Let $Y = max(X, 0)$, then median of $Y$ is _____.












0












$begingroup$



Let $X$ be a Gaussian random variable with mean $0$ and variance $σ^2$. Let $Y = max(X, 0)$ where $max(a, b)$ is the maximum of $a$ and $b$. The median of $Y$ is _____.




My try:



Somewhere it explain as: Here, half of the values of Y are to the left of the mean X = 0 and the remaining half of the values of Y lies to the right of the mean X = 0. hence,The median of Y = 0.



Can you please explain in other way?






probability probability-distributions random-variables means median







share|cite|improve this question









$endgroup$

















    0












    $begingroup$



    Let $X$ be a Gaussian random variable with mean $0$ and variance $σ^2$. Let $Y = max(X, 0)$ where $max(a, b)$ is the maximum of $a$ and $b$. The median of $Y$ is _____.




    My try:



    Somewhere it explain as: Here, half of the values of Y are to the left of the mean X = 0 and the remaining half of the values of Y lies to the right of the mean X = 0. hence,The median of Y = 0.



    Can you please explain in other way?






    probability probability-distributions random-variables means median







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$



      Let $X$ be a Gaussian random variable with mean $0$ and variance $σ^2$. Let $Y = max(X, 0)$ where $max(a, b)$ is the maximum of $a$ and $b$. The median of $Y$ is _____.




      My try:



      Somewhere it explain as: Here, half of the values of Y are to the left of the mean X = 0 and the remaining half of the values of Y lies to the right of the mean X = 0. hence,The median of Y = 0.



      Can you please explain in other way?






      probability probability-distributions random-variables means median







      share|cite|improve this question









      $endgroup$





      Let $X$ be a Gaussian random variable with mean $0$ and variance $σ^2$. Let $Y = max(X, 0)$ where $max(a, b)$ is the maximum of $a$ and $b$. The median of $Y$ is _____.




      My try:



      Somewhere it explain as: Here, half of the values of Y are to the left of the mean X = 0 and the remaining half of the values of Y lies to the right of the mean X = 0. hence,The median of Y = 0.



      Can you please explain in other way?






      probability probability-distributions random-variables means median




      probability probability-distributions random-variables means median






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Sep 22 '17 at 10:50









      Mithlesh UpadhyayMithlesh Upadhyay

      2,90982864




      2,90982864






















          3 Answers
          3






          active

          oldest

          votes


















          2












          $begingroup$

          You have for $X$ a continuous random variable with median $0$




          • $P(X le 0)=frac12$

          • $P(X gt 0)=frac12$


          so, since $X le 0 implies Y=0$ and $Xgt 0 implies Y gt 0$,




          • $P(Y =0)=frac12$

          • $P(Y gt 0)=frac12$


          so, moving towards the definition of a median,




          • $P(Y le 0) ge frac12$

          • $P(Y ge 0) ge frac12$


          so the median of $Y$ is $0$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for nice explanation.
            $endgroup$
            – Mithlesh Upadhyay
            Sep 22 '17 at 11:24










          • $begingroup$
            @Henry-What could be the density function of Y in this case?
            $endgroup$
            – user3767495
            Jul 10 '18 at 12:27










          • $begingroup$
            @user3767495 $Y$ will have the same density as $X$ for positive values, plus a point of positive probability $frac12$ (not density) at $0$
            $endgroup$
            – Henry
            Jul 10 '18 at 13:00



















          2












          $begingroup$

          The median of $Y$ is defined as the least value $m$ such that $mathsf P(Yleqslant m)geqslanttfrac 12$.



          Since $Y=max{0,X}$, then: $mathsf P(Yleqslant 0){~=~mathsf P(Y=0)\~=~ mathsf P(Xleqslant 0)\~=~tfrac 12}$



          ... since $X$ is symmetrically distributed about $0$.



          Thus $0$ is the median of $Y$.



          That is all there is too it.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for nice explanation.
            $endgroup$
            – Mithlesh Upadhyay
            Sep 22 '17 at 11:24



















          -1












          $begingroup$

          As mean =0 and variance σ2 represents the standard normal distribution.And the graph of Standard normal distribution is symmetric And in symmetric we have,
          Mean=Median=Mode.



          As given that mean is 0.



          Hence Median and mode will be 0






          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









            2












            $begingroup$

            You have for $X$ a continuous random variable with median $0$




            • $P(X le 0)=frac12$

            • $P(X gt 0)=frac12$


            so, since $X le 0 implies Y=0$ and $Xgt 0 implies Y gt 0$,




            • $P(Y =0)=frac12$

            • $P(Y gt 0)=frac12$


            so, moving towards the definition of a median,




            • $P(Y le 0) ge frac12$

            • $P(Y ge 0) ge frac12$


            so the median of $Y$ is $0$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks for nice explanation.
              $endgroup$
              – Mithlesh Upadhyay
              Sep 22 '17 at 11:24










            • $begingroup$
              @Henry-What could be the density function of Y in this case?
              $endgroup$
              – user3767495
              Jul 10 '18 at 12:27










            • $begingroup$
              @user3767495 $Y$ will have the same density as $X$ for positive values, plus a point of positive probability $frac12$ (not density) at $0$
              $endgroup$
              – Henry
              Jul 10 '18 at 13:00
















            2












            $begingroup$

            You have for $X$ a continuous random variable with median $0$




            • $P(X le 0)=frac12$

            • $P(X gt 0)=frac12$


            so, since $X le 0 implies Y=0$ and $Xgt 0 implies Y gt 0$,




            • $P(Y =0)=frac12$

            • $P(Y gt 0)=frac12$


            so, moving towards the definition of a median,




            • $P(Y le 0) ge frac12$

            • $P(Y ge 0) ge frac12$


            so the median of $Y$ is $0$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks for nice explanation.
              $endgroup$
              – Mithlesh Upadhyay
              Sep 22 '17 at 11:24










            • $begingroup$
              @Henry-What could be the density function of Y in this case?
              $endgroup$
              – user3767495
              Jul 10 '18 at 12:27










            • $begingroup$
              @user3767495 $Y$ will have the same density as $X$ for positive values, plus a point of positive probability $frac12$ (not density) at $0$
              $endgroup$
              – Henry
              Jul 10 '18 at 13:00














            2












            2








            2





            $begingroup$

            You have for $X$ a continuous random variable with median $0$




            • $P(X le 0)=frac12$

            • $P(X gt 0)=frac12$


            so, since $X le 0 implies Y=0$ and $Xgt 0 implies Y gt 0$,




            • $P(Y =0)=frac12$

            • $P(Y gt 0)=frac12$


            so, moving towards the definition of a median,




            • $P(Y le 0) ge frac12$

            • $P(Y ge 0) ge frac12$


            so the median of $Y$ is $0$






            share|cite|improve this answer









            $endgroup$



            You have for $X$ a continuous random variable with median $0$




            • $P(X le 0)=frac12$

            • $P(X gt 0)=frac12$


            so, since $X le 0 implies Y=0$ and $Xgt 0 implies Y gt 0$,




            • $P(Y =0)=frac12$

            • $P(Y gt 0)=frac12$


            so, moving towards the definition of a median,




            • $P(Y le 0) ge frac12$

            • $P(Y ge 0) ge frac12$


            so the median of $Y$ is $0$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Sep 22 '17 at 11:17









            HenryHenry

            98.5k476163




            98.5k476163












            • $begingroup$
              Thanks for nice explanation.
              $endgroup$
              – Mithlesh Upadhyay
              Sep 22 '17 at 11:24










            • $begingroup$
              @Henry-What could be the density function of Y in this case?
              $endgroup$
              – user3767495
              Jul 10 '18 at 12:27










            • $begingroup$
              @user3767495 $Y$ will have the same density as $X$ for positive values, plus a point of positive probability $frac12$ (not density) at $0$
              $endgroup$
              – Henry
              Jul 10 '18 at 13:00


















            • $begingroup$
              Thanks for nice explanation.
              $endgroup$
              – Mithlesh Upadhyay
              Sep 22 '17 at 11:24










            • $begingroup$
              @Henry-What could be the density function of Y in this case?
              $endgroup$
              – user3767495
              Jul 10 '18 at 12:27










            • $begingroup$
              @user3767495 $Y$ will have the same density as $X$ for positive values, plus a point of positive probability $frac12$ (not density) at $0$
              $endgroup$
              – Henry
              Jul 10 '18 at 13:00
















            $begingroup$
            Thanks for nice explanation.
            $endgroup$
            – Mithlesh Upadhyay
            Sep 22 '17 at 11:24




            $begingroup$
            Thanks for nice explanation.
            $endgroup$
            – Mithlesh Upadhyay
            Sep 22 '17 at 11:24












            $begingroup$
            @Henry-What could be the density function of Y in this case?
            $endgroup$
            – user3767495
            Jul 10 '18 at 12:27




            $begingroup$
            @Henry-What could be the density function of Y in this case?
            $endgroup$
            – user3767495
            Jul 10 '18 at 12:27












            $begingroup$
            @user3767495 $Y$ will have the same density as $X$ for positive values, plus a point of positive probability $frac12$ (not density) at $0$
            $endgroup$
            – Henry
            Jul 10 '18 at 13:00




            $begingroup$
            @user3767495 $Y$ will have the same density as $X$ for positive values, plus a point of positive probability $frac12$ (not density) at $0$
            $endgroup$
            – Henry
            Jul 10 '18 at 13:00











            2












            $begingroup$

            The median of $Y$ is defined as the least value $m$ such that $mathsf P(Yleqslant m)geqslanttfrac 12$.



            Since $Y=max{0,X}$, then: $mathsf P(Yleqslant 0){~=~mathsf P(Y=0)\~=~ mathsf P(Xleqslant 0)\~=~tfrac 12}$



            ... since $X$ is symmetrically distributed about $0$.



            Thus $0$ is the median of $Y$.



            That is all there is too it.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thanks for nice explanation.
              $endgroup$
              – Mithlesh Upadhyay
              Sep 22 '17 at 11:24
















            2












            $begingroup$

            The median of $Y$ is defined as the least value $m$ such that $mathsf P(Yleqslant m)geqslanttfrac 12$.



            Since $Y=max{0,X}$, then: $mathsf P(Yleqslant 0){~=~mathsf P(Y=0)\~=~ mathsf P(Xleqslant 0)\~=~tfrac 12}$



            ... since $X$ is symmetrically distributed about $0$.



            Thus $0$ is the median of $Y$.



            That is all there is too it.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thanks for nice explanation.
              $endgroup$
              – Mithlesh Upadhyay
              Sep 22 '17 at 11:24














            2












            2








            2





            $begingroup$

            The median of $Y$ is defined as the least value $m$ such that $mathsf P(Yleqslant m)geqslanttfrac 12$.



            Since $Y=max{0,X}$, then: $mathsf P(Yleqslant 0){~=~mathsf P(Y=0)\~=~ mathsf P(Xleqslant 0)\~=~tfrac 12}$



            ... since $X$ is symmetrically distributed about $0$.



            Thus $0$ is the median of $Y$.



            That is all there is too it.






            share|cite|improve this answer











            $endgroup$



            The median of $Y$ is defined as the least value $m$ such that $mathsf P(Yleqslant m)geqslanttfrac 12$.



            Since $Y=max{0,X}$, then: $mathsf P(Yleqslant 0){~=~mathsf P(Y=0)\~=~ mathsf P(Xleqslant 0)\~=~tfrac 12}$



            ... since $X$ is symmetrically distributed about $0$.



            Thus $0$ is the median of $Y$.



            That is all there is too it.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Sep 22 '17 at 11:23


























            community wiki





            2 revs
            Graham Kemp













            • $begingroup$
              Thanks for nice explanation.
              $endgroup$
              – Mithlesh Upadhyay
              Sep 22 '17 at 11:24


















            • $begingroup$
              Thanks for nice explanation.
              $endgroup$
              – Mithlesh Upadhyay
              Sep 22 '17 at 11:24
















            $begingroup$
            Thanks for nice explanation.
            $endgroup$
            – Mithlesh Upadhyay
            Sep 22 '17 at 11:24




            $begingroup$
            Thanks for nice explanation.
            $endgroup$
            – Mithlesh Upadhyay
            Sep 22 '17 at 11:24











            -1












            $begingroup$

            As mean =0 and variance σ2 represents the standard normal distribution.And the graph of Standard normal distribution is symmetric And in symmetric we have,
            Mean=Median=Mode.



            As given that mean is 0.



            Hence Median and mode will be 0






            share|cite|improve this answer









            $endgroup$


















              -1












              $begingroup$

              As mean =0 and variance σ2 represents the standard normal distribution.And the graph of Standard normal distribution is symmetric And in symmetric we have,
              Mean=Median=Mode.



              As given that mean is 0.



              Hence Median and mode will be 0






              share|cite|improve this answer









              $endgroup$
















                -1












                -1








                -1





                $begingroup$

                As mean =0 and variance σ2 represents the standard normal distribution.And the graph of Standard normal distribution is symmetric And in symmetric we have,
                Mean=Median=Mode.



                As given that mean is 0.



                Hence Median and mode will be 0






                share|cite|improve this answer









                $endgroup$



                As mean =0 and variance σ2 represents the standard normal distribution.And the graph of Standard normal distribution is symmetric And in symmetric we have,
                Mean=Median=Mode.



                As given that mean is 0.



                Hence Median and mode will be 0







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 28 '18 at 16:18









                GeeklovenerdsGeeklovenerds

                529




                529






























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