Solving recursive integral equation from Markov transition probability How do I solve something like:
$$f(x) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty e^{\frac{-(y - x/2)^2}{2}}f(y)\:\mathrm{d}y$$
for $f(x)$?
Is there also a general formula that this falls under? The closest thing I found was the Fredholm integral equation, but those (I believe) assume that the eigenfunction is linear.
In case it helps, here's my motivation for this problem: I'm trying to find the stationary distribution of a continuous state, discrete time Markov process. I came up with this transition function:
$$P(s_i = x | s_{i-1} = y) = p(x,y) = \frac{1}{\sqrt{2 \pi}} e^{\frac{-(x - y/2)^2}{2}}$$
The idea is that the probability function always brings the mean closer towards $0$ by using a Normal distribution with mean $\mu = y/2$. I then tried to solve the stationary distributions as follows
$$\pi_x = \int_{-\infty}^{\infty}p(y, x)\pi_y \:\mathrm{d}y$$
At this point, I'm stuck. (I got the original equation by setting $\pi_x = f(x)$ for clarity.) This might deserve its own questions, but does solving for stationary distributions have its own technique?

We are given a discrete step continuous state transition system from state $y$ to state $x$ as
$$M[y, x] = \frac 1 {\sqrt{2\pi}} e^{-\frac 12(x - y/2)^2}$$
which has the following characteristics 


*

*If you are at $x$, your destination is a normal curve from $-\infty$ to $\infty$ centered at $x/2$ with a variance of $1$


So you can see that in this system, you are statistically going to approach $0$ with some variance.  Define $*$ as
$$(M*N)[y, x] = \int_{-\infty}^{\infty} M[y, u]\cdot N[u, x] du$$
The steady state transition matrix is given by $M * M * M \dots = M^{\infty}$.
We can tell right away that 
$$\begin{align}
M^2[y, x] &= \int_{-\infty}^{\infty} M[y, u]\cdot M[u, x] du \\
          &= \int_{-\infty}^{\infty}
 \left(\frac 1 {\sqrt{2\pi}} e^{-\frac 12(u - y/2)^2}\right) 
 \left(\frac 1 {\sqrt{2\pi}} e^{-\frac 12(x - u/2)^2}\right) du \\
          &= \sqrt{\frac{2}{5\pi}} e^{-\frac 1{40}y^2 - \frac 15 xy + \frac 25 x^2}
\end{align}$$
Which is a normal curve peaking at $x = y/4$.  Similarly,
$$M^4[y, x] = \frac 45 \sqrt{\frac{10}{17\pi}}e^{-\frac 1{680}y^2 + \frac 4{85} xy - \frac {32}{85} x^2}$$
which is a normal curve peaking at $x = y/16$.  So each step of the transition does indeed move the distribution closer to a normal curve peaking at $x = 0$.  Define ansatz :
$$M^{\infty}[y, x] = N[y, x] = Re^{-Ax^2}$$
with the constraint that $N$ is stochastic so $\int_{-\infty}^{\infty} N[y, x] dx = 1 \quad \Rightarrow \quad \sqrt{\pi}R = \sqrt{A}$.  We wish to find:
$$\begin{align}
N &= N*M \\
Re^{-Ax^2} &= \int_{-\infty}^{\infty} Re^{-Au^2} \cdot \frac 1 {\sqrt{2\pi}} e^{-\frac 12(x - u/2)^2} du \\
\sqrt{\frac{A}{\pi}}e^{-Ax^2} &= \sqrt{\frac{A}{\pi}} \cdot \frac 1 {\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-Au^2} \cdot e^{-\frac 12(x - u/2)^2} du \\
\sqrt{2\pi}e^{-Ax^2} &= \sqrt{\frac{8\pi}{8A + 1}} e^{-4Ax^2/(8A + 1)} \\
A &= \frac 38
\end{align}$$
So your final steady state is going to be
$$M^\infty[y, x] = \sqrt{\frac{3}{8\pi}}e^{-\frac 38 x^2}$$
