Product of random numbers.

Question

Find the constant $p$ such that the product of any (positive) number $N_0$ multiplied by successive random numbers between $0$ and $p$ will, on average, neither diverge to infinity nor converge to zero.

(This is not for homework or anything. This is a problem I "invented" and solved, and now want to see how real mathematicians would think about this. Take your time.)

(Thanks md2perpe for fixing up my tag.)

(Thanks Forester for making the variables render nicely.)

Answer 1

Here is my attempt. I leave you to fill missing details.

Call $X_n$ the random variable which is uniformly distributed in the interval $(0,p)$. Then $$\prod_{n=1}^N X_n \mbox{ is bounded and stays away from zero} \Longleftrightarrow \sum_{n=1}^N \log X_n \mbox{ is bounded }$$ The random variables $\log X_n$ has values in the interval $(- \infty , \log p)$, and it is not uniformly distributed. Clearly, a necessary condition of $$\sum_{n=1}^N \log X_n \mbox{ is bounded }$$ is that the mean of $\log X_n$ is zero.

Hence we need $$\int_{0}^{ p} \log t \ \mathrm d t =0 $$ this condition is equivalent to $$p \log p -p =0$$

This equation in the variable $p >0$ is easily solved and the only positive solution is $p=e$.

Answer 2

On reflection, I'm not sure the question is correct as stated.

If one takes the arithmetic average of the answer, even with p=e, this will tend to become very large - because you will get the odd case where you are multiplying by numbers bigger than 1 consistently. You only need a few such numbers for the arithmetic average to blow up.

Perhaps as you multiply by more and more such numbers, the average will come down(?) but I’m not sure.

Alternatively, the question could be reformulated to be about some kind of geometric average?

Answer 3

Thanks everyone. It turns out $p=e$ is exactly right, which all hinges on using the logarithm, and then transforming an infinite sum into an integral. My steps were pretty much the same:

Let $\tilde{N}$ denote the product after $j$ iterations. Then, we have $$\frac{\tilde{N}}{N_0} = \left(p \cdot r_1\right) \cdot \left(p \cdot r_2\right) \cdot \left(p \cdot r_3\right) \cdot \cdots \:,$$ where $p$ is constant, and each $r_j$ is a random number between $0$ and $1$. Next, take the natural log of both sides, turning the product on the right into sum:$$\ln\left(\frac{\tilde{N}}{N_0}\right) = \ln\left(p \cdot r_1\right) + \ln\left(p \cdot r_2\right) + \ln\left(p \cdot r_3\right) + \cdots$$

To satisfy the question as stated, we have the left side (on average) resolving to zero, with the number of terms on the right approaching infinity. This effectively makes $r$ a continuous variable: after many many iterations, we have sampled all of the real numbers between zero and one. Point is, we may replace the sum by an integral: $$0 \approx \ln\left(\frac{\tilde{N}}{N_0}\right) = \int_0^1 \left(\ln p + \ln r\right) dr \:,$$ and the answer falls right out. Only $p=e$ satisfies the above.