Question

Given \(f(x ; \theta)=1 / \theta, 00\), formally compute the reciprocal of $$ n E\left\\{\left[\frac{\partial \log f(X: \theta)}{\partial \theta}\right]^{2}\right\\} $$ Compare this with the variance of \((n+1) Y_{n} / n\), where \(Y_{n}\) is the largest observation of a random sample of size \(n\) from this distribution. Comment.

Short answer

The reciprocal of the expectation of the square of the derivative of the logarithm of \(f(x;\theta)\) with respect to \(\theta\) is \(\theta\). The variance of \((n+1)Y_n/n\) where \(Y_n\) is the largest observation in a random sample of size \(n\) is \((n+1)^2/n^2\). The two expressions do not coincide.

Step-by-step solution

Computing derivative

The first step is to compute the derivative of the logarithm of \(f(x;\theta)\) with respect to \(\theta\). The function \(f(x;\theta) = 1/ \theta\) has the domain \(0 < x < \theta\) and its logarithm is \(\log f(x;\theta) = log(1/\theta) = -log \theta\). The derivative of this with respect to \(\theta\) is \(-1/\theta\). The square of the derivative is \(-1/\theta^2\).

Computing expectation

The next step is to compute the expectation \(E\left\{ \left[ -1/\theta \right]^2 \right\}\). This is given by \(\int_0^\theta 1/ \theta^2 dx\). Since the integrand does not depend on \(x\), the integral is the integrand times \(\theta\) which results in \(1/\theta\). The reciprocal of this is \(\theta\).

Find the variance

The final step is to compute the variance of \((n+1)Y_n/n\) where \(Y_n\) is the largest observation in a random sample of size \(n\). The variance of \((n+1)Y_n/n = (n+1)Y_n/n^2\delta\) is given by the expectation of the square of \(\frac{(n+1)^2 Y_n^2}{n^2}\) both of which are equal to unity. Thus, the variance of \((n+1)Y_n/n\) is \((n+1)^2/n^2\).