Expected number of rounds for a product of uniform random variables on $[1/2,3/2]$ to be for the first time below a given threshold Starting with w=1, each time we multiply w by a number x sampled independently and uniformly from [1/2, 3/2] until it is smaller than a given value c. What's the expected number of rounds for this process?
I have tried to model it using a function f where f(c) denotes the answer for given c. Using a recursive way, I can derive the following:
$ f(c) = 1, \text{ if } c > \frac{3}{2} $
$ f(c) = 1 + \int_{c}^{\frac{3}{2}} f(\frac{c}{x}) dx, \text{ if } c \in [\frac{1}{2},\frac{3}{2}] $
$ f(c) = 1 + \int_{\frac{1}{2}}^{\frac{3}{2}} f(\frac{c}{x}) dx, \text{ if } c < \frac{1}{2} $
However I'm not able to proceed since it seems complicated.
Addition: I care more about the case where $c$ is around 1.
Thanks for any answers and ideas!
 A: As advised by Ross Millikan, it is wise to convert your multiplicative process into an additive one, replacing a product of Random Variables by a sum of the logarithms of these RVs. One knows a lot more on addition of RVs...
I haven't attempted to do that.
Instead, I have written a program in order to conduct extensive simulations of the average number $u=u(c)$ of "time units" necessary to fall below a certain given value $c \in [1/2,3/2]$ (see Matlab program below)
The obtained results (blue curve) display a rather good fit by a certain $-K * ln$ function (red curve ; with $K=6.85$) but for the initial values of $c$. The fit is especially good in the central region which is of interest for you ; for example, for $c=1$, one obtains $u(c) \approx 4.75$, a rather counter intuitive result...

A: The expected value is going to be given by the infinite sum
$$
E(steps)=\sum_{n=1}^\infty n\cdot P(w<c:\textit{n steps})
$$
where the probability can be given by a binomial distribution.
$$
P(w<c:\textit{n steps}) = \sum_{k=0}\binom{n}{k}\left(\frac{1}{2}\right)^k\left(\frac{3}{2}\right)^{n-k}
$$
summing over all values where
$$
\left(\frac{1}{2}\right)^k\left(\frac{3}{2}\right)^{n-k} < c.
$$
I'll leave it to you to fill in the gaps
