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Variance of binomial distribution - 1 trial

  1. Jul 31, 2012 #1
    1. The problem statement, all variables and given/known data

    For n trials, [itex]S_n[/itex] can be seen as the sum of n independent single trials [itex]X_i[/itex], i = 1,2,...,n, with [itex]\mathbb{E}[X_i][/itex]=p and Var[itex][X_i]=p(1-p)[/itex].


    2. What I don't understand

    I don't understand why Var[itex][X_i]=p(1-p)[/itex].

    We know that: Var[itex][X_i]=\mathbb{E}[(X_i - \mathbb{E}[X_i])^2] = \mathbb{E}[X_i^2 - 2X_i\mathbb{E}[X_i] + \mathbb{E}[X_i]^2] = \mathbb{E}[X_i^2] - \mathbb{E}[X_i]^2[/itex].

    Taking [itex]\mathbb{E}[X_i]^2[/itex], we have [itex]\mathbb{E}[X_i]^2=p^2[/itex].

    Taking [itex]\mathbb{E}[X_i^2][/itex], we have [itex]\mathbb{E}[X_i^2]=\mathbb{E}[\prod_{i=1}^2X_i] = \prod_{i=1}^2 \mathbb{E}[X_i] = \mathbb{E}[X_i]^2 = p^2[/itex].

    So Var[itex][X_i]= p^2 - p^2 = 0 \not= p(1-p)[/itex], which contradicts what my lecture notes say.
     
  2. jcsd
  3. Aug 1, 2012 #2

    Ray Vickson

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    No, [itex] E X_i^2 \neq p^2.[/itex] In fact, [itex]E f(X_i) = \sum_{k} P\{X_i = k\} f(k)[/itex] for any function f, so we get [itex] E X_i^2 = p.[/itex]

    RGV
     
  4. Aug 1, 2012 #3
    Enlightening post, thank you.

    I guess my error was assuming that X_i was independent in:
    [itex]\mathbb{E}[X_i^2]=\mathbb{E}[\prod_{i=1}^2X_i] = \prod_{i=1}^2 \mathbb{E}[X_i] = \mathbb{E}[X_i]^2 = p^2[/itex].


    ?


    But they're the same event so that would be non-sensical....
     
  5. Aug 1, 2012 #4
    Use the definition of variance of a discrete variable.
     
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