Why is the expression $fracI(x>theta)I(x>theta)$ independent of $theta$?

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This question has been asked before but I can't really grasp the explanations.



Let $f(xmid theta)=fracI(x>theta)I(x>theta)$ for some particular value of $x$. I want to show that $f$ is independent of $theta$, but I cannot reconcile this with the fact that if $theta<X$, then $f=1$, and if $theta>X$, then $f=frac00$.
Could someone give me an intuitive and full explanation for why $ f=frac00=1$ in this scenario?




real-analysis probability real-numbers




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  • Note $fracajfit184mgajfit184mg=1.$
    – Dzoooks
    Aug 18 at 2:59















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down vote

favorite
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This question has been asked before but I can't really grasp the explanations.



Let $f(xmid theta)=fracI(x>theta)I(x>theta)$ for some particular value of $x$. I want to show that $f$ is independent of $theta$, but I cannot reconcile this with the fact that if $theta<X$, then $f=1$, and if $theta>X$, then $f=frac00$.
Could someone give me an intuitive and full explanation for why $ f=frac00=1$ in this scenario?




real-analysis probability real-numbers




share|cite|improve this question






















  • Note $fracajfit184mgajfit184mg=1.$
    – Dzoooks
    Aug 18 at 2:59













up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





This question has been asked before but I can't really grasp the explanations.



Let $f(xmid theta)=fracI(x>theta)I(x>theta)$ for some particular value of $x$. I want to show that $f$ is independent of $theta$, but I cannot reconcile this with the fact that if $theta<X$, then $f=1$, and if $theta>X$, then $f=frac00$.
Could someone give me an intuitive and full explanation for why $ f=frac00=1$ in this scenario?




real-analysis probability real-numbers




share|cite|improve this question














This question has been asked before but I can't really grasp the explanations.



Let $f(xmid theta)=fracI(x>theta)I(x>theta)$ for some particular value of $x$. I want to show that $f$ is independent of $theta$, but I cannot reconcile this with the fact that if $theta<X$, then $f=1$, and if $theta>X$, then $f=frac00$.
Could someone give me an intuitive and full explanation for why $ f=frac00=1$ in this scenario?





real-analysis probability real-numbers





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edited Aug 18 at 3:13









Michael Hardy

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asked Aug 18 at 2:58









DavidS

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  • Note $fracajfit184mgajfit184mg=1.$
    – Dzoooks
    Aug 18 at 2:59

















  • Note $fracajfit184mgajfit184mg=1.$
    – Dzoooks
    Aug 18 at 2:59
















Note $fracajfit184mgajfit184mg=1.$
– Dzoooks
Aug 18 at 2:59





Note $fracajfit184mgajfit184mg=1.$
– Dzoooks
Aug 18 at 2:59











2 Answers
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The value $0/0$ is somewhat arbitrary. We may assume $0/0 = 1$, in which case $f(x|theta)$ is indeed independent of $theta$. But, if we pick $0/0 = 0$, then obviously it depends on $theta$.



Your source probably assumes $0/0 = 1$.






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  • The book does assume that $0/0=1$, but I don't feel comfortable using this assumption without understanding it.
    – DavidS
    Aug 18 at 3:20






  • 1




    0/0 is really indeterminate, but we can give it a value for notation convenience. For example in your example we have without assumptions, $I(x > theta) f(x | theta) = I(x > theta)$. Letting $0/0 = 1$ lets us write is down more compactly as $f(x | theta) = 1$.
    – Guillaume F.
    Aug 18 at 3:30

















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I wouldn't say $0/0=1$ in this situation.



Suppose it is asserted that $f(xmidtheta) = dfracg(theta)h(theta)$ does not depend on $theta.$ As long as there is no value of $theta$ for which $h(theta)=0,$ that statement is not problematic. And it is equivalent to saying there is some constant $c,$ not depending on $theta,$ for which $h(theta) f(xmidtheta) = ccdot g(theta)$ for all values of $theta.$ And that last statement is probably what was intended.






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    2 Answers
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    2 Answers
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    up vote
    0
    down vote













    The value $0/0$ is somewhat arbitrary. We may assume $0/0 = 1$, in which case $f(x|theta)$ is indeed independent of $theta$. But, if we pick $0/0 = 0$, then obviously it depends on $theta$.



    Your source probably assumes $0/0 = 1$.






    share|cite|improve this answer




















    • The book does assume that $0/0=1$, but I don't feel comfortable using this assumption without understanding it.
      – DavidS
      Aug 18 at 3:20






    • 1




      0/0 is really indeterminate, but we can give it a value for notation convenience. For example in your example we have without assumptions, $I(x > theta) f(x | theta) = I(x > theta)$. Letting $0/0 = 1$ lets us write is down more compactly as $f(x | theta) = 1$.
      – Guillaume F.
      Aug 18 at 3:30














    up vote
    0
    down vote













    The value $0/0$ is somewhat arbitrary. We may assume $0/0 = 1$, in which case $f(x|theta)$ is indeed independent of $theta$. But, if we pick $0/0 = 0$, then obviously it depends on $theta$.



    Your source probably assumes $0/0 = 1$.






    share|cite|improve this answer




















    • The book does assume that $0/0=1$, but I don't feel comfortable using this assumption without understanding it.
      – DavidS
      Aug 18 at 3:20






    • 1




      0/0 is really indeterminate, but we can give it a value for notation convenience. For example in your example we have without assumptions, $I(x > theta) f(x | theta) = I(x > theta)$. Letting $0/0 = 1$ lets us write is down more compactly as $f(x | theta) = 1$.
      – Guillaume F.
      Aug 18 at 3:30












    up vote
    0
    down vote










    up vote
    0
    down vote









    The value $0/0$ is somewhat arbitrary. We may assume $0/0 = 1$, in which case $f(x|theta)$ is indeed independent of $theta$. But, if we pick $0/0 = 0$, then obviously it depends on $theta$.



    Your source probably assumes $0/0 = 1$.






    share|cite|improve this answer












    The value $0/0$ is somewhat arbitrary. We may assume $0/0 = 1$, in which case $f(x|theta)$ is indeed independent of $theta$. But, if we pick $0/0 = 0$, then obviously it depends on $theta$.



    Your source probably assumes $0/0 = 1$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Aug 18 at 3:19









    Guillaume F.

    349211




    349211











    • The book does assume that $0/0=1$, but I don't feel comfortable using this assumption without understanding it.
      – DavidS
      Aug 18 at 3:20






    • 1




      0/0 is really indeterminate, but we can give it a value for notation convenience. For example in your example we have without assumptions, $I(x > theta) f(x | theta) = I(x > theta)$. Letting $0/0 = 1$ lets us write is down more compactly as $f(x | theta) = 1$.
      – Guillaume F.
      Aug 18 at 3:30
















    • The book does assume that $0/0=1$, but I don't feel comfortable using this assumption without understanding it.
      – DavidS
      Aug 18 at 3:20






    • 1




      0/0 is really indeterminate, but we can give it a value for notation convenience. For example in your example we have without assumptions, $I(x > theta) f(x | theta) = I(x > theta)$. Letting $0/0 = 1$ lets us write is down more compactly as $f(x | theta) = 1$.
      – Guillaume F.
      Aug 18 at 3:30















    The book does assume that $0/0=1$, but I don't feel comfortable using this assumption without understanding it.
    – DavidS
    Aug 18 at 3:20




    The book does assume that $0/0=1$, but I don't feel comfortable using this assumption without understanding it.
    – DavidS
    Aug 18 at 3:20




    1




    1




    0/0 is really indeterminate, but we can give it a value for notation convenience. For example in your example we have without assumptions, $I(x > theta) f(x | theta) = I(x > theta)$. Letting $0/0 = 1$ lets us write is down more compactly as $f(x | theta) = 1$.
    – Guillaume F.
    Aug 18 at 3:30




    0/0 is really indeterminate, but we can give it a value for notation convenience. For example in your example we have without assumptions, $I(x > theta) f(x | theta) = I(x > theta)$. Letting $0/0 = 1$ lets us write is down more compactly as $f(x | theta) = 1$.
    – Guillaume F.
    Aug 18 at 3:30










    up vote
    0
    down vote













    I wouldn't say $0/0=1$ in this situation.



    Suppose it is asserted that $f(xmidtheta) = dfracg(theta)h(theta)$ does not depend on $theta.$ As long as there is no value of $theta$ for which $h(theta)=0,$ that statement is not problematic. And it is equivalent to saying there is some constant $c,$ not depending on $theta,$ for which $h(theta) f(xmidtheta) = ccdot g(theta)$ for all values of $theta.$ And that last statement is probably what was intended.






    share|cite|improve this answer
























      up vote
      0
      down vote













      I wouldn't say $0/0=1$ in this situation.



      Suppose it is asserted that $f(xmidtheta) = dfracg(theta)h(theta)$ does not depend on $theta.$ As long as there is no value of $theta$ for which $h(theta)=0,$ that statement is not problematic. And it is equivalent to saying there is some constant $c,$ not depending on $theta,$ for which $h(theta) f(xmidtheta) = ccdot g(theta)$ for all values of $theta.$ And that last statement is probably what was intended.






      share|cite|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        I wouldn't say $0/0=1$ in this situation.



        Suppose it is asserted that $f(xmidtheta) = dfracg(theta)h(theta)$ does not depend on $theta.$ As long as there is no value of $theta$ for which $h(theta)=0,$ that statement is not problematic. And it is equivalent to saying there is some constant $c,$ not depending on $theta,$ for which $h(theta) f(xmidtheta) = ccdot g(theta)$ for all values of $theta.$ And that last statement is probably what was intended.






        share|cite|improve this answer












        I wouldn't say $0/0=1$ in this situation.



        Suppose it is asserted that $f(xmidtheta) = dfracg(theta)h(theta)$ does not depend on $theta.$ As long as there is no value of $theta$ for which $h(theta)=0,$ that statement is not problematic. And it is equivalent to saying there is some constant $c,$ not depending on $theta,$ for which $h(theta) f(xmidtheta) = ccdot g(theta)$ for all values of $theta.$ And that last statement is probably what was intended.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 18 at 3:20









        Michael Hardy

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