Question

Statistics help. Please.
Problem 8.44 text book, part c
Let Y have probability density function \(f_y(y) = \dfrac{2(\theta -y)}{\theta^2}\) when \(0 < y<\theta\)
a) Show \(F_y(y) = \dfrac{2y}{\theta}-\dfrac{y^2}{\theta^2}\) when \(0 10 years ago

Answer 1

From b) I got \(F_U(u) = 2u-u^2\)
Hence for c) I need find a on \(P(a<u<b)= 0.9\)
\(=P(u<b) - P(u<a)=0.9\)
\(F_U(a) = 2a -a^2\) , then I stuck. If I assign \(P(u <a) = 0.05\) then \(2a-a^2 =0.05\) which gives me a = 2.024.
If I assign \(P(u<a) = 0.025\), then a = 1.98
so on.
I don't know exactly what I have to do to get correct lower confidence. In other words, I don't know which % I should apply to get a.

Answer 2

@Zarkon

Answer 3

It just depends on what "lower confidence limit" means:

which side?

Answer 4

I think -without reading your book- that it means: \(P(U>q) = 0.9\)
Then,
\[
P\left(\frac{Y}{\theta} \le q\right) = 0.1\\
\]
You must solve \(F_U(u) = 2u-u^2 = 0.1\). -> \(u=0.0513\). So,
\[
P(Y \le \theta q) = 0.1
\]
equivalently,
\[
P(\frac{Y}q \le \theta) = 0.1
\]

Answer 5

On \(P(a<u<b) =0.9\) ,we find a, b then do some more calculation to get the lower and upper limits,
If I imitate exactly what the book says, then
\(2b-b^2 =0.975\) and it gives me b = 1.158 or b= 0.84188
suppose b = 1.158 and a = 0.053, then
\(P(0.053<u<1.158) = 0.9\\P(0.053<\dfrac{Y}{\theta} < 1.158=0.9\)
Hence the CI is \([\dfrac{Y}{1.158},\dfrac{Y}{0.053}]\)
Then, lower limit is \(\dfrac{Y}{1.158}\)
but it is weak to me. Since I can change the point a, b any time.

Answer 6

The "book" considers confidence intervals of the form [a,b]. If `c) Use b to find a 90% lower confidence limit for \(\theta\) ` means `find the lower limit of a 90% confidence interval`, then you are right.

But I thought it might means `find a confidence interval such that we just want to be 90% sure that theta is above some value`. -> inteval=(q, infinity).

My interpretation is a bit of a stretch.. I think you are right. You must look at the value q such that \(F_U(q)=0.05\), then \(P(Y/\theta < q) = P(Y/q<\theta)\), and the lower bound is \(Y/q\).

The solution to \(F(u)=0.05\) is 0.02532. (it is logical that it is smaller than the value \(q\) such that \(F(q)=0.1\). (see why?)

Answer 7

(0.1 /2 = 0.05 , not 0.025.)

Answer 8

Yes, I saw my mistake. Thanks for that,
but we need find b, right? and we assume that \(P (u < b) = 0.95\) so that \(2b-b^2 =0.95\)
that give me b = 1.223 or b = 0.776

Answer 9

Then \(Y/\theta > 0.776\) which gives lower limit is Y/0.776

Answer 10

0.776 is right, because the maximual value of \(U=Y/\theta\) is \(1\).

0.776 would be the upper limit of the confidence interval.

Answer 11

while Y/1.223 < Y /0.776

Answer 12

why don't we take Y/1.223?

Answer 13

that's not the argument. This is not true: \(P(U<b) = 0.95\). (it is equal to 1)

Answer 14

(if b =1.223)

Answer 15

I am sorry, I don't get what you mean

Answer 16

\(P(U\le 1) = 1\) , so \(P(U\le 1.223)=1\)

Answer 17

oh, got it. Thank you