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 .
Using the transformation method to get the pdf of U:
.
.
Note: The above gives the pdf for U when 0 < u < 1 (why?). Otherwise .
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 .
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Now you require two numbers and such that .
So solve and using the pdf for U found above. Then:
.
Now you need data .....