I'm having trouble proving a result I think should be true for the insurance payout model.
Say you have $U_n = x + c*n - X_n$
where $X_n = X_{n-1} + Y_n$ and the $Y_i$ are all iid $Y_i \sim \text{Exp}(1)$ and $X_0 = 0$
How do I show that if $c\le E[Y_i]=1$ then there is a 100% probability that there exists an $m$ such that $U_m<0$. I.e. The probability of ruin is 1.
My first thought was defining a function $f(x)$ such that $f(x)$ is the probability of ruin for the process given that the starting capital is $x$ (as in the definition for $U_n$).
First I can observe that $f(x)=1$ for all $x<0$. Then I can observe that the probability of ruin for $f(x)$ has a probability given by
\begin{align}
f(x)&=\int_0^{\infty}e^{-u}f(x+1-u) du\\
f(x)&=\int_0^{x+1}e^{-u}f(x+1-u) du + \int_{x+1}^{\infty}e^{-u} du
\end{align}
where in the second step I used the fact that $f(x)=1$ for all $x<0$.
I then yield
\begin{align}
f(x)&=\int_0^{x+1}e^{-u}f(x+1-u) du + e^{-x-1}\\
f'(x)&=e^{-(x+1)}f(x+1-(x+1)) - e^{-(0)}f(x+1-(0)) - e^{-x-1}\\
f'(x)&=e^{-x-1}f(0) - f(x+1) - e^{-x-1}\\
f(x+1)&=e^{-x-1}f(0) - f'(x) - e^{-x-1}\\
f(x)&=e^{-x}f(0) - f'(x-1) - e^{-x}\\
f(x)&=e^{-x}(f(0)-1) - f'(x-1)
\end{align}
I can solve this final equation for $0<x<1$ by using the fact that $f'(x-1)=0$ for $0<x<1$
I can then solve this equation for $1<x<2$ by using the fact that I can find the derivative of the equation for $0<x<1$
I can keep doing this. But this gives me quite nonsensical results. What am I doing wrong and how else can I solve this problem?
Any pointers/help greatly appreciated