0
$\begingroup$

I am trying to get a handle on finding pivotal quantities to use in confidence intervals. I came across a question regarding a uniform distribution:

Suppose that we take a sample of size $n = 1$ from a uniform distribution defined on the interval $ [0,θ]$, where $θ$ is unknown. Find a 95% lower confidence bound for $θ$.

Because Y is uniform on [0,θ], the method of transformation can be used to show that $U = Y/θ$ is uniformly distributed over $[0, 1]$. That is, $f_U (u) = 1, $$ 0≤u≤1.$

I thought the p.d.f. for this would have been $\frac{1}{\theta}$, so I am confused where the $U=\frac{Y}{\theta}$ came from. Maybe this is trivial, but it is new to me. Can someone explain how to actually arrive at this?

$\endgroup$
  • $\begingroup$ Pdf of $Y$ is $f_Y(y)=\frac{1}{\theta}\mathbf1_{0<y<\theta}$, which implies pdf of $U=Y/\theta$ is $f_U(u)=\mathbf1_{0<u<1}$. This can be shown from distribution functions: $P(U\le u)=P(Y/\theta\le u)=...$. If you are confused about why to consider $U$, then note that $U$ is a pivotal quantity, which can be used for deriving a confidence interval for $\theta$. $\endgroup$ – StubbornAtom Feb 12 at 7:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.