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
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.
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 $\displaystyle f(u) = \frac{dF}{du}$.
----------------------------------------------------------------------------------
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 .....