I've encountered the following equation:
$p_t = \pi_t + c \int_{-\infty}^te^{-\gamma(t-\tau)}dp_\tau, \quad 0<\gamma, 0<c<1$.
It is claimed that it can be rewritten as follows:
$p_t = \pi_t + \frac{c}{1-c}(\pi_t-\bar\pi_t)$,
where,
$\bar\pi_t = \lambda\int_{-\infty}^te^{-\lambda(t-\tau)}\pi_\tau d\tau, \quad \lambda = \frac{\gamma}{1-c}$
Do you have an idea why this is the case?
Any help is appreciated,
Let's first differentiate the first equation with respect to $t$: \begin{align} \dot{p}_t&=\dot{\pi}_t+c\dot{p}_t-\gamma c\int_{-\infty}^te^{-\gamma(t-\tau)}dp_{\tau} \\ &=\dot{\pi}_t+c\dot{p}_t-\gamma(p_t-\pi_t). \tag{1} \end{align} Rearranging terms, we obtain $$ (1-c)\dot{p}_t+\gamma p_t=\dot{\pi}_t+\gamma\pi_t \implies \dot{p}_t+\lambda p_t=\frac{\dot{\pi}_t}{1-c}+\lambda\pi_t. \tag{2} $$ The general solution to the above ODE is $$ p_t=e^{-\lambda t}\left\{A+\int_{-\infty}^t e^{\lambda\tau}\left(\frac{\dot{\pi}_{\tau}}{1-c}+\lambda\pi_{\tau}\right)d\tau \right\}. \tag{3} $$ Integrating by parts the first term of the integral yields \begin{align} p_t&=e^{-\lambda t}\left\{A+\frac{e^{\lambda t}\pi_t}{1-c}+\int_{-\infty}^t e^{\lambda\tau}\left(-\frac{\lambda}{1-c}+\lambda\right)\pi_{\tau}\,d\tau \right\} \\ &=Ae^{-\lambda t}+\frac{\pi_t}{1-c}-\frac{c\lambda}{1-c}\int_{-\infty}^t e^{-\lambda(t-\tau)}\pi_{\tau}\,d\tau \\ &=Ae^{-\lambda t}+\pi_t+\frac{c}{1-c}\left(\pi_t-\lambda\int_{-\infty}^t e^{-\lambda(t-\tau)}\pi_{\tau}\,d\tau \right). \tag{4} \end{align} The first term on the RHS of $(4)$ can be eliminated if we impose the condition that $p=0$ if $\pi=0$. Then, using the definition of $\bar{\pi}_t$, we finally arrive at $$ p_t=\pi_t+\frac{c}{1-c}(\pi_t-\bar{\pi}_t).\quad{\square} \tag{5} $$
$\int_{-\infty}^t e^{-\gamma(t-\tau)}dp_\tau = \int_{-\infty}^t e^{-\gamma(t-\tau)}\dot p _\tau d\tau$
but I haven't found anything that solves my problem
– Mik92 May 15 '22 at 10:15