B Probability a randomly chosen natural number, which is not smaller than 𝑥, is greater than x?

[littlemathquark]

TL;DR Summary

What is the probability that a randomly chosen natural number, which is not smaller than a given natural number 𝑥, is greater than x?

What is the probability that a randomly chosen natural number, which is not smaller than a given natural number x, is greater than x?

I think it must be $1$ because there are infinitely numbers that greater than $x$ but how can I proof this? Thank you.

 

[Dale]

What is the probability mass function?

 

[littlemathquark]

I don't know probability mass function. İs the answer exactly $1$?

 

[martinbn]

What is the probability measure? (What do you mean by randomly chosen natural number? )

 

[Dale]

You have to state the probability mass function for the question to have an answer. The answer is not $1$, but rather $1-P(X=x)$ where $P(X)$ is the probability mass function and for this problem it is specified that $\sum_{X<x}P(X)=0$.

For example, if $P$ is the Poisson distribution with mean $\mu$ and $x=0$ then the probability is $1-P(0)=1-e^{-\mu}$.

 

[littlemathquark]

I assume a uniform distribution, I mean each outcome has an equal chance.

 

[Dale]

I assume a uniform distribution, I mean each outcome has an equal chance.

Is there such a distribution? Can you write down the formula for that?

 

[littlemathquark]

Are you saying that it is impossible to select each natural number with equal probability because the set of natural numbers N is infinite?

 

[martinbn]

Are you saying that it is impossible to select each natural number with equal probability because the set of natural numbers N is infinite?

Isn't that obvious?

 

[FactChecker]

What is the probability that a randomly chosen natural number,

When you say "random" do you mean with equal probability?

which is not smaller than a given natural number x, is greater than x?

I think it must be $1$ because there are infinitely numbers that greater than $x$

Infinity is difficult to work with. It is better to set some upper limit, like $x+m$. Given that the randomly chosen number, $n \ge x$, all natural numbers from $x$ to $x+m$ have equal probability of $1/(m+1)$. Then you can make $m$ as large as you want.

but how can I proof this?

For any $\epsilon \gt 0$, you can make $m$ large enough so the probability that $n = x$ is less than $\epsilon$. See how far you can get with that.

 

[littlemathquark]

Is there such a distribution? Can you write down the formula for that?

We can calculate the probability of a real number chosen between 0 and 5 being between 3 and 4 (even though the elements of the numerator and denominator are not finite). I asked the question in this context

 

[Dale]

Are you saying that it is impossible to select each natural number with equal probability because the set of natural numbers N is infinite?

That is the usual answer. I am open to other answers, but the point is that you have to start with actually constructing your $P(X)$ function. Infinity is tricky. When dealing with infinity you cannot afford to be sloppy.

We can calculate the probability of a real number chosen between 0 and 5 being between 3 and 4 (even though the elements of the numerator and denominator are not finite). I asked the question in this context

So here again, you need to specify the probability density function $P(X)$. Here it is a little easier because you can say “uniform probability” and that gives a well defined function.

I am not sure what you mean by “even though the elements of the numerator and denominator are not finite”. Everything is finite in that case (for a uniform distribution). Are you thinking of some other distribution that is not finite on that interval?

 

[littlemathquark]

That is the usual answer. I am open to other answers, but the point is that you have to start with actually constructing your $P(X)$ function. Infinity is tricky. When dealing with infinity you cannot afford to be sloppy.

So here again, you need to specify the probability density function $P(X)$. Here it is a little easier because you can say “uniform probability” and that gives a well defined function.

I am not sure what you mean by “even though the elements of the numerator and denominator are not finite”. Everything is finite in that case (for a uniform distribution). Are you thinking of some other distribution that is not finite on that interval?

“even though the elements of the numerator and denominator are not finite”
I mean there are infinite points (real number) in (0,5) interval and in (3,4) interval.

 

[Dale]

“even though the elements of the numerator and denominator are not finite”
I mean there are infinite points (real number) in (0,5) interval and in (3,4) interval.

Sure, but that isn't particularly critical. What is important is that $P(X)$ is finite and continuous so $\int_{3}^{4} P(X) \ dX$ is finite also.

The problem with your original problem isn't the infinite number of points. My example of the Poisson distribution also has an infinite number of points. The problem is the value of $P(X)$ itself. For the Poisson distribution every $P(X)$ is finite, and the sum over all of the infinite points is $1$.

 

[littlemathquark]

When you say "random" do you mean with equal probability?

Infinity is difficult to work with. It is better to set some upper limit, like $x+m$. Given that the randomly chosen number, $n$, is greater than $x$, all natural numbers from $x$ to $x+m$ have equal probability of $1/(m+1)$. Then you can make $m$ as large as you want.

For any $\epsilon \gt 0$, you can make $m$ large enough so the probability that $n = x$ is less than $\epsilon$. See how far you can get with that.

Yes, I mean equal probability.

 

[Dale]

I am open to other answers, but the point is that you have to start with actually constructing your P(X) function. Infinity is tricky. When dealing with infinity you cannot afford to be sloppy.

Let me provide an example what I mean by "I am open to other answers".

One thing that you could do is to use hyperreal numbers. Then you could define the infinite hyperreal $\omega$ as the number of integers greater than or equal to $x$. That is a perfectly valid hyperreal which is infinite. Then $P(X)$ would be defined as $P(X)=\epsilon=1/\omega$ for all $X\ge x$ and $P(X)=0$ otherwise. This is uniform for $X\ge x$, and sums to $1$.

With this definition of $P(X)$ then the answer to your question is not $1$ but rather $1-\epsilon$ where $\epsilon$ is the specific infinitesimal hyperreal defined above. This is not a sloppy use of hyperreals, but it opens up another problem.

That is that we are now allowing probabilities to be hyperreal numbers instead of real numbers. Hyperreals have the benefit that all first-order logic statements about real numbers are also valid statements about hyperreals. So you would need to look through all of the proofs about probabilities that you are using for any given exercise and you would need to see if that proof uses only first-order logic statements. If so, then you are fine to use that proof, but if it uses second-order logic statements then you would need to check to see if the proof still holds for the hyperreals.

That is no small task, but it must be done if you want to take this approach. Instead, the usual approach is to stick with the real numbers and simply say there is no such real number $\epsilon$.