Załóżmy, że mamy próbki dwóch niezależnych zmiennych losowych Bernoulliego, i .
Jak udowodnimy, że
?
Załóżmy, że .
distributions
sampling
bernoulli-distribution
Stary człowiek na morzu.
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Odpowiedzi:
Puta=θ1(1−θ1)√n1√ , b=θ2(1−θ2)√n2√ ,
A=(X¯1−θ1)/a ,
B=(X¯2−θ2)/b . We have
A→dN(0,1), B→dN(0,1) .
In terms of characteristic functions it means
SinceA and B are independent,
This proof is incomplete. Here we need some estimates for uniform convergence of characteristic functions. However in the case under consideration we can do explicit calculations. Putp=θ1, m=n1 .
Note that similar calculations may be done for arbitrary (not necessarily Bernoulli) distributions with finite second moments, using the expansion of characteristic function in terms of the first two moments.
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Proving your statement is equivalent to proving the (Levy-Lindenberg) Central Limit Theorem which states
If{Zi}ni=1 is a sequence of i.i.d random variable with finite mean E(Zi)=μ and finite variance V(Zi)=σ2 then
HereZ¯=∑iZi/n that is the sample variance.
Then it is easy to see that if we put
and
(There's a last passage, and you have to adjust this a bit for the general case wheren1≠n2 but I have to go now, will finish tomorrow or you can edit the question with the final passage as an exercise )
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