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







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












$begingroup$


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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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









      $endgroup$













        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "65"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: false,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: null,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f383332%2fdoes-mathbbpx-a-mathbbpfx-fa%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        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






























                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Cross Validated!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid …



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f383332%2fdoes-mathbbpx-a-mathbbpfx-fa%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                -

                                Popular posts from this blog

                                Список кардиналов, возведённых папой римским Каликстом III

                                Image
                                Папа Каликст III Кардиналы, возведённые Папой римским Каликстом III  — 9 прелатов и мирян были возведены в сан кардинала на двух Консисториях за трёхлетний понтификат Каликста III. Самой крупной консисторией были Консистория от 17 декабря 1456 года, на которой было возведено шесть кардиналов. Папа Каллист III хотел назвать отца Жан Сорета, O.Carm., епископом или кардиналом, но он отказался, потому что хотел посвятить себя своему ордену и продвижению религиозного рвения и соблюдения устава монашеского ордена. Консистория от 17 сентября 1456 года | Луис Хуан дель Мила, племянник Его Святейшества, епископ Сегорбе (королевство Арагон); Жайме Португальский, администратор архиепархии Лиссабона (королевство Португалия); Родриго де Борха-и-Борха, племянник Его Святейшества, ризничий Валенсии (Папская область). Консистория от 17 декабря 1456 года | Ринальдо Пишичелло, архиепископ Неаполя (Неаполитанское королевство); Хуан де Мелья, епископ Саморы (королевство Ка
                                Read more

                                Deduzione

                                Image
                                .mw-parser-output .nota-disambigua{clear:both;margin-bottom:.5em;border:1px solid #CCC;padding-left:4px}.mw-parser-output .nota-disambigua i{vertical-align:middle} Disambiguazione – Se stai cercando il teorema, vedi Teorema di deduzione . Disambiguazione – Se stai cercando il concetto di deduzione in economia, vedi deduzione fiscale . Il ragionamento deduttivo umano assimilato a un meccanismo Il metodo deduttivo o deduzione è il procedimento razionale che fa derivare una certa conclusione da premesse più generiche, dentro cui quella conclusione è implicita. Il termine significa letteralmente «condurre da», perché proviene dal latino "de" (traducibile con da , preposizione indicante provenienza, o moto di discesa dall'alto verso il basso) e "ducere" ( condurre ). Questo metodo parte da postulati e princìpi primi e, attraverso una serie di rigorose concatenazioni logiche, procede verso determinazioni più particolari attinenti alla realtà tangi
                                Read more

                                Mysql.sock missing - “Can't connect to local MySQL server through socket”

                                Image
                                .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 Operating System - OSX 10.11.2 MAMP version - 3.5 I'm following along with this guide, setting up wordpress with Xammp on PC and MAMP on Mac, where I'm in step "Mac - 4" is asked to input: "In the Terminal run: /Applications/MAMP/Library/bin/mysqladmin -u root -p password {Your Windows MySQL password here}" However, this (with out bracket text) results in error message: error: 'Can't connect to local MySQL server through socket '/Applications/MAMP/tmp/mysql/mysql.sock' (2)' Check that mysqld is running and that the socket: '/Applications/MAMP/tmp/mysql/mysql.sock'
                                Read more