I am reviewing the following example and am confused on a portion of it
Let X1,...,Xn be a random sample from the continuous uniform distribution on the interval from $0$ to $\theta$
Using the CDF method, it can be shown $Y_n = max(X_1,...,X_n)$ has the pdf given by
$ f(y;\theta) = \frac{n}{\theta ^n}y^{n-1}, 0 < y < \theta$
and $0$ otherwise
Find a valid $100(1-\alpha)$ confidence interval for $\theta$
How I have approached it,
The statistic (or data) is given $Y_n = max(X_1,...,X_n)$
I then am trying to identify a function of the statistic. The example's worked out solution says let $V = \frac {Y}{\theta}$, the sample mean
They then solve for $Y$ and plug $V\theta$ into their PDF and get the following
$ f(y;\theta) = \frac{n}{\theta ^n}(v\theta)^{n-1} *\theta$
where does the last $\theta$ term come from here? I thought after choosing the above function you would end up with...
$f(y;\theta) = \frac{n}{\theta ^n}(v\theta)^{n-1}$
after this they simplify and take the intergal from 0 to k with the integral set equal to
$1-\alpha$ and solve for $k$
The resultant answer to the question is then
$0 < \frac{Y}{(1-\alpha)^\frac {1}{n}} < \theta$
I understand the algebra once you have plugged in your function but I am not understanding how the function $V = \frac{y}{\theta}$ is chosen and then how after plugging it in they arrive at $ f(y;\theta) = \frac{n}{\theta ^n}(v\theta)^{n-1} *\theta$
I have tried to solve the problem without the extra $\theta$ and did not get the correct answer.
You know for sure that $\theta \ge Y_n$, so a reasonable confidence interval might be $[Y_n, t]$ for some $t> Y_n$. By scaling invariance, it seems reasonable to take $t = c Y_n$ where $c$ is a constant $> 1$. The probability that the interval $[Y_n, c Y_n]$ contains $\theta$, i.e. that $Y_n \ge \theta/c$, is $1 - c^{-n}$ (i.e. the complement of this event is that all $X_n < \theta/c$). To make this probability be $1-\alpha$, you want $\alpha = c^{-n}$, i.e. $c = \alpha^{-1/n}$.