Question operation on indicator function [uniform distribution]

Question

I do not understand a particular step in deriving the likelihood function for a continuous uniform distribution $U[0,\theta]$. This has to do with a particular operation on the indicator function.

My problem arises in Step 4. What is the rationale in this step? What is the performed operation and why is it allowed? My tutorial teacher said the inverse is taken of all terms but how do we in that case come up with the interval $[0,1]$. She said to use 1/0 = $\infty$ but if I recall correctly this is not true (division by zero is undefined).

Answer 1

It is allowed because if $y>0, \theta>0$ then $\frac{\theta}{y_{(n)}}\geq1\Rightarrow \theta\geq y_{(n)}\Rightarrow 1\geq\frac{y_{(n)}}{\theta}>0$

So $I_{[1,\infty]}\bigg(\frac{\theta}{y_{(n)}}\bigg)=I_{(0,1]}\bigg(\frac{y_{(n)}}{\theta}\bigg)$ or $I_{[1,\infty]}\bigg(\frac{\theta}{y_{(n)}}\bigg)=I_{[0,1]}\bigg(\frac{y_{(n)}}{\theta}\bigg)$ a.e. (almost everywhere).

Answer 2

Because for positive $n$, if $n< x\leqslant 1$ then $1\leqslant x^{-1}<n^{-1}$ so therefore taking the limits: $$\lim_{n\to 0^+}\mathbf I_{[n;1]}(x)=\lim_{m\to\infty}\mathbf I_{[1;m]}(x^{-1})\qquad (m=n^{-1})$$

Which we can abbreviate as: $$\mathbf I_{[0;1]}(x)=\mathbf I_{[1;\infty)}(x^{-1})$$

That is all.