Does $mathbb{P}(X < a) = mathbb{P}(f(X) < f(a))$?












4












$begingroup$


If $f(x)$ is a monotonic increasing function, then does $mathbb{P}(X < a) = mathbb{P}(f(X) < f(a))$? My intuition says it's true but I cannot prove the case nor find the name of the theorem.






probability mathematical-statistics function







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$endgroup$








  • 4




    $begingroup$
    By "monotonic increasing" do you mean strictly increasing (if $a < b$ then $f(a) < f(b)$) or non-decreasing (if $a < b$ then $f(a) leq f(b)$)?
    $endgroup$
    – Artem Mavrin
    Dec 17 '18 at 5:15
















4












$begingroup$


If $f(x)$ is a monotonic increasing function, then does $mathbb{P}(X < a) = mathbb{P}(f(X) < f(a))$? My intuition says it's true but I cannot prove the case nor find the name of the theorem.






probability mathematical-statistics function







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    By "monotonic increasing" do you mean strictly increasing (if $a < b$ then $f(a) < f(b)$) or non-decreasing (if $a < b$ then $f(a) leq f(b)$)?
    $endgroup$
    – Artem Mavrin
    Dec 17 '18 at 5:15














4












4








4





$begingroup$


If $f(x)$ is a monotonic increasing function, then does $mathbb{P}(X < a) = mathbb{P}(f(X) < f(a))$? My intuition says it's true but I cannot prove the case nor find the name of the theorem.






probability mathematical-statistics function







share|cite|improve this question











$endgroup$




If $f(x)$ is a monotonic increasing function, then does $mathbb{P}(X < a) = mathbb{P}(f(X) < f(a))$? My intuition says it's true but I cannot prove the case nor find the name of the theorem.






probability mathematical-statistics function




probability mathematical-statistics function






share|cite|improve this question















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edited Dec 17 '18 at 7:10









kjetil b halvorsen

29.3k980213




29.3k980213










asked Dec 17 '18 at 4:22









Linsu HanLinsu Han

213




213








  • 4




    $begingroup$
    By "monotonic increasing" do you mean strictly increasing (if $a < b$ then $f(a) < f(b)$) or non-decreasing (if $a < b$ then $f(a) leq f(b)$)?
    $endgroup$
    – Artem Mavrin
    Dec 17 '18 at 5:15














  • 4




    $begingroup$
    By "monotonic increasing" do you mean strictly increasing (if $a < b$ then $f(a) < f(b)$) or non-decreasing (if $a < b$ then $f(a) leq f(b)$)?
    $endgroup$
    – Artem Mavrin
    Dec 17 '18 at 5:15








4




4




$begingroup$
By "monotonic increasing" do you mean strictly increasing (if $a < b$ then $f(a) < f(b)$) or non-decreasing (if $a < b$ then $f(a) leq f(b)$)?
$endgroup$
– Artem Mavrin
Dec 17 '18 at 5:15




$begingroup$
By "monotonic increasing" do you mean strictly increasing (if $a < b$ then $f(a) < f(b)$) or non-decreasing (if $a < b$ then $f(a) leq f(b)$)?
$endgroup$
– Artem Mavrin
Dec 17 '18 at 5:15










3 Answers
3






active

oldest

votes


















6












$begingroup$

Consider the set of $x$, call it $S$, where $x<a$. You seek for the probability, $P(S)$. Any expression that lead to $S$ produces the exact same probability, $b$, irrespective of its decleration. If $f(x)$ is a monotonic (strictly) increasing function, $x<a$ directly implies $f(x)<f(a)$ and vice versa, i.e. if $f(x)<f(a)$, then $x<a$.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    If $f$ is strictly increasing then you have:



    $$begin{equation} begin{aligned}
    { X < a }
    &= { omega in Omega | X(omega) < a } \[6pt]
    &= { omega in Omega | f(X(omega)) < f(a) } \[6pt]
    &= { omega in Omega | f(X)(omega) < f(a) } \[6pt]
    &= { f(X) < f(a) }, \[6pt]
    end{aligned} end{equation}$$



    which means that $mathbb{P}(X<a) = mathbb{P}(f(X)<f(a))$. If $f$ is only non-decreasing then you cannot derive this result but you can derive an analogous result with non-strict inequality.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      No.

      Assuming that you call monotonically increasing a function that is non-decreasing (for all $x$ and $y$ such that $xleq y$, one has $f ( x ) leq f ( y )$), consider $X$ following a uniform distribution on $[0, 1]$, $f=0$ and $a=1$.

      Then, $P(X<a) = P(X<1) = 1 neq 0 = P(0<0) = P(f(X) < f(a))$.

      Your assumption is true for strictly increasing functions. If f is strictly increasing, $ {X<a} = {f(X)<f(a)}$.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        6












        $begingroup$

        Consider the set of $x$, call it $S$, where $x<a$. You seek for the probability, $P(S)$. Any expression that lead to $S$ produces the exact same probability, $b$, irrespective of its decleration. If $f(x)$ is a monotonic (strictly) increasing function, $x<a$ directly implies $f(x)<f(a)$ and vice versa, i.e. if $f(x)<f(a)$, then $x<a$.






        share|cite|improve this answer









        $endgroup$


















          6












          $begingroup$

          Consider the set of $x$, call it $S$, where $x<a$. You seek for the probability, $P(S)$. Any expression that lead to $S$ produces the exact same probability, $b$, irrespective of its decleration. If $f(x)$ is a monotonic (strictly) increasing function, $x<a$ directly implies $f(x)<f(a)$ and vice versa, i.e. if $f(x)<f(a)$, then $x<a$.






          share|cite|improve this answer









          $endgroup$
















            6












            6








            6





            $begingroup$

            Consider the set of $x$, call it $S$, where $x<a$. You seek for the probability, $P(S)$. Any expression that lead to $S$ produces the exact same probability, $b$, irrespective of its decleration. If $f(x)$ is a monotonic (strictly) increasing function, $x<a$ directly implies $f(x)<f(a)$ and vice versa, i.e. if $f(x)<f(a)$, then $x<a$.






            share|cite|improve this answer









            $endgroup$



            Consider the set of $x$, call it $S$, where $x<a$. You seek for the probability, $P(S)$. Any expression that lead to $S$ produces the exact same probability, $b$, irrespective of its decleration. If $f(x)$ is a monotonic (strictly) increasing function, $x<a$ directly implies $f(x)<f(a)$ and vice versa, i.e. if $f(x)<f(a)$, then $x<a$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 17 '18 at 5:38









            gunesgunes

            3,4471111




            3,4471111

























                3












                $begingroup$

                If $f$ is strictly increasing then you have:



                $$begin{equation} begin{aligned}
                { X < a }
                &= { omega in Omega | X(omega) < a } \[6pt]
                &= { omega in Omega | f(X(omega)) < f(a) } \[6pt]
                &= { omega in Omega | f(X)(omega) < f(a) } \[6pt]
                &= { f(X) < f(a) }, \[6pt]
                end{aligned} end{equation}$$



                which means that $mathbb{P}(X<a) = mathbb{P}(f(X)<f(a))$. If $f$ is only non-decreasing then you cannot derive this result but you can derive an analogous result with non-strict inequality.






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  If $f$ is strictly increasing then you have:



                  $$begin{equation} begin{aligned}
                  { X < a }
                  &= { omega in Omega | X(omega) < a } \[6pt]
                  &= { omega in Omega | f(X(omega)) < f(a) } \[6pt]
                  &= { omega in Omega | f(X)(omega) < f(a) } \[6pt]
                  &= { f(X) < f(a) }, \[6pt]
                  end{aligned} end{equation}$$



                  which means that $mathbb{P}(X<a) = mathbb{P}(f(X)<f(a))$. If $f$ is only non-decreasing then you cannot derive this result but you can derive an analogous result with non-strict inequality.






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    If $f$ is strictly increasing then you have:



                    $$begin{equation} begin{aligned}
                    { X < a }
                    &= { omega in Omega | X(omega) < a } \[6pt]
                    &= { omega in Omega | f(X(omega)) < f(a) } \[6pt]
                    &= { omega in Omega | f(X)(omega) < f(a) } \[6pt]
                    &= { f(X) < f(a) }, \[6pt]
                    end{aligned} end{equation}$$



                    which means that $mathbb{P}(X<a) = mathbb{P}(f(X)<f(a))$. If $f$ is only non-decreasing then you cannot derive this result but you can derive an analogous result with non-strict inequality.






                    share|cite|improve this answer









                    $endgroup$



                    If $f$ is strictly increasing then you have:



                    $$begin{equation} begin{aligned}
                    { X < a }
                    &= { omega in Omega | X(omega) < a } \[6pt]
                    &= { omega in Omega | f(X(omega)) < f(a) } \[6pt]
                    &= { omega in Omega | f(X)(omega) < f(a) } \[6pt]
                    &= { f(X) < f(a) }, \[6pt]
                    end{aligned} end{equation}$$



                    which means that $mathbb{P}(X<a) = mathbb{P}(f(X)<f(a))$. If $f$ is only non-decreasing then you cannot derive this result but you can derive an analogous result with non-strict inequality.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 17 '18 at 6:36









                    BenBen

                    22.8k224108




                    22.8k224108























                        1












                        $begingroup$

                        No.

                        Assuming that you call monotonically increasing a function that is non-decreasing (for all $x$ and $y$ such that $xleq y$, one has $f ( x ) leq f ( y )$), consider $X$ following a uniform distribution on $[0, 1]$, $f=0$ and $a=1$.

                        Then, $P(X<a) = P(X<1) = 1 neq 0 = P(0<0) = P(f(X) < f(a))$.

                        Your assumption is true for strictly increasing functions. If f is strictly increasing, $ {X<a} = {f(X)<f(a)}$.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          No.

                          Assuming that you call monotonically increasing a function that is non-decreasing (for all $x$ and $y$ such that $xleq y$, one has $f ( x ) leq f ( y )$), consider $X$ following a uniform distribution on $[0, 1]$, $f=0$ and $a=1$.

                          Then, $P(X<a) = P(X<1) = 1 neq 0 = P(0<0) = P(f(X) < f(a))$.

                          Your assumption is true for strictly increasing functions. If f is strictly increasing, $ {X<a} = {f(X)<f(a)}$.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            No.

                            Assuming that you call monotonically increasing a function that is non-decreasing (for all $x$ and $y$ such that $xleq y$, one has $f ( x ) leq f ( y )$), consider $X$ following a uniform distribution on $[0, 1]$, $f=0$ and $a=1$.

                            Then, $P(X<a) = P(X<1) = 1 neq 0 = P(0<0) = P(f(X) < f(a))$.

                            Your assumption is true for strictly increasing functions. If f is strictly increasing, $ {X<a} = {f(X)<f(a)}$.






                            share|cite|improve this answer









                            $endgroup$



                            No.

                            Assuming that you call monotonically increasing a function that is non-decreasing (for all $x$ and $y$ such that $xleq y$, one has $f ( x ) leq f ( y )$), consider $X$ following a uniform distribution on $[0, 1]$, $f=0$ and $a=1$.

                            Then, $P(X<a) = P(X<1) = 1 neq 0 = P(0<0) = P(f(X) < f(a))$.

                            Your assumption is true for strictly increasing functions. If f is strictly increasing, $ {X<a} = {f(X)<f(a)}$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 17 '18 at 7:03









                            Charles SioutiCharles Siouti

                            111




                            111






























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