Ans 1.
Xis are n random variables which follow binomial distribution.
Goal: to show u=sum(Xi) is a sufficient statistic using factorization theorem
Solution:
Joint Binomial distribution, since they are independently distributed, they can be simply multiplied:
(p^x1 * (1-p)^n-x1 )* (p^x2 * (1-p)^n-x2 )* …… ..
So,
Likelihood function L(p|x) becomes =(p^sum(xi) * (1-p)^n-sum(xi )
Now g(u,p)= =(p^sum(xi) * (1-p)^n-sum(xi )
With u=sum(xi) and h(x)=1
So,using factorization theorem, since h(x) is independent of p, sum(xi) is a sufficient statistic.
Proved
Ans 2.
Xis are n random variables which follow negative binomial distribution.
Goal: find a sufficient statistic for p
Joint Binomial distribution, since they are independently distributed, they can be simply multiplied:
(p^r * (1-p)^x1-r )* (p^r * (1-p)^x2-r )* …… .. and the joint probability of X1 = x1,X2 = x2, ...,Xn = xn and T =sum(xi)=t is f(x
1
, x
2
, ...., x n
, t; p) = p^ r
(1 − p)^t
-r where t=sum(xi)
g(t; p) = (nCt)*p^ r
(1 − p)^ t−r for t = 0, 1, ..., n f(x1,x2,…xn|T=t;p)= f(x1,x2,,,xn,t;p)/g(t;p)= 1/ (nCt)
So, t= sum(xi) is a sufficient statistic for p for negative binomial distribution.