$f_Y(y)= \dfrac{2(\theta -y)}{\theta^2}, 0

Question

$f_Y(y)= \dfrac{2(\theta -y)}{\theta^2}, 0<y<\theta$
Pivotal quantity is $Y/\theta$.
Use the pivotal quantity to find a 90% lower confidence limit for $\theta$
Please, help

loser66 · Answer

I got $F_U(u) = 2u-u^2$
I understand the problem like this: I need find a, b, such that P(a<U<b) =0.90 
and after finding out b, we can get the lower confidence. But it is not that. 
My Prof used directly $P(U < b)=0.90$
It confused me.

reemii · Answer

I see. It all depends on how you understand `a 90% lower confidence limit for θ`.
If this means `P(U < b)=0.90` or `P(U > b)=0.90` or the lower limit of the interval [a,b] such that `P(a < U < b)=0.90`.
You have to ask the professor or look in the course what it means and just use that all the time. Honestly I would not know, I would ask the professor..

loser66 · Answer

Thanks a lot. I will ask my prof also.

reemii · Answer

In my opinion it would rather be `P(Y > b) = 0.9` ... :)
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