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Can anyone help me with this problem, I can't figure out how to solve it...

Let $X$ be a random variable which can take an infinite and countable set of values. Prove that $X$ cannot be uniformly distributed among these values.

Thanks!

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Consider a geometric random variable as a counterexample. –  Sasha Dec 10 '12 at 18:34
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I think that the question is the opposite: Let $X$ be a random variable which can take an infinite and countable set of values. Prove that $X$ cannot be uniformly distributed among these values. –  madprob Dec 10 '12 at 19:05
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@madprod you're right, thank you. I translated it worng, sorry! I edited. –  user52516 Dec 10 '12 at 19:58

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Let $X$ be a random variable which assumes values in a countable infinite set $Q$. We can prove there is no uniform distribution on $Q$.

Assume there exists such a uniform distribution, that is, there exists $a \geq 0$ such that $P(X=q) = a$ for every $q \in Q$.

Observe that, since $Q$ is countable, by countable additivity of $P$,

$1 = P(X \in Q) = \sum_{q \in Q}{P(X = q)} = \sum_{q \in Q}{a}$

Observe that if $a=0$, $\sum_{q \in Q}{a}=0$. Similarly, if $a>0$, $\sum_{q \in Q}{a} = \infty$. Contradiction.

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Horga, observe that if $Q$ is infinite but non-countable, there might exist a uniform distribution on $Q$. For example, take $Q = [0,1]$. Can you see why the proof doesn't work for this case? Can you find an infinite non-countable set which admits no uniform distribution? Answering these questions might help you to understand what is really going on. –  madprob Dec 10 '12 at 20:26
    
What is the difference between infinite but non-countable and countably infinite sample space?? The prove above is a complete contradiction prove for countably infinite sample space, right? –  Timothy Leung Oct 2 '13 at 20:49
    
Yes. It is a contradiction proof for a countably infinite sample space. A countably infinite sample space $\Omega$ is such that there exists a bijective function between $\Omega$ and $\mathbb{N}$, the natural numbers. A non-countable infinite sample space is such that there exists no such bijective function. –  madprob Jan 30 at 1:16

If each point has probability $p\ne 0$, then there is some integer $n$ such that $np\gt 1$. So if $x_1,\dots,x_n$ are some of the distinct values taken on by $X$, the probability that $X=x_1$ or $X=x_2$ or $\dots$ or $X=x_n$ is $np$. Since $np\gt 1$, this is impossible.

On the other hand, if $p=0$, then by countable additivity the probability assigned to any subset of the values taken on by $X$ must be $0$, contradicting the fact that the sum of the probabilities over the range of $X$ must be $1$.

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