# Method of Transformation in Statistics

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?

• 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 $\theta$. – StubbornAtom Feb 12 at 7:40