Results 1 to 4 of 4

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

  1. #1
    Junior Member
    Joined
    Jun 2008
    Posts
    38
    Thanks
    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...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by RanDom View Post
    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)
    Last edited by Moo; April 26th 2009 at 12:48 AM. Reason: forgot to end a sentence
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by RanDom View Post
    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jun 2008
    Posts
    38
    Thanks
    1
    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?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Discrete versus Continuous
    Posted in the Statistics Forum
    Replies: 1
    Last Post: September 3rd 2011, 02:15 AM
  2. continuous or discrete??
    Posted in the Statistics Forum
    Replies: 1
    Last Post: November 8th 2010, 09:27 AM
  3. Discrete or continuous?
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: March 10th 2010, 05:33 PM
  4. Please help: continuous or discrete?
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 3rd 2009, 02:10 AM
  5. discrete and continuous
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: September 12th 2009, 04:34 PM

Search Tags


/mathhelpforum @mathhelpforum