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Alex787 [66]
3 years ago
8

We have n = 100 many random variables xi 's, where the xi 's are independent and identically distributed bernoulli random variab

les with p = 0.5 (e(xi)=p and var(xi)=p(1-p)).
a. what distribution does pn i=1 xi follow exactly (sum bernoulli random varaibles)? state the type of distribution and what the parameter is
Mathematics
1 answer:
Alex777 [14]3 years ago
8 0
Recall that for a random variable X following a Bernoulli distribution \mathrm{Ber}(p), we have the moment-generating function (MGF)

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

and also recall that the MGF of a sum of i.i.d. random variables is the product of the MGFs of each distribution:

M_{X_1+\cdots+X_n}(t)=M_{X_1}(t)\times\cdots\times M_{X_n}(t)

So for a sum of Bernoulli-distributed i.i.d. random variables X_i, we have

M_{\sum\limits_{i=1}^nX_i}(t)=\displaystyle\prod_{i=1}^n(1-p+pe^t)=(1-p+pe^t)^n

which is the MGF of the binomial distribution \mathcal B(n,p). (Indeed, the Bernoulli distribution is identical to the binomial distribution when n=1.)
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