I need to solve this equation to find the cumulative distribution function of a random variable $$f(t) = e^{\lambda ( \int_{0}^{t}e^{-x} f(t-x) dx -1)}$$ where $f$ defined on $[0, \infty)$ is countinuous, increasing and satisfies $f(\infty)=1$
Does anyone can recommend a method or a book that can lead to solve this kind of equations? Thanks in advance.
The integral expression is a convolution, which is commutative, so you can write.
$$ \int_0^t e^{-x}f(t-x) dx = e^{-t}*f(t)_{t\in[0,\infty)} = \int_0^t e^{-(t-x)}f(x) dx $$
Note from the equation that $f(0) = e^{-\lambda}$
Take the log, and differentiate both sides
$$ \frac{f'(t)}{f(t)} = \lambda\left[f(t) -\int_0^t e^{-(t-x)}f(x)dx \right] = \lambda f(t) -\ln(f(t)) - \lambda $$
Substitute $g(t) = -\ln(f(t))$ and rewrite
$$ g'(t) + g(t) = \lambda\big(1-e^{-g(t)}\big) $$
This is now an eigenvalue problem with boundary conditions $g(0)=\lambda, g(\infty)=0$
Since the equation is separable, you may write the inverse function as
$$ t = g^{-1}(y) = \int_{\lambda}^y \frac{1}{\lambda(1-e^{-s})-s}ds $$
Which appears to satisfy the other B.C. when $\lambda\in(0,1]$