This is on my professor's notes: $$\frac {1_{0<y<x<1} (x,y)} {1_{0,1}(x)}=1_{0,x}(y).$$ I can provide more context if you want but I believe this is sufficient for my question.

I am ultimately wondering why would $x$ not need to be in $(0,1)$ in the RHS? Is there a general way to derive this?

The reason I tagged is because I want to know what happens if ${1_{0,1}(x)}=0$? Would it then be undefined?

  • $\begingroup$ As far as I'm concerned, you're right: if $1_{0,1}(x) = 0$ then the quotient is undefined, so the expression can't be generally true. Your professor might have a convention that $\frac{0}{0} = 1$ in this situation, or similar. $\endgroup$ – Patrick Stevens Apr 8 '18 at 8:05
  • $\begingroup$ no he doesn't which is weird but if the convetion is 0/0=1 then thats why we dont have x between (0,1) on the RHS? since they can cancel out? $\endgroup$ – james black Apr 8 '18 at 8:18
  • $\begingroup$ maybe the function on the LHS has been defined (only) on $\{(x,y):0<x,y<1\}$? $\endgroup$ – user52227 Apr 8 '18 at 9:01
  • $\begingroup$ the full question is $f_{x,y}$ is defined on $0<y<x<1$ so we calculate $f_{y given x}=f_{x,y}/f_x$ and if we integrate, we would get f_x defined on (0,1); then we derive the result, which i dont think would be defined on anything? $\endgroup$ – james black Apr 8 '18 at 9:04
  • $\begingroup$ its a bivariate probability density based on nomral distribution but that shouldnt matte $\endgroup$ – james black Apr 8 '18 at 9:05

Division by $0$ is undefined. This means that for $x\neq 0,1$ your LHS is undefined, no matter your conventions or other definitions.

Also, you cannot just define $\frac00=1$, for this will get you into trouble. For instance it would imply that $$\frac00\cdot0=1\cdot0 =0,$$ and that $$\frac00\cdot0=\frac{0\cdot0}0=\frac00=1.$$ This implies $0=1$. This is either a contradiction or a very hard constraint on the algebraic set $x$ and $y$ live in.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.