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)

### and is zero elsewhere.

### Further, we know that T = sum(xi) has the negative binomial

### Distribution

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.