# Rayleigh distribution and unbiased estimator

• June 19th 2010, 12:57 PM
losm1
Rayleigh distribution and unbiased estimator
$X_1 ... X_n$ is random sample from Rayleigh distribution $f(x;\Theta)$

1. Show that $E(X^2) = 2\Theta$ and than construct unbiased estimator of parameter $\Theta$ based on $\sum_{k=1}^n X_k^2$

2. Estimate parameter $\Theta$ from following $n=10$ observations:
Code:

16.88 10.23 4.59 6.66 13.68 14.23 19.87 9.40 6.51 10.95
---

1. I have just plugged in 2 theta in Rayleigh's variance formula and it evaluates to true, but I'm not sure about correct way of constructing unbiased estimator $\hat{\Theta}$

$Var(X)=E(X^2) - E^2(X)$

$Var(X)=2\Theta - E^2(X)$

$\frac{4-\pi}{2}\Theta=2\Theta - \sqrt{\Theta\cdot\frac{\pi}{2}}^2$

$4\Theta - \pi\Theta = 2(2\Theta - \frac{\pi\Theta}{2})$

$\Theta = \Theta$

2. I need help with this one
• June 19th 2010, 02:05 PM
mr fantastic
Quote:

Originally Posted by losm1
$X_1 ... X_n$ is random sample from Rayleigh distribution $f(x;\Theta)$

1. Show that $E(X^2) = 2\Theta$ and than construct unbiased estimator of parameter $\Theta$ based on $\sum_{k=1}^n X_k^2$

2. Estimate parameter $\Theta$ from following $n=10$ observations:
Code:

16.88 10.23 4.59 6.66 13.68 14.23 19.87 9.40 6.51 10.95
---

1. I have just plugged in 2 theta in Rayleigh's variance formula and it evaluates to true, but I'm not sure about correct way of constructing unbiased estimator $\hat{\Theta}$

$Var(X)=E(X^2) - E^2(X)$

$Var(X)=2\Theta - E^2(X)$

$\frac{4-\pi}{2}\Theta=2\Theta - \sqrt{\Theta\cdot\frac{\pi}{2}}^2$

$4\Theta - \pi\Theta = 2(2\Theta - \frac{\pi\Theta}{2})$

$\Theta = \Theta$

2. I need help with this one

Use the unbiased estimator found in part 1!
• June 20th 2010, 02:07 AM
losm1
First, how do I know that it is unbiased?
Second, basically I should just calculate $E(X^2)$ with given dataset?

Thanks
• June 20th 2010, 02:26 AM
SpringFan25
An estimate is unbiased if its expected value is equal to the true value of the parameter being estimated. So, you can confirm the estimate is unbiased by taking its expectation.

$\hat{\Theta} = \frac{\sum{x^2}}{2n}$
$E(\hat{\Theta}) = E \left(\frac{\sum{x^2}}{2n} \right)$
$E(\hat{\Theta}) = 0.5n^{-1}1 E \left(\sum{x^2} \right)$
$E(\hat{\Theta}) = 0.5n^{-1} \left(\sum{E(x_i^2)} \right)$
$E(\hat{\Theta}) = 0.5n^{-1} \left(\sum{E(X^2)} \right)$
$E(\hat{\Theta}) = 0.5n^{-1} * n{E(X^2) \right)$
$E(\hat{\Theta}) = 0.5 E(X^2) \right)$
$E(\hat{\Theta}) = \Theta$