Consider the family of probability density functions {h(z ; θ): θ∈Ω}, where h(z ; θ)=1 / θ, 0<z<

step 1 Alright, so here in this problem, in part A, let integration from 0 to theta u of x times 1 over

Consider the family of probability density functions {h(z ;...

Consider the family of probability density functions $\{h(z ; \theta): \theta \in \Omega\}$, where $h(z ; \theta)=1 / \theta, 0<z<\theta$, zero elsewhere. (a) Show that the family is complete provided that $\Omega=\{\theta: 0<\theta<\infty\}$. Hint: $\quad$ For convenience, assume that $u(z)$ is continuous and note that the derivative of $E[u(Z)]$ with respect to $\theta$ is equal to zero also. (b) Show that this family is not complete if $\Omega=\{\theta: 1<\theta<\infty\}$. Hint: $\quad$ Concentrate on the interval $0<z<1$ and find a nonzero function $u(z)$ on that interval such that $E[u(Z)]=0$ for all $\theta>1$.

Consider the family of probability density functions $\{h(z ; \theta): \theta \in \Omega\}$, where $h(z ; \theta)=1 / \theta, 0<z<\theta$, zero elsewhere.
(a) Show that the family is complete provided that $\Omega=\{\theta: 0<\theta<\infty\}$. Hint: $\quad$ For convenience, assume that $u(z)$ is continuous and note that the derivative of $E[u(Z)]$ with respect to $\theta$ is equal to zero also.
(b) Show that this family is not complete if $\Omega=\{\theta: 1<\theta<\infty\}$. Hint: $\quad$ Concentrate on the interval $0<z<1$ and find a nonzero function $u(z)$ on that interval such that $E[u(Z)]=0$ for all $\theta>1$.

Key Concepts

Construction of Counterexamples

To show that a family is not complete for a given parameter space, one may construct a nontrivial function that, despite being nonzero on a subset of the support, integrates to zero for every parameter value in the space. This technique typically involves finding such a function on an interval where the restriction of the parameter space allows slack in the imposed conditions, thereby showing that the completeness property does not hold.

Differentiation Under the Integral Sign in Completeness Proofs

One common technique to establish completeness uses differentiation under the integral sign with respect to the parameter. By differentiating the expectation of an arbitrary function with respect to the parameter and applying the condition that the expectation is zero for all parameter values, one can derive conditions that force the original function to be zero. This method is frequently used in proofs of completeness in statistical inference.

Role of the Parameter Space

The parameter space specifies the set of all possible values for the parameter in the family of distributions. In completeness proofs, the extent of the parameter space is critical. A larger parameter space, such as (0, ?), can sometimes force a function to vanish identically because the condition of having zero expectation for all parameter values imposes stronger restrictions. Conversely, a restricted parameter space can allow nontrivial functions to have zero expectation across all parameter values, thereby violating completeness.

Completeness of a Family of Distributions

Completeness is a property of a family of probability distributions which states that if a measurable function has expectation zero for every parameter value in the family, then that function must be almost surely zero. This concept is crucial in ensuring uniqueness in unbiased estimation and in establishing that no nontrivial function of the data exists that is orthogonal to all functions of the parameter. It is a foundational concept in the theory of statistical inference.

Uniform Distribution and Its Density

The example uses a uniform distribution on the interval (0, ?), with density 1/?. Understanding the properties of the uniform distribution—especially its dependence on the parameter ?—is essential. In this context, examining how the expected value of a function of the variable behaves as a function of ? provides insights into whether the density function family is complete or not.

Please give Ace some feedback