It is sometimes relatively easy to establish consistency

Chapter 9, Problem 26E

(choose chapter or problem)

It is sometimes relatively easy to establish consistency or lack of consistency by appealing directly to Definition 9.2, evaluating \(P\left(\left|\widehat{\theta}_{n}-\theta\right| \leq \varepsilon\right)\)directly, and then showing that \(lim _{n \rightarrow \infty} P\left(\left|\hat{\theta}_{2}-\theta\right| \leq \varepsilon\right) = 1\). Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) denote a random sample of size \(n\) from a uniform distribution on the interval \((0, \theta)\). If \(Y_{(n)}=\max \left(Y_{1}, Y_{2}, \ldots, Y_{n}\right)\) we showed in Exercise 6.74 that the probability distribution function of \(Y_{(n)}\) is given by

                                                       \(F_{(n)}(y)=\left\{\begin{array}{cc}0, & y<0 \\(y / \theta)^{n}, & 0 \leq y \leq \theta, \\1, & y>\theta .]\end{array}\right.\)

a  For each \(n \geq 1\) and every \(\varepsilon>\theta\), it follows that \(P\left(\left|Y_{(n)}-\theta\right| \leq \varepsilon\right)=P\left(\theta-\varepsilon \leq Y_{(n)} \leq \theta+\varepsilon\right.\). If \(\varepsilon>\theta\), verify that \(P\left(\theta-\varepsilon \leq Y_{(n)} \leq \theta+\varepsilon\right)=1\) and that, for every positive \(\varepsilon>\theta\), we obtain  \(P\left(\theta-\varepsilon \leq Y_{(n)} \leq \theta+\varepsilon\right)=1-[(\theta-\varepsilon) / \theta]^{n}\)

b  Using the result from part (a), show that \(Y_{(n)}\) is a consistent estimator for \(\theta\) by showing that, for every \(\varepsilon>0, \lim _{n \rightarrow \infty} P\left(\left|Y_{(n)}-\theta\right| \leq \varepsilon\right)=1\).

Equation Transcription:

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Text Transcription:

P(hat{theta}_{n} - theta| leq varepsilon)

lim _{n rightarrow infty} P(|hat{theta}_{2} -theta| leq varepsion) = 1

Y_{1}, Y_{2}, \ldots, Y_{n}

n

(0, theta)

Y_(n) = max (Y_1, Y_2, …., Y_n)

Y_n

F_{(n)}(y) = {\begin{array}{cl}0, & y<0 \\(y/theta)^{n}, & 0 leq y leq theta \\1, & y > theta.]

n geq 1

varepsilon > theta

P(|Y_{(n)} - theta| leq varepsilon) = P(theta - varepsilon leq Y_{(n)} leq theta + varepsilon.

varepsilon > theta

P(theta - varepsilon leq Y_{(n)} leq theta + varepsilon) = 1

varepsilon > theta

P(theta - varepsilon leq Y_{(n)} leq theta + varepsilon) = 1 -[(theta - varepsilon) / theta]^n

Y_n

theta

varepsilon > 0, lim _{n rightarrow infty} P(|Y_{(n)} - theta| leq varepsilon) = 1