I have the following problem. Assume that $X_t$ are all i.i.d. random variable uniformly distributed between 0.5 and 1.5, and that $\phi$ is a little larger than 1.
$$Y_t= \min(X_t, Y_{t-1} \phi)$$
Which of course can be written as
$$Y_t= \min(X_t, X_{t-1} \phi, Y_{t-2} \phi^2)$$
and therefore
$$Y_t= \min (X_t, X_{t-1} \phi, X_{t-2} \phi^2,X_{t-3} \phi^3\ldots)$$
Is there an elegant way to describe the probability distribution? Or at least calculate expected mean and variance?
$$P(Y_n>y) = P(X_n>y, X_{n-1}\phi>y, \ldots, X_1\phi^{n-1}>y)$$
$$P(Y_n>y) = \prod_{i=1}^{n}P\left(X_i>\max\left(\frac{y}{\phi^{n-i}},0.5\right)\right) = \prod_{i=1}^{n}\left(1-\frac{\max\left(\frac{y}{\phi^{n-i}},0.5\right)-0.5}{1.5-0.5}\right) \\ = \prod_{i=1}^{n}\left(1.5-\max\left(\frac{y}{\phi^{n-i}},0.5\right)\right)$$
Edit: I would just like to attach the following with this answer which might be helpful for OP (or for me to review some useful thoughts related to this question).
If $X_t \sim U(0,1)$, then,
$$P(Y_n>y) = \prod_{i=1}^{n}P\left(X_i>\frac{y}{\phi^{n-i}}\right) = \prod_{i=0}^{n-1}(1-yq^i) = (y;q)_n$$
where $q = \frac{1}{\phi}$ and $(y;q)_n$ is q-Pochhammer symbol.
$$\log P(Y_n > y) = \sum_{i=0}^{n-1}\log(1-yq^i)$$
Taking derivative wrt $y$ gives,
$$\frac{f_{Y_n}(y)}{1-F_{Y_n}(y)} = \sum_{i=0}^{n-1}\frac{q^i}{1-yq^i}$$
which is the hazard function $h_{Y_n}(y) = \frac{f_{Y_n}(y)}{1-F_{Y_n}(y)}$.
