Let's define a sequence of numbers between 0 and 1. The first term, $r_1$ will be chosen uniformly randomly from $(0, 1)$, but now we iterate this process choosing $r_2$ from $(0, r_1)$, and so on, so $r_3\in(0, r_2)$, $r_4\in(0, r_3)$... The set of all possible sequences generated this way contains the sequence of the reciprocals of all natural numbers, which sum diverges; but it also contains all geometric sequences in which all terms are less than 1, and they all have convergent sums. The question is: does $\sum_{n=1}^{\infty} r_n$ converge in general? (I think this is called almost sure convergence?) If so, what is the distribution of the limits of all convergent series from this family?

  • 3
    $\begingroup$ Not sure if my question itself makes sense: But how about a modified question where r1 is chosen uniformly randomly from (-1,+1) and we iterate by choosing r2 from (-1 x abs(r1),+1 x abs(r1)) Would that series converge too? $\endgroup$ – curious_cat Feb 5 '17 at 17:24
  • 2
    $\begingroup$ Your second series almost surely converges absolutely (i. e. $|r_1| + $|r_2|$ + \cdots$ converges) if and only if your first series almost surely converges. Combining that with Byron Schmuland's answer, the modified series almost surely converges absolutely, and so a fortiori almost surely converges. $\endgroup$ – Michael Lugo Feb 5 '17 at 18:11

Let $(u_i)$ be a sequence of i.i.d. uniform(0,1) random variables. Then the sum you are interested in can be expressed as $$S_n=u_1+u_1u_2+u_1u_2u_3+\cdots +u_1u_2u_3\cdots u_n.$$ The sequence $(S_n)$ is non-decreasing and certainly converges, possibly to $+\infty$.

On the other hand, taking expectations gives $$E(S_n)={1\over 2}+{1\over 2^2}+{1\over 2^3}+\cdots +{1\over 2^n},$$ so $\lim_n E(S_n)=1.$ Now by Fatou's lemma, $$E(S_\infty)\leq \liminf_n E(S_n)=1,$$ so that $S_\infty$ has finite expectation and so is finite almost surely.

  • 2
    $\begingroup$ @ByronSchmuland $S_\infty$ will have the same distribution as $X(1+S_\infty)$ where $X$ is uniformly distributed from 0 to 1. Since that is a recursive expression it does not directly tell you the distribution, but it can be used to sanity check any distribution you came up with. $\endgroup$ – kasperd Feb 5 '17 at 18:35
  • 12
    $\begingroup$ ...And the distribution of $S_\infty$ is uniquely determined by its Laplace transform $$L(s)=E(e^{-sS_\infty})$$ which is the unique solution of the differential equation $$\left(sL(s)\right)'=e^{-s}L(s)\qquad L(0)=1$$ solved by $$L(s)=\exp\left(-\int_0^s\frac{1-e^{-t}}tdt\right)=\exp\left(-\int_0^se^{-t}\ln\left(\frac{s}t\right)dt\right)$$ $\endgroup$ – Did Feb 5 '17 at 19:45
  • 2
    $\begingroup$ @polfosol if your previously drawn number is $a$, a uniformly distributed number between $a$ and $0$ is $a \cdot u$, where $u$ is uniformly distributed between $0$ and $1$. The rest follows by recurrence ($u_1=u$, $u_2 = u_1 \cdot u$...) $\endgroup$ – Davidmh Feb 6 '17 at 16:21
  • 3
    $\begingroup$ Can someone tell me what i.i.d. uniformity means? $\endgroup$ – Aalok Feb 7 '17 at 13:28
  • 5
    $\begingroup$ @zakoda i.i.d. means "independent identical distribution". Uniform is specifically what kind of distribution (a uniform one where the probability density function is constant over the whole range). $\endgroup$ – Kyle Feb 7 '17 at 16:18

The probability $f(x)$ that the result is $\in(x,x+dx)$ is given by $$f(x) = \exp(-\gamma)\rho(x)$$ where $\rho$ is the Dickman function as @Hurkyl pointed out below. This follows from the the delay differential equation for $f$, $$f^\prime(x) = -\frac{f(x-1)}{x}$$ with the conditions $$f(x) = f(1) \;\rm{for}\; 0\le x \le1 \;\rm{and}$$ $$\int\limits_0^\infty f(x) = 1.$$ Derivation follows

From the other answers, it looks like the probability is flat for the results less than 1. Let us prove this first.

Define $P(x,y)$ to be the probability that the final result lies in $(x,x+dx)$ if the first random number is chosen from the range $[0,y]$. What we want to find is $f(x) = P(x,1)$.

Note that if the random range is changed to $[0,ay]$ the probability distribution gets stretched horizontally by $a$ (which means it has to compress vertically by $a$ as well). Hence $$P(x,y) = aP(ax,ay).$$

We will use this to find $f(x)$ for $x<1$.

Note that if the first number chosen is greater than x we can never get a sum less than or equal to x. Hence $f(x)$ is equal to the probability that the first number chosen is less than or equal to $x$ multiplied by the probability for the random range $[0,x]$. That is, $$f(x) = P(x,1) = p(r_1<x)P(x,x)$$

But $p(r_1<x)$ is just $x$ and $P(x,x) = \frac{1}{x}P(1,1)$ as found above. Hence $$f(x) = f(1).$$

The probability that the result is $x$ is constant for $x<1$.

Using this, we can now iteratively build up the probabilities for $x>1$ in terms of $f(1)$.

First, note that when $x>1$ we have $$f(x) = P(x,1) = \int\limits_0^1 P(x-z,z) dz$$ We apply the compression again to obtain $$f(x) = \int\limits_0^1 \frac{1}{z} f(\frac{x}{z}-1) dz$$ Setting $\frac{x}{z}-1=t$, we get $$f(x) = \int\limits_{x-1}^\infty \frac{f(t)}{t+1} dt$$ This gives us the differential equation $$\frac{df(x)}{dx} = -\frac{f(x-1)}{x}$$ Since we know that $f(x)$ is a constant for $x<1$, this is enough to solve the differential equation numerically for $x>1$, modulo the constant (which can be retrieved by integration in the end). Unfortunately, the solution is essentially piecewise from $n$ to $n+1$ and it is impossible to find a single function that works everywhere.

For example when $x\in[1,2]$, $$f(x) = f(1) \left[1-\log(x)\right]$$

But the expression gets really ugly even for $x \in[2,3]$, requiring the logarithmic integral function $\rm{Li}$.

Finally, as a sanity check, let us compare the random simulation results with $f(x)$ found using numerical integration. The probabilities have been normalised so that $f(0) = 1$.

Comparison of simulation with numerical integral and exact formula for $x\in[1,2]$

The match is near perfect. In particular, note how the analytical formula matches the numerical one exactly in the range $[1,2]$.

Though we don't have a general analytic expression for $f(x)$, the differential equation can be used to show that the expectation value of $x$ is 1.

Finally, note that the delay differential equation above is the same as that of the Dickman function $\rho(x)$ and hence $f(x) = c \rho(x)$. Its properties have been studied. For example the Laplace transform of the Dickman function is given by $$\mathcal L \rho(s) = \exp\left[\gamma-\rm{Ein}(s)\right].$$ This gives $$\int_0^\infty \rho(x) dx = \exp(\gamma).$$ Since we want $\int_0^\infty f(x) dx = 1,$ we obtain $$f(1) = \exp(-\gamma) \rho(1) = \exp(-\gamma) \approx 0.56145\ldots$$ That is, $$f(x) = \exp(-\gamma) \rho(x).$$ This completes the description of $f$.

  • $\begingroup$ Can you explain your reasoning to multiply the probability that the first pick is less than $x$ by the probability that you land between $x$ and $x+dx$ starting at $x$? I don't understand that. Wouldn't the probability depend on what the first pick is, necessitating an integral like the one you use later on? $\endgroup$ – user142299 Feb 6 '17 at 0:32
  • 1
    $\begingroup$ $$P(x,x) = 1/x \int_0^x P(x-z,z) dz$$ $$P(x(<1),1)=1/1\int_0^xP(x-z,z)dz$$ $$P(x(<1),1) =xP(x,x)=P(1,1)$$ $\endgroup$ – Raziman T V Feb 6 '17 at 6:16
  • 8
    $\begingroup$ This is (proportional to) the Dickman function! $\endgroup$ – Hurkyl Feb 6 '17 at 7:54
  • $\begingroup$ @Hurkyl Wow, that's cool. Just added it to the answer. $\endgroup$ – Raziman T V Feb 6 '17 at 8:21

Just to confirm the simulation by @curious_cat, here is mine:


It's a histogram, but I drew it as a line chart because the bin sizes were quite small ($0.05$ in length, with 10 million trials of 5 iterations).

Note: vertical axis is frequency, horizontal axis is sum after 5 iterations. I found a mean of approximately $0.95$.

  • $\begingroup$ The shape of that graph does not look like I would have expected it to. It looks flat from 0 to 1 (I assume the variance seen there is entirely due to only having sampled a finite number of times). I would have expected it to start decreasing right away. And it looks like the derivative is discontinuous at 1 and nowhere else. I would have expected the derivative to be continuous. And given that the distribution can be expressed recursively I can't quite see how a single discontinuity in the derivative wouldn't cause an infinite number of discontinuities. $\endgroup$ – kasperd Feb 5 '17 at 18:44
  • $\begingroup$ That doesn't mean I think you are wrong. Just that I am surprised. I will now try to simulate it myself to see if I can reproduce your result. $\endgroup$ – kasperd Feb 5 '17 at 18:45
  • 4
    $\begingroup$ I have proved why it has to be constant for $x<1$, please check my answer. $\endgroup$ – Raziman T V Feb 5 '17 at 19:15
  • $\begingroup$ I have done a simulation myself which produced a graph similar to yours except from an anomaly at 0 due to using discrete random numbers in the simulation. So it looks like the shape of the graph is pretty much correct. $\endgroup$ – kasperd Feb 5 '17 at 19:36
  • 1
    $\begingroup$ For an infinite sum, the mean should be $1$ since it satisfies $\mu=\frac12 +\frac12\mu$ where $\frac12$ is the mean of a uniform random variable on $[0,1]$. For five terms in the sum it should be $\frac{2^5-1}{2^5}=\frac{31}{32}=0.96875$ $\endgroup$ – Henry Feb 9 '17 at 14:13

I just ran a quick simulation and I get a mean sum of one (standard deviation of 0.7)

enter image description here

enter image description here

Caveat: Not sure I coded it all right! Especially since I didn't test convergence.

  • 4
    $\begingroup$ How did you run a quick simulation of an infinite sequence? $\endgroup$ – JiK Feb 6 '17 at 12:24
  • 3
    $\begingroup$ @JiK Right. I didn't. $\endgroup$ – curious_cat Feb 6 '17 at 12:39
  • 2
    $\begingroup$ Chuck Norris could have run a quick simulation of an infinite sequence. After all, he counted to infinity. Twice. $\endgroup$ – eipi10 Feb 8 '17 at 5:49
  • $\begingroup$ For what it's worth, I did the simulation in excel and let it cook for half a day. My histogram was essentially the same as posted above: flat from 0 to 1, then an exponential like decay. This leads to another question involving histograms, which I shall submit to the community... Recently the Israel prime minister dismissed his security staff. He was meeting Chuck Norris, and knew the Norris could personally handle any security problem. $\endgroup$ – richard1941 Feb 8 '17 at 19:12

protected by J. M. is not a mathematician Mar 11 '18 at 15:09

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.