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operationsres
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Homework Statement
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.