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












4














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
















4














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




















  • 4




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














4












4








4







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















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






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edited Dec 17 at 7:10









kjetil b halvorsen

28.4k980208




28.4k980208










asked Dec 17 at 4:22









Linsu Han

213




213








  • 4




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














  • 4




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








4




4




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




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










3 Answers
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6














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





























    3














    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





























      1














      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





















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        3 Answers
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        3 Answers
        3






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        active

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        active

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        6














        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


























          6














          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
























            6












            6








            6






            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












            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 at 5:38









            gunes

            2,642111




            2,642111

























                3














                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


























                  3














                  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
























                    3












                    3








                    3






                    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












                    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 at 6:36









                    Ben

                    21.6k224103




                    21.6k224103























                        1














                        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


























                          1














                          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
























                            1












                            1








                            1






                            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












                            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 at 7:03









                            Charles Siouti

                            111




                            111






























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