Solution Found!
The Rayleigh density function is given by ReferenceA
Chapter 9, Problem 34E(choose chapter or problem)
The Rayleigh density function is given by
\(f(y)=\left\{\begin{array}{cl}\left(\frac{2 y}{\theta}\right) e^{-y^{2} / \theta}, & y>0 \\0 & \text { elsewhere }\end{array}\right.\)
In Exercise 6.34(a), you established that \(Y^{2}\)has an exponential distribution with mean \(\theta\). If \(Y_{1}, Y_{2}, \ldots, Y_{n}\)denote a random sample from a Rayleigh distribution, show that \(W_{n}=\frac{1}{n} \sum_{i=1}^{n} Y_{i}^{2}\) is a consistent estimator for \(\theta\).
Equation Transcription:
{
Text Transcription:
f(x) = {(frac{2 y}{theta}) e^{-y^{2} /theta}, & y > 0 0 & elsewhere }
Y^2
theta
Y_1, Y_2, …., Y_n
W_n = frac{1}{n} sum_{i=1}^{n} Y_{i}^{2}
theta
Questions & Answers
QUESTION:
The Rayleigh density function is given by
\(f(y)=\left\{\begin{array}{cl}\left(\frac{2 y}{\theta}\right) e^{-y^{2} / \theta}, & y>0 \\0 & \text { elsewhere }\end{array}\right.\)
In Exercise 6.34(a), you established that \(Y^{2}\)has an exponential distribution with mean \(\theta\). If \(Y_{1}, Y_{2}, \ldots, Y_{n}\)denote a random sample from a Rayleigh distribution, show that \(W_{n}=\frac{1}{n} \sum_{i=1}^{n} Y_{i}^{2}\) is a consistent estimator for \(\theta\).
Equation Transcription:
{
Text Transcription:
f(x) = {(frac{2 y}{theta}) e^{-y^{2} /theta}, & y > 0 0 & elsewhere }
Y^2
theta
Y_1, Y_2, …., Y_n
W_n = frac{1}{n} sum_{i=1}^{n} Y_{i}^{2}
theta
ANSWER:Step 1 of 4
Here, we are given that the Rayleigh density function is given by
Let
Taking derivatives on both sides, we get
Therefore the probability density function of is given by
Therefore, is the exponential distribution