Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Standard deviation and Bernoulli distribution

  1. Feb 3, 2006 #1
    Let us assume that X has Bernoulli distribution, with P(X = 1) = p and P(X = 0) = q = 1 - p. Of course, E(X) = p and Var(X) = pq. Now, since pq < 1, standard deviation is bigger than variance.

    I have got the following question:

    Does this fact make standard deviations and theorems based on standard deviation (like Chebyshev's inequality) unusable in this case?
     
  2. jcsd
  3. Feb 3, 2006 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    I don't see why it should make any difference. Could you elaborate on what's your concern?
     
  4. Feb 4, 2006 #3

    EnumaElish

    User Avatar
    Science Advisor
    Homework Helper

    I don't see why it would. You don't need Bernoulli for this property (sigma < sigma^2), as there is no rule that says the variance of, say, a normal dist. has to be > 1.
     
  5. Feb 4, 2006 #4

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Read the theorem -- it tells you exactly when you can use it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?