two parts

for F(y) = 2(sigma-y)/sigma^2 sigma > y > 0

0 elsewhere

show that y/sigma is a pivotal quantity

use the pivotal quantity to construct a 80% confidence interval for sigma

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- Nov 24th 2008, 11:49 AMbatemanlpivotal quantities
two parts

for F(y) = 2(sigma-y)/sigma^2 sigma > y > 0

0 elsewhere

show that y/sigma is a pivotal quantity

use the pivotal quantity to construct a 80% confidence interval for sigma - Nov 24th 2008, 05:48 PMmr fantastic
Have you tried reading these threads:

http://www.mathhelpforum.com/math-he...tribution.html

http://www.mathhelpforum.com/math-he...-function.html

http://www.mathhelpforum.com/math-he...-interval.html

http://www.mathhelpforum.com/math-he...-interval.html

If this doesn't help, I'll try to find the time later to say something specific about your question. - Nov 24th 2008, 07:52 PMmr fantastic
Let $\displaystyle U = \frac{Y}{\sigma}$.

Using the transformation method to get the pdf of U:

$\displaystyle u = \frac{y}{\sigma} \Rightarrow y = \sigma u$.

$\displaystyle f_U(u) = f_Y(y) \, \frac{dy}{du} = \frac{2(\sigma - y)}{\sigma^2} \, \sigma = \frac{2(\sigma - \sigma u)}{\sigma^2} \, \sigma = 2 - 2u$.

Note: The above gives the pdf for U when 0 < u < 1 (why?). Otherwise $\displaystyle f_U(u) = 0$.

So the two conditions for U to be a pivotal quantity are met.

You can also easily get the pdf of U by first getting the cdf of U and then using $\displaystyle f(u) = \frac{dF}{du}$.

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Now you require two numbers $\displaystyle \alpha$ and $\displaystyle \beta$ such that $\displaystyle \Pr(\alpha < U < \beta) = 0.8$.

So solve $\displaystyle \Pr(U > \beta) = 0.1$ and $\displaystyle \Pr(U < \alpha) = 0.1$ using the pdf for U found above. Then:

$\displaystyle \Pr(\alpha < U < \beta) = 0.8$

$\displaystyle \Rightarrow \Pr\left(\alpha < \frac{Y}{\sigma} < \beta\right) = 0.8$

$\displaystyle \Rightarrow \Pr\left(\beta < \frac{\sigma}{Y} < \alpha\right) = 0.8$

$\displaystyle \Rightarrow \Pr\left(\beta Y < \sigma < \alpha Y\right) = 0.8$.

Now you need data .....