Question

Let \(Y_{1}

Short answer

The distribution function for \(Z\) is \(H_{n}(z ; \theta) = 1- \left(1-\frac{z}{n \theta}\right)^{n}\) for \(0 < z < n\theta\). The limiting distribution of \(Z\) as \(n\) approaches infinity is an exponential distribution with mean \(\theta\).

Step-by-step solution

Change of variables

Make a change of variables for \(Y_{n} = \theta - Z/n\). Then find the Jacobian of this transformation, which is \(-1/n\).

Compute the density function \(h_{n}(z ; \theta) = g\left(\theta - \frac{z}{n} ; \theta\right) \times |J|\).

Substitute the value of \(Y_{n}\) into the density function \(g\), then obtain \(h_{n}(z ; \theta)\) by multiplying by the absolute value of the Jacobian. Hence we obtain \(h_{n}(z ; \theta)= n\left(\theta-\frac{z}{n}\right)^{n-1} \times \frac{1}{n} = (\theta-\frac{z}{n})^{n-1}\). This is valid for \(0 < z < n\theta\).

Compute the distribution function \(H_{n}(z ; \theta) = \int_{0}^{z} h_{n}(x ; \theta) dx\).

Integrate \(h_{n}(z ; \theta)\) with respect to \(x\) from \(0\) to \(z\). Applying the change of variables \(v=\theta-x/n\), we get \(H_{n}(z ; \theta)=\int_{\theta}^{\theta-z/n} v^{n-1} dv = 1- \left(1-\frac{z}{n \theta}\right)^{n}\) for \(0 < z < n\theta\).

Compute the limit \(\lim _{n \rightarrow \infty} H_{n}(z ; \theta)\).

As \(n\) goes to infinity, this limit becomes \(\lim _{n \rightarrow \infty}\left[1- \left(1-\frac{z}{n \theta}\right)^{n}\right] = 1- e^{- z/\theta}\) for \(z > 0\). This is the cumulative distribution function of an exponential distribution with mean \(\theta\).