1. ## Standard Uniform Distribution

Hi, how to prove that U~uniform[0,1] is equivalent to (i.e. if and only if) [nU]~uniform{0,1,2,...,n-1} for all positive integers n. [] is the floor function. I totally have no idea how to start, so guidance would be greatly appreciated. Also, why is the range from 0 to n-1? If you have [0,1] in the beginning and you multiply by n, shouldn't you have from 0 to n? I am so confused.

2. ## Re: Standard Uniform Distribution

What would P([nU] = n) = P(U = 1) be?

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geniusacamel (10-14-2016)

4. ## Re: Standard Uniform Distribution

Originally Posted by Dason
What would P([nU] = n) = P(U = 1) be?
0 because U is continuous? In con. dist, P of any point is 0.

5. ## Re: Standard Uniform Distribution

Anyway, from your question, it has given me an idea.

[nU] = n-1 if and only if n-1 <= nU < n if and only if (n-1)/n < U < 1. And probability is 1 - n/n + 1/n = 1/n.
[nU] = n-2 if and only if n-2 <= nU < n-1 if and only if (n-2)/n < U < (n-1)/n. Probability is 1/n again.
Continue with n-3, n-4 etc. Is it correct?

6. ## Re: Standard Uniform Distribution

So is my above solution good enough or more is needed? Anyway, I need help for a further question:

U_n = [nU]/n. Compute P(U_(n+1) > U_n) and P(U_(n+1) = U_n) for each n in N (natural number) to show that the convergence is not monotone.

I have no idea how to do it at all and I will show you my attempt and I won't be surprised if I get it totally on the wrong track.

So here is my attempt:

[(n+1)U]/(n+1) > [nU]/n when k/(n+1) <= U < k/n , where k is an integer from 1 to n
[(n+1)U]/(n+1) = [nU]/n when U < 1/(n+1) or when U=1

So P(U_(n+1) > U_n) = summation (k/n - k/(n+1)), k from 1 to n = 1/2
P(U_(n+1) = U_n) = 1/(n+1)

Assuming that my probabilities are correct (please let me know if I am wrong), how do I show that the convergence is not monotone?

7. ## Re: Standard Uniform Distribution

I think a better way to do this is to show convergence in probability - which is weaker than convergence almost surely.

So, consider a sample space with a Uniform probability measure P. That is, the probability associated with the interval is .

Let be a sequence of intervals of the form for and where , where

,,..., .

Next, define a sequence of random variables on this sample space as for
.

Let and note that is 1 only on the interval such that
,
where is a sequence that converges to 0, .

It follows that for any ,

,

so then it necessarily follows that as .

8. ## Re: Standard Uniform Distribution

Originally Posted by Dragan
I think a better way to do this is to show convergence in probability - which is weaker than convergence almost surely.

So, consider a sample space with a Uniform probability measure P. That is, the probability associated with the interval is .

Let be a sequence of intervals of the form for and where , where

,,..., .

Next, define a sequence of random variables on this sample space as for
.

Let and note that is 1 only on the interval such that
,
where is a sequence that converges to 0, .

It follows that for any ,

,

so then it necessarily follows that as .
How do you compute the intervals of m(n): How do you get ,,..., ? And how do you show that the convergence is not monotone?

9. ## Re: Standard Uniform Distribution

Well, to be more clear, you compute the intervals like this: m(1)=[0, 1], m(2)=[0, 1/2], m(3)=[1/2, 1], m(4)=[0, 1/3], m(5)=(1/3, 2/3), m(6)=[2/3, 1] - see the pattern.

What I am showing is convergence in probability. It will not converge almost surely. As I said, convergence in probability is weaker than convergence almost surely.

In terms of whether your question regarding monotonicity: Are you being asked to use the Lebesque's Monotone Convergence Theorem???...I guess, I'm just not sure why you asking this.

10. ## Re: Standard Uniform Distribution

Well, the problem asked me to find those probabilities and to use them to show that the convergence is not monotone.

11. ## Re: Standard Uniform Distribution

Originally Posted by geniusacamel
Well, the problem asked me to find those probabilities and to use them to show that the convergence is not monotone.
Okay, so their is convergence in probability (and not convergence almost surely) - but the entire sequence of intervals m(n) is not (strictly) monotone:

m(1)=[0, 1], m(2)=[0, 1/2], m(3)=[1/2, 1], m(4)=[0, 1/3], m(5)=(1/3, 2/3), m(6)=[2/3, 1], m(7)=[0, 1/4], m(8)=[1/4, 1/2], m(9)=[1/2, 3,4], m(10)=[3/4, 1].

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