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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!

Jean Marie
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Marco_L_T
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  • If $c$ is very small you can use the normal approximation. Find the mean and variance of the distribution of $\log x$ and you are asking for the number of samples to get the sum less than $\log c$ – Ross Millikan May 19 '22 at 21:00
  • Could you tell the origin of the problem ? I see now how it can be treated rigorously using a discrete approximation and a Markov approach ; but the initial ("continuous") version looks to me rather difficult... – Jean Marie May 20 '22 at 12:55
  • I have taken the liberty to modify your title in order to attract more people. Do you agree ? – Jean Marie May 21 '22 at 08:13

2 Answers2

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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...

enter image description here

Jean Marie
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  • Thanks a lot for the simulation! – Marco_L_T May 20 '22 at 01:53
  • A rather similar issue with a solution here. See as well there – Jean Marie May 21 '22 at 05:27
  • Remark: in the case you consider transforming your multiplicative issue into an additive issue, if we call $U$ the uniformly distributed Random Variable (RV) on $[1/2,3/2]$, the RV $V:=\ln(U)$ has pdf $f(x)=\ln(2)2^x$ for $x \in [\ln(1/2),\ln(3/2)]$ (and $0$ elsewhere). – Jean Marie May 21 '22 at 08:06
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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

wjmccann
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  • Caution: the values taken are not ${1/2,3/2}$ but the whole real interval $[1/2,3/2]$ – Jean Marie May 19 '22 at 21:00
  • Thanks for your answer! However, as said by Jean, x is uniform on the real interval $[\frac{1}{2}, \frac{3}{2}]$ instead of a binary distribution. – Marco_L_T May 19 '22 at 21:12
  • Besides, your formula $P(w<c)...$ will give rise to probabilities $>1$ because you are considering that $p=1/2$ and $q=3/2$ instead of $q=1-p=1/2$... – Jean Marie May 20 '22 at 07:21