$\textbf{Problem}$ Let $Y_1,Y_2,\cdots$ be independent and identically distributed with \begin{align*} &P[Y_n=0]=\alpha\\ &P[Y_n>y]=(1-\alpha)e^{-y} \end{align*}
Define the random variables $X_n, n\geq 0$ by \begin{align*} &X_0=0\\ &X_{n+1}=\alpha X_n+Y_{n+1} \end{align*}
Prove that \begin{align*} &P[X_n=0]=\alpha^n\\ &P[X_n>x]=(1-\alpha^n)e^{-x} \end{align*}
$\textbf{Attempt}$ Firstly, I found $X_n=\alpha^{n-1}Y_1+\cdots+\alpha Y_{n-1}+Y_n$.
If $X_n=0$ , then $Y_1=Y_2=\cdots =Y_n=0 $ since $Y_i \geq 0 $ for all $1\leq i \leq n$.
Thus, I got $P[X_n=0]=\alpha^n$.
However, I stuck how to $P[X_n>x]$ calculate...
Any help is appreciated... Thank you!
$\textbf{Update}$ I'll use induction. If n=1, we can easily check that $P[X_n>x]=(1-\alpha^n)e^{-x}$
\begin{align*} P[X_{n+1}>x] &= P[\alpha X_n+Y_{n+1}>x]\\ &=P[Y_{n+1}=0]P[\alpha X_{n}>x]+\int_{0}^x P[Y_{n+1}=t] P[\alpha X_n>x-t]dt +\int_x^{\infty} P[Y_{n+1}=t]dt \\ &(\textrm{because $ \alpha X_n +Y_{n+1}>x $ if $ Y_{n+1}>x $)}\\ &=\alpha(1-\alpha^n)e^{-\frac{x}{\alpha}}+\int_{0}^x (1-\alpha)e^{-t} (1-\alpha^n)e^{-\frac{x-t}{\alpha}}dt +\int_{x}^{\infty} (1-\alpha)e^{-t}dt \\ &=\alpha(1-\alpha^n)e^{-\frac{x}{\alpha}}+\alpha(1-\alpha^n)(e^{-x}-e^{-\frac{x}{\alpha}})+(1-\alpha)e^{-x}\\ &=(1-\alpha^{n+1}) e^{-x} \end{align*}
I'm not sure my proof is right.....
Could you check my error in my proof?
An alternative strategy notes that $Y_n$ has characteristic function $\alpha+\frac{1-\alpha}{1-it}=\frac{1-\alpha it}{1-it}$, so $X_n$ has characteristic function $\prod_{j=1}^n\frac{1-\alpha^{n-j+1}it}{1-i\alpha^{n-j}t}=\frac{1-\alpha^nit}{1-it}$ as required.