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how to prove binomial through Bernoulli indicators? is it x-bernoulli (P) y=x1,x2,...,xn. where xi is the independent variable bernoulli gives y-Bin(n,p)?

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What is it that you want to prove about the Binomial through indicator random variables? –  André Nicolas May 2 '13 at 18:05
    
Do you mean to say that if $\{x_i\}_{i=1}^n$ are iid Bernoulli with success rate $p$, then $y = \sum_{k=1}^n x_k \sim \mathcal{B}(n,p)$? That would be correct, but what would you like to prove? –  gt6989b May 2 '13 at 18:14
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1 Answer

let X1,X2,...,Xn are iid Bernoulli random variables with parameter p then Xi has probability distribution:

f(xi;p)=p^xi * (1-p)^(1-xi) ; xi=0 or 1

& the moment generating function of Xi is

M_Xi(t)=(1-p)+pe^t

let Y=X1+X2+...+Xn

the moment generating function of Y is

M_Y(t)=M_X1+X2+...+Xn(t)=M_X1(t) * M_X2(t)*...*M_Xn(t)

                    ={(1-p)+pe^t}*{(1-p)+pe^t}*...*{(1-p)+pe^t}

                    ={(1-p)+pe^t}^n

which is the MGF of Binomial distribution

so y-Bin(n,p)

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