(a) The likelihood function for a random sample X1, X2, ..., Xn from the distribution with pdf f(x;θ) is given by:
L(θ|x1, x2, ..., xn) = ∏i=1^n f(xi;θ)
For the one-parameter exponential family of distributions, the pdf is given by:
f(x;θ) = exp[A(θ)B(x) + C(x) + D(θ)]
Therefore, the likelihood function can be written as:
L(θ|x1, x2, ..., xn) = exp[∑i=1^n A(θ)B(xi) + ∑i=1^n C(xi) + nD(θ)]
(b) If the likelihood function can be expressed as the product of a function which depends on θ and which depends on the data only through a statistic T(x1, x2, ..., xn), and a function that does not depend on θ, then T is a sufficient statistic for θ.
In the one-parameter exponential family of distributions, we can write the likelihood function as:
L(θ|x1, x2, ..., xn) = exp[nA(θ)B(T) + nC(T) + nD(θ)]
where T = T(x1, x2, ..., xn) is a statistic that depends on the data only and not on θ.
Comparing this to the general form, we see thatthe function that depends on θ is exp[nA(θ)B(T) + nD(θ)], and the function that does not depend on θ is exp[nC(T)]. Therefore, T is a sufficient statistic for θ.
To show that B(x) is a sufficient statistic for θ in the one-parameter exponential family, we need to show that the likelihood function can be written in the form:
L(θ|x1, x2, ..., xn) = h(x1, x2, ..., xn)g(B(x1), B(x2), ..., B(xn);θ)
where h(x1, x2, ..., xn) is a function that does not depend on θ, and g(B(x1), B(x2), ..., B(xn);θ) is a function that depends on θ only through B(x1), B(x2), ..., B(xn).
Starting with the likelihood function from part (a):
L(θ|x1, x2, ..., xn) = exp[∑i=1^n A(θ)B(xi) + ∑i=1^n C(xi) + nD(θ)]
Let's define:
h(x1, x2, ..., xn) = exp[∑i=1^n C(xi)]
g(B(x1), B(x2), ..., B(xn);θ) = exp[∑i=1^n A(θ)B(xi) + nD(θ)]
Now we can rewrite the likelihood function as:
L(θ|x1, x2, ..., xn) = h(x1, x2, ..., xn)g(B(x1), B(x2), ..., B(xn);θ)
which shows that B(x1), B(x2), ..., B(xn) is a sufficient statistic for θ in the one-parameter exponential family.
(c) If the sample consists of iid observations from the Uniform distribution on the interval (0, θ), then the pdf of each observation is:
f(x;θ) = 1/θ for 0 < x < θ
The likelihood function for a random sample X1, X2, ..., Xn from this distribution is:
L(θ|x1, x2, ..., xn) = ∏i=1^n f(xi;θ) = (1/θ)^n for 0 < X1, X2, ..., Xn < θ
To find a sufficient statistic for θ, we need to express the likelihood function in the form:
L(θ|x1, x2, ..., xn) = h(x1, x2, ..., xn)g(T(x1, x2, ..., xn);θ)
where T(x1, x2, ..., xn) is a statistic that depends on the data only and not on θ.
Since the likelihood function only depends on the maximum value of the sample, we can define T(x1, x2, ..., xn) = max(X1, X2, ..., Xn) as the maximum of the observed values.
The likelihood function can then be written as:
L(θ|x1, x2, ..., xn) = (1/θ)^n * I(x1, x2, ..., xn ≤ θ)
where I(x1, x2, ..., xn ≤ θ) is the indicator function that equals 1 if all the observed values are less than or equal to θ, and 0 otherwise.
We can see that the likelihood function depends on θ only through the term 1/θ, and the function I(x1, x2, ..., xn ≤ θ) depends on the data only and not on θ. Therefore, T(x1, x2, ..., xn) = max(X1, X2, ..., Xn) is a sufficient statistic for θ in the Uniform distribution on the interval (0, θ).