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    If X_1 ,..., X_n are a sequence of i.i.d. random variables then why is the last equality of the line below true?

    \displaystyle \text{Var}(\overline{X}_n) = \text{Var} \left( \frac{1}{n} \sum_{i=1}^n X_i \right) = \frac{1}{n} \text{Var} (X)

    Does \displaystyle \text{Var} \left( \frac{1}{n} \sum_{i=1}^n X_i \right) = \frac{1}{n^2} \text{Var} \left( \sum_{i=1}^n X_i \right) ? But then how do I deal with the sum as variance isn't linear?
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      Use Var(aX) = a^2Var(X).

      So \displaystyle  \text{Var} \left( \frac{1}{n} \sum_{i=1}^n X_i \right) = \frac{1}{n^2} \text{Var} \left( \sum_{i=1}^n X_i \right) = \frac{1}{n^2} \sum_{i=1}^n \text{Var}(X_i) by independence, but this is just n copies of Var(X_1) = Var(X) (by identicality), so you get what you want.
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      (Original post by Glutamic Acid)
      Use Var(aX) = a^2Var(X).

      So \displaystyle  \text{Var} \left( \frac{1}{n} \sum_{i=1}^n X_i \right) = \frac{1}{n^2} \text{Var} \left( \sum_{i=1}^n X_i \right) = \frac{1}{n^2} \sum_{i=1}^n \text{Var}(X_i) by independence, but this is just n copies of Var(X_1) = Var(X) (by identicality), so you get what you want.
      I didn't think \displaystyle \text{Var} \left( \sum_{i=1}^n X_i \right) = \sum_{i=1}^n \text{Var}(X_i) would be true as variance isn't linear like expectation but obviously so.
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        (Original post by TheEd)
        I didn't think \displaystyle \text{Var} \left( \sum_{i=1}^n X_i \right) = \sum_{i=1}^n \text{Var}(X_i) would be true as variance isn't linear like expectation but obviously so.
        It is if the random variables are independent. Otherwise not necessarily so.
       
       
       
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