# Math Help - A conceptual difficulty... discrete vs continuous

1. ## A conceptual difficulty... discrete vs continuous

Studying analysis for the first time has made me realise I don't understand what discrete and continuous are as well as I thought I did! I've arrived at what seems to be a couple of paradoxes to me... I'd be be grateful if someone could help me resolve them

A discrete random variable is defined as one that can take a finite or countably infinite number of values... Consider choosing a natural number at random... the probability of choosing any particular one of them is zero since there are an infinite number of them and they are all equally likely... But then the expected value is zero using the discrete RV formula (since the natural numbers are countably infinite)... ? This doesn't seem right!

Also, the irrational numbers are uncountable... so I suppose that choosing one at random must be a continuous random variable... but this is very counter-intuitive since there is a rational number between any two irrationals...

2. Originally Posted by RanDom
Studying analysis for the first time has made me realise I don't understand what discrete and continuous are as well as I thought I did! I've arrived at what seems to be a couple of paradoxes to me... I'd be be grateful if someone could help me resolve them

A discrete random variable is defined as one that can take a finite or countably infinite number of values... Consider choosing a natural number at random... the probability of choosing any particular one of them is zero since there are an infinite number of them and they are all equally likely... But then the expected value is zero using the discrete RV formula (since the natural numbers are countably infinite)... ? This doesn't seem right!
No it's not right ^^'
The thing is that they're not equally likely !
A discrete distribution is not necessarily a uniform distribution (this is what you meant in your bold sentence)

You can define :
$\mathbb{P}(X=i)=\frac{1}{2^{i+1}}$

And we indeed have $\sum_{i=0}^\infty \mathbb{P}(X=i)=1$, so it's a random variable.

Does this example shed some light on your mistake ?

Also, the irrational numbers are uncountable... so I suppose that choosing one at random must be a continuous random variable... but this is very counter-intuitive since there is a rational number between any two irrationals...
What does it mean : "choosing one is a continuous random variable" ?

While working with continuous rv's, the probability of taking a given value, even if it's an integer, or an irrational, is 0.

(because the Lebesgue measure of a singleton is 0... but I guess you didn't learn that yet)

3. Originally Posted by RanDom
Studying analysis for the first time has made me realise I don't understand what discrete and continuous are as well as I thought I did! I've arrived at what seems to be a couple of paradoxes to me... I'd be be grateful if someone could help me resolve them

A discrete random variable is defined as one that can take a finite or countably infinite number of values... Consider choosing a natural number at random... the probability of choosing any particular one of them is zero since there are an infinite number of them and they are all equally likely... But then the expected value is zero using the discrete RV formula (since the natural numbers are countably infinite)... ? This doesn't seem right!

Also, the irrational numbers are uncountable... so I suppose that choosing one at random must be a continuous random variable... but this is very counter-intuitive since there is a rational number between any two irrationals...
You can't choose an integer at random (from the uniform distribution over the integers) because there is no such distribution.

Same for the rationals (and irrationals on an unbounded interval).

CB

4. Ahhh, I think I see. The discrete uniform distribution only permits a finite number of outcomes... cool

(and irrationals on an unbounded interval).
Could you elabrate on what you mean by that though?