I have the following equation to solve for $f(y)$: \begin{equation} f(y)=(y-N\langle f(y)\rangle)^{-1} \end{equation} Where $N$ is a fixed integer, and $y$ is a random variable. $\langle \cdot\rangle$ denotes an average over the distribution of $y$.
If $y$ is deterministic then $\langle f(y)\rangle=f(y)$ and the equation simply becomes: \begin{equation} f(y)=(y-N f(y))^{-1}\implies -Nf(y)^2+f(y)y-1=0 \end{equation}
However when $y$ actually follows a random distribution, it spices things up. In the most simple case, when $y$ can only take two values $a$ or $b$ how can I solve the equation? i.e
when $P(Y=y)=p\delta(a-y)+(1-p)\delta(b-y)$ with $p\leq 1$
how can we solve : \begin{equation} f(y)=(y-N\langle f(y)\rangle)^{-1} \end{equation}
You can solve it for a $y$ that takes a finite number of values, say $y_1,y_2,\ldots, y_r$. Put $p_k=P(Y=y_k)$ and let $\lambda\in{\mathbb R}$. Let $X$ be the random variable defined by $X=\frac{1}{Y-N\lambda}$.
We will have a solution to our problem iff ${\mathbb{E}}(X)=\lambda$, or in other words,
$$ \sum_{k=1}^{r} \frac{p_k}{y_k-N\lambda} = \lambda \tag{1} $$
Assuming the denominator is nonzero, this is equivalent to
$$ y_r = N\lambda +\frac{p_r}{\lambda - \sum_{k=1}^{r-1} \frac{p_k}{y_k-N\lambda} } \tag{2} $$