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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


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


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    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


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    Read the theorem -- it tells you exactly when you can use it.
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