Solution Found!
Let X1,X2, . . . ,Xn be a random sample from N(?1, ?2).
Chapter 6, Problem 9E(choose chapter or problem)
Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from \(N\left(\theta_{1}, \theta_{2}\right)\). Show that the sufficient statistics \(Y_{1}=\bar{X}\) and \(Y_{2}=S^{2}\) are independent of the statistic.
\(Z=\sum_{i=1}^{n-1} \frac{\left(X_{i+1}-X_{i}\right)^{2}}{S^{2}}\)
because \(Z\) has a distribution that is free of \(\theta_{1}\) and \(\theta_{2}\).
HINT: Let \(w_{i}=\left(x_{i}-\theta_{1}\right) / \sqrt{\theta_{2}}, i=1,2, \ldots, n\), in the multivariate integral representing \(E\left[e^{z^{z}}\right]\).
Equation Transcription:
Text Transcription:
X_1,X_2,...,X_n
N(theta_1,theta_2)
Y_1=bar X
Y_2=S^2
Z=sum_i=1^n-1 (X_i+1-X_i)^2/S^2
Z
theta_1
theta_2
w_i=(x_i-theta_1)/sqrt theta_2,i=1,2,...,n
E[e^tZ]
Questions & Answers
QUESTION:
Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from \(N\left(\theta_{1}, \theta_{2}\right)\). Show that the sufficient statistics \(Y_{1}=\bar{X}\) and \(Y_{2}=S^{2}\) are independent of the statistic.
\(Z=\sum_{i=1}^{n-1} \frac{\left(X_{i+1}-X_{i}\right)^{2}}{S^{2}}\)
because \(Z\) has a distribution that is free of \(\theta_{1}\) and \(\theta_{2}\).
HINT: Let \(w_{i}=\left(x_{i}-\theta_{1}\right) / \sqrt{\theta_{2}}, i=1,2, \ldots, n\), in the multivariate integral representing \(E\left[e^{z^{z}}\right]\).
Equation Transcription:
Text Transcription:
X_1,X_2,...,X_n
N(theta_1,theta_2)
Y_1=bar X
Y_2=S^2
Z=sum_i=1^n-1 (X_i+1-X_i)^2/S^2
Z
theta_1
theta_2
w_i=(x_i-theta_1)/sqrt theta_2,i=1,2,...,n
E[e^tZ]
ANSWER:
Step 1 of 3
The joint pdf of the random variablesis,