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The title is my question and the reason for asking it is the following.

Define a set $\mathbb N (≤ n) \equiv$ {1,2,3, … , n} and define a “random selection” to be a selection in which each member of $\mathbb N (≤ n)$ has an equal probability, $p$, of being selected.

As $\mathbb N (≤ n)$ is the whole population from which the selection will be made then the probability of selecting some member of $\mathbb N (≤ n)$ is $np = 1$.

Since $p \to 0$ as $n \to ∞$ then the probabilty of any member of $\mathbb N $ being randomly selected is $p = 0$.

Is this correct?

Duplicate Question ?

I agree my question is similar to the other question but it is not similar enough to be called a duplicate. For example, the definition of “random selection” I use is different.

Answer from snarski

That is mostly correct. What is true is that there is no distribution on $\mathbb N$ that associates to each $n \in \mathbb N$ an equal probability of being selected. On the other hand, that doesn't mean you can't select a random nonnegative integer "at random", because "at random" doesn't imply the distribution is uniform.

Reply to snarski

Thanks for your answer but I have some questions.

I don’t understand the statement about there being “no distribution that associates to each $n \in \mathbb N$ an equal probability of being selected”. Doesn’t the argument in my question show that there is such a distribution, namely $p = 0$ for each member of $\mathbb N$ ?

Also I don’t understand how a member of $\mathbb N$ could be randomly selected unless each member of $\mathbb N$ had the same probability of selection?

The conclusion of the argument in my question seems reasonable because it implies it is impossible to make a random selection of a single member of $\mathbb N$. This agrees with the practical impossibility of an algorithm that randomly selected a single member of an infinite set because an infinite set as input could not be processed by such an algorithm in a finite time.

User334732 comment

Let $f(n)=\frac{1} {6}$ for all $n \in \mathbb N$. Let the order be 1,2,3,4,5... Then start with 1 and roll a dice. If you get a 6, stop and you have selected number 1 as your answer. If you don't, then go to $n$=2 and roll the dice again. Continue until you roll a 6.

Reply to user334732

With this procedure the probability of selecting $n$ is $(\frac{5} {6})^{(n-1)} \frac{1}{6}$ so smaller numbers have a greater probability of being selected than larger numbers. This selection does not seem to be “random”. Specifically it does not satisfy the definition of “random selection” I use in my question which is all $n \in \mathbb N$ must have an equal probability of being selected.

Comment from Vincent

Thanks for the clarification. What Snarski is saying is: 1) You are correct that there is no "random selection" according to your definition 2) your definition is somewhat non-standard. Usually (at least among mathematicians, the general population might lean more toward your definition) 'random' (roughly) means that probabilities are involved, not that they are necessary all equal. So yes, a random process for selecting integers that gives each equal, positive probability is impossible, but the next best thing, a random process that gives all numbers at least some positive probability, is.

user334732's comment is an example of that latter, 'next best' thing, and a very beautiful example on top of that because it is very hands on.

Reply to Vincent

It seems we agree on the impossibility of a random selection of a single member of $\mathbb N$ given the definition of "random selection" in my question.

You mention that my definition is "somewhat non-standard" and while my definition may be favored by the "general population" usually to mathematicians, " 'random' (roughly) means that probabilities are involved, not that they are necessary all equal".

The definitions of a "random sample", at least the ones I have seen, say a "random sample" has only two defining properties:

P1. each member of the population has an equal chance of selection, and

P2. each selection is independent of the other selections.

My question asks about the random selection of a single member of $\mathbb N$, that is, a random sample of size one from the population $\mathbb N$ , so property P2 is not applicable and the general definition specializes to property P1 which is the definition in my question.

If you know of a definition of "random selection" that differs from or extends the P1 + P2 definition above then can you supply a reference ? I would be interested to read about it.

I do not share your opinion of the procedure suggested by u334732 although I appreciate u334732 taking the time to attempt to answer my question.

Problem 1 with u334732's suggestion is that the procedure does not give an equal probability of selection for each member of $\mathbb N$ since the probability of selecting $n$ is $(\frac{5} {6})^{(n-1)} \frac{1}{6}$. Hence the procedure does not comply with the definition of "random selection" in my question. I mentioned this problem in a previous reply.

Problem 2 is that the procedure may never select a member of $\mathbb N$ because the probability of not selecting a member of $\mathbb N$ that is ≤ n is $(\frac{5} {6})^{n}$ which is > 0 for all $n \in \mathbb N$ hence there is a non-zero probability that no member of $\mathbb N$ will be selected by any finite number of applications of the procedure.

I think these problems reflect the impossibility of randomly selecting a single member of $\mathbb N$ and there would be some problem with any suggested procedure.

Xiu LL
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  • Yes this is correct. You can remedy the problem by taking a random selection that does not give the same probability of being selected to each natural number, e.g. a Poisson distribution. – Vincent Dec 12 '18 at 10:30
  • Progress through the integers along a wellorder and combine that with a probability $p_x=f(x)$ of stopping on any given element (given that one arrives at it). – it's a hire car baby Dec 12 '18 at 10:46
  • @user334732 - Can you give more detail about how your procedure actually selects a member of $\mathbb N$. – Xiu LL Dec 12 '18 at 18:27
  • @Vincent - Can you give more detail about how your suggestion would actually select a member of $\mathbb N$. – Xiu LL Dec 12 '18 at 18:29
  • No matter re the wellorder, just arrange all the elements on $\Bbb N$ in a sequence. Then give every number a probability less than $1$. Now start with the first integer $n$ in your sequence and stop there with its probability $f(n)$ – it's a hire car baby Dec 12 '18 at 18:30
  • @user334732 - Can you tell me how you define a "wellorder"? – Xiu LL Dec 12 '18 at 18:32
  • Let $f(n)=1/6$ for all $n\in\Bbb N$ – it's a hire car baby Dec 12 '18 at 18:34
  • Let the order be $1,2,3,4,5...$ – it's a hire car baby Dec 12 '18 at 18:34
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    Then start with $1$ and roll a dice. If you get a 6, stop and you have selected number $1$ as your answer. If you don't, then go to $n=2$ and roll the dice again. Continue until you roll a $6$. – it's a hire car baby Dec 12 '18 at 18:35
  • Thanks for the clarification. What Snarski is saying is: 1) You are correct that there is no "random selection" according to your definition 2) your definition is somewhat non-standard. Usually (at least among mathematicians, the general population might lean more toward your definition) 'random' (roughly) means that probabilities are involved, not that they are necessary all equal. So yes, a random process for selecting integers that gives each equal, positive probability is impossible, but the next best thing, a random process that gives all numbers at least some positive probability, is. – Vincent Dec 12 '18 at 22:44
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    user334732's comment is an example of that latter, 'next best' thing, and a very beautiful example on top of that because it is very hands on. – Vincent Dec 12 '18 at 22:45
  • @XiuLL $f(n)$ can be any function from $\Bbb N\to[0,1)$ so for example $f(1)=0.000000001, f(2)=0.0000000001, f(3)=0.5,\ldots$ then using the standard order $3$ is more likely to occur than $1$ or $2$. Also I said choose an order for the integers. You can choose $1,000,000,10,1,2,3,4$ if you want and then even with $f(n)=1/2^n$, the probability of $1,000,000$ would be much greater than the probability of say $p(100)$ – it's a hire car baby Dec 15 '18 at 19:28

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That is mostly correct. What is true is that there is no distribution on $\mathbb{N}$ that associates to each $n \in \mathbb{N}$ an equal probability of being selected. On the other hand, that doesn't mean you can't select a random nonnegative integer "at random", because "at random" doesn't imply the distribution is uniform.

snar
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