Understanding $P[X gt alpha] = P[e^theta X gt e^theta alpha]$

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In Chernof Bound, it is expressed $P[X gt alpha] = P[e^theta X gt e^theta alpha]$ for given $M_X(theta) = E[e^theta X]$.



I am not getting how we can equal $P[X gt alpha] = P[e^theta X gt e^theta alpha]$.



Can you clarify this?




probability random-variables




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    Assuming $theta$ is positive, we have $P(X>alpha)=P(theta X>thetaalpha)$. Now use the fact that $e^x$ is an increasing function of $x$.
    – StubbornAtom
    Aug 28 at 17:35







  • 2




    If $f$ is increasing, then $[X > alpha] = [f(X) > f(alpha)]$.
    – copper.hat
    Aug 28 at 17:55














up vote
0
down vote

favorite












In Chernof Bound, it is expressed $P[X gt alpha] = P[e^theta X gt e^theta alpha]$ for given $M_X(theta) = E[e^theta X]$.



I am not getting how we can equal $P[X gt alpha] = P[e^theta X gt e^theta alpha]$.



Can you clarify this?




probability random-variables




share|cite|improve this question


















  • 2




    Assuming $theta$ is positive, we have $P(X>alpha)=P(theta X>thetaalpha)$. Now use the fact that $e^x$ is an increasing function of $x$.
    – StubbornAtom
    Aug 28 at 17:35







  • 2




    If $f$ is increasing, then $[X > alpha] = [f(X) > f(alpha)]$.
    – copper.hat
    Aug 28 at 17:55












up vote
0
down vote

favorite









up vote
0
down vote

favorite











In Chernof Bound, it is expressed $P[X gt alpha] = P[e^theta X gt e^theta alpha]$ for given $M_X(theta) = E[e^theta X]$.



I am not getting how we can equal $P[X gt alpha] = P[e^theta X gt e^theta alpha]$.



Can you clarify this?




probability random-variables




share|cite|improve this question














In Chernof Bound, it is expressed $P[X gt alpha] = P[e^theta X gt e^theta alpha]$ for given $M_X(theta) = E[e^theta X]$.



I am not getting how we can equal $P[X gt alpha] = P[e^theta X gt e^theta alpha]$.



Can you clarify this?





probability random-variables





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 28 at 19:43









BruceET

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asked Aug 28 at 17:27









Malintha

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  • 2




    Assuming $theta$ is positive, we have $P(X>alpha)=P(theta X>thetaalpha)$. Now use the fact that $e^x$ is an increasing function of $x$.
    – StubbornAtom
    Aug 28 at 17:35







  • 2




    If $f$ is increasing, then $[X > alpha] = [f(X) > f(alpha)]$.
    – copper.hat
    Aug 28 at 17:55












  • 2




    Assuming $theta$ is positive, we have $P(X>alpha)=P(theta X>thetaalpha)$. Now use the fact that $e^x$ is an increasing function of $x$.
    – StubbornAtom
    Aug 28 at 17:35







  • 2




    If $f$ is increasing, then $[X > alpha] = [f(X) > f(alpha)]$.
    – copper.hat
    Aug 28 at 17:55







2




2




Assuming $theta$ is positive, we have $P(X>alpha)=P(theta X>thetaalpha)$. Now use the fact that $e^x$ is an increasing function of $x$.
– StubbornAtom
Aug 28 at 17:35





Assuming $theta$ is positive, we have $P(X>alpha)=P(theta X>thetaalpha)$. Now use the fact that $e^x$ is an increasing function of $x$.
– StubbornAtom
Aug 28 at 17:35





2




2




If $f$ is increasing, then $[X > alpha] = [f(X) > f(alpha)]$.
– copper.hat
Aug 28 at 17:55




If $f$ is increasing, then $[X > alpha] = [f(X) > f(alpha)]$.
– copper.hat
Aug 28 at 17:55










1 Answer
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The fundamental observations are that
beginalign*
& X > a \
iff & theta X > theta a qquad text(assuming $theta > 0$) \
iff & e^theta X > e^theta a
endalign*
because $e^x$ is an increasing function. Hence, the events $X > a$ and $e^theta X > e^theta a$ are the same event.



It may be instructive to compare a statement that would be proved similarly: $mathbb P(X > 3) = mathbb P(5X + 2 > 17)$. You're permitted to work algebraically on the underlying inequalities, which is what's happening here.






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    The fundamental observations are that
    beginalign*
    & X > a \
    iff & theta X > theta a qquad text(assuming $theta > 0$) \
    iff & e^theta X > e^theta a
    endalign*
    because $e^x$ is an increasing function. Hence, the events $X > a$ and $e^theta X > e^theta a$ are the same event.



    It may be instructive to compare a statement that would be proved similarly: $mathbb P(X > 3) = mathbb P(5X + 2 > 17)$. You're permitted to work algebraically on the underlying inequalities, which is what's happening here.






    share|cite|improve this answer
























      up vote
      2
      down vote



      accepted










      The fundamental observations are that
      beginalign*
      & X > a \
      iff & theta X > theta a qquad text(assuming $theta > 0$) \
      iff & e^theta X > e^theta a
      endalign*
      because $e^x$ is an increasing function. Hence, the events $X > a$ and $e^theta X > e^theta a$ are the same event.



      It may be instructive to compare a statement that would be proved similarly: $mathbb P(X > 3) = mathbb P(5X + 2 > 17)$. You're permitted to work algebraically on the underlying inequalities, which is what's happening here.






      share|cite|improve this answer






















        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        The fundamental observations are that
        beginalign*
        & X > a \
        iff & theta X > theta a qquad text(assuming $theta > 0$) \
        iff & e^theta X > e^theta a
        endalign*
        because $e^x$ is an increasing function. Hence, the events $X > a$ and $e^theta X > e^theta a$ are the same event.



        It may be instructive to compare a statement that would be proved similarly: $mathbb P(X > 3) = mathbb P(5X + 2 > 17)$. You're permitted to work algebraically on the underlying inequalities, which is what's happening here.






        share|cite|improve this answer












        The fundamental observations are that
        beginalign*
        & X > a \
        iff & theta X > theta a qquad text(assuming $theta > 0$) \
        iff & e^theta X > e^theta a
        endalign*
        because $e^x$ is an increasing function. Hence, the events $X > a$ and $e^theta X > e^theta a$ are the same event.



        It may be instructive to compare a statement that would be proved similarly: $mathbb P(X > 3) = mathbb P(5X + 2 > 17)$. You're permitted to work algebraically on the underlying inequalities, which is what's happening here.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 28 at 22:01









        Aaron Montgomery

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