Question

Show that each of the following families is not complete by finding at least one nonzero function \(u(x)\) such that \(E[u(X)]=0\), for all \(\theta>0\). (a) $$ f(x ; \theta)=\left\\{\begin{array}{ll} \frac{1}{2 \theta} & -\theta

Short answer

Distribution (a) is not complete because for \(u(x) = x\), \(E[u(X)]=0\), but \(u(x) ≠ 0\). Likewise, distribution (b) is not complete as for \(u(x) = x^3\), \(E[u(X)]=0\), but \(u(x) ≠ 0\).

Step-by-step solution

Finding function for distribution (a)

Consider the uniform density function given in the problem. One can pick the function \(u(x) = x\). Then, one can calculate the expectation \(E[u(X)] = E[X]\) which is the mean of a uniform distribution: \(E[X] = \int_{-\theta}^{\theta}x*\frac{1}{2\theta}dx = 0\). However, \(u(x) = x\) is not equal to zero for all x ∈ (-θ, θ), hence distribution (a) is not complete.

Finding function for distribution (b)

The given distribution is a normal distribution \(N(0,\theta)\). In this case, one can take the function \(u(x) = x^3\). The expectation \(E[u(X)] = E[X^3]\) for the normal distribution becomes zero when the power of x is an odd number. Thus, \(E[X^3] = 0\), but \(x^3\) is not equal to zero for all \(x ≠ 0\). Hence, the distribution (b) is also not complete.