  stochastic processes, e.g. SDE) and I have a difficulty at one point.
Suppose I have an Ornstein-Uhlenbeck process
$$ \dot{x}(t) + ax(t) - \sqrt{D} \eta(t) = 0$$
which has  probability density
$$ P(x) =  \int \mathcal{D} \tilde{x} e^{S[x,\tilde{x}]} ,$$
$$S[x,\tilde{x}]=\int \left( \tilde{x}(t)\bigl(\dot{x}(t)+ax(t)-y\delta(t-t_0)\bigr) -\frac{D}{2} \tilde{x}^2(t) \right) dt $$
The variable $\tilde{x}(t)$ is called auxiliary or 'response' variable (for reasons that I don't understand, unfortunately).
In several sources I saw written that linear response has the form $\langle x(t),\tilde{x}(t')\rangle .$ 
Question: why does the (quite artificial) auxilary variable has something to do with linear response? In my understanding the words 'linear response' mean the size of the system $x(t)$ response to a perturbation, so I don't understand both 1) for what reason $\tilde{x}$ appears there 2) why the formula has a form of the second moment, instead of someting like $\langle \frac{\delta x(t)}{\delta\eta(t)} \rangle$.
More precisely I am trying to understand this article.
 A: Quick rush to convert the following Ornstein-Uhlenbeck process (in the presence of a external field to probe position of the particle) :
$$\frac{d}{dt}x(t)+ax(t)+h(t)-\sqrt{D}\eta(t)=0$$
with gaussian white noise with mean and variance :
$\langle\eta(t)\rangle=0$ and $\langle \eta(t) \eta(t')\rangle=\delta(t-t')$ to a path integral is :
$$P[\{x(t)\}]=\int D[\{x'(t)\}]\langle\delta\big[x(t)-x'[\eta(t)]\big]\underbrace{\mathcal{J}\delta\big[\frac{d}{dt}x'(t)+ax'(t)+h(t)-\sqrt{D}\eta(t)\big]}_{\text{Constraining the paths to those satisfying langevin equation }}^{}\rangle\big|_{\eta(t)},$$ with the functional Jacobian within Ito convention being equal to one ($\mathcal{J}=1$).
Now, represent the second $\delta$ functional using an auxillary field $\tilde{x}(t)$ and get :
$$P[\{x(t)\}]=\int \int D[\{x'(t)\}] D[\{\tilde{x}(t')\}]\langle\delta\big[x(t)-x'[\eta(t)]\big]e^{i\int dt'\tilde{x}(t')\big[\frac{d}{dt'}x'(t')+ax'(t')+h(t')-\sqrt{D}\eta(t')\big]}\rangle\big|_{\eta(t)}.$$
Perform average over the noise and perform the $x'$ functional integral to get :
$$P[\{x(t)\}]=\int D[\{\tilde{x}(t)\}]e^{i\int dt \tilde{x}(t)\big[\frac{d}{dt}x(t)+ax(t)+h(t)\big]-\frac{D}{2}\int dt\tilde{x}(t)^2}.$$
Now coming to main point of the question posed above, consider average of position defined as :
$$\langle x(t) \rangle=\int D[\{x(t')\}]x(t)P[\{x(t')\}].$$ As response of position to external field is defined as :
$$R(t,t')=\frac{\delta\langle x(t) \rangle}{\delta h(t')}\big|_{h(t')=0}.$$ Which upon using the above defined functional integral definition of probability functional in the presence of external field and setting field to zero gives :
$$R(t,t')=i\langle x(t) \tilde{x}(t')|_{h(t')=0}.$$
Note : Extra $i$ is due to the choice OP made in integrating $\tilde{x}(t')$ along imaginary axis by analytic continuation.
