Imagine $X_1$,…,$X_n$ are iid uniformly distributed and $X=max(a_1X_1,..,a_nX_n)$,...












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$begingroup$


Imagine $X_1$,...,$X_n$ are i.i.d. uniformly distributed on the interval [0,1] and $X=max(a_1X_1,..,a_nX_n)$ and $Y=max(b_1X_1,..,b_nX_n)$ for some constants $a_1,...,a_n$, $b_1,...,b_n$.(All real, positive numbers)



What is the joint pdf (or cdf) of X and Y?



My idea: We have $f_{X,Y}(x,y)=f_{Xmid Y}(x)f_Y(y)$, Now the latter term can be calculated very easily because we have $F_Y(x)=prod_{i=1}^n F_{b_iX_i}(x)$. So now we need to calculate the conditional probability;



However, I do not know how to continue here;



Any idea?










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    0












    $begingroup$


    Imagine $X_1$,...,$X_n$ are i.i.d. uniformly distributed on the interval [0,1] and $X=max(a_1X_1,..,a_nX_n)$ and $Y=max(b_1X_1,..,b_nX_n)$ for some constants $a_1,...,a_n$, $b_1,...,b_n$.(All real, positive numbers)



    What is the joint pdf (or cdf) of X and Y?



    My idea: We have $f_{X,Y}(x,y)=f_{Xmid Y}(x)f_Y(y)$, Now the latter term can be calculated very easily because we have $F_Y(x)=prod_{i=1}^n F_{b_iX_i}(x)$. So now we need to calculate the conditional probability;



    However, I do not know how to continue here;



    Any idea?










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Imagine $X_1$,...,$X_n$ are i.i.d. uniformly distributed on the interval [0,1] and $X=max(a_1X_1,..,a_nX_n)$ and $Y=max(b_1X_1,..,b_nX_n)$ for some constants $a_1,...,a_n$, $b_1,...,b_n$.(All real, positive numbers)



      What is the joint pdf (or cdf) of X and Y?



      My idea: We have $f_{X,Y}(x,y)=f_{Xmid Y}(x)f_Y(y)$, Now the latter term can be calculated very easily because we have $F_Y(x)=prod_{i=1}^n F_{b_iX_i}(x)$. So now we need to calculate the conditional probability;



      However, I do not know how to continue here;



      Any idea?










      share|cite|improve this question











      $endgroup$




      Imagine $X_1$,...,$X_n$ are i.i.d. uniformly distributed on the interval [0,1] and $X=max(a_1X_1,..,a_nX_n)$ and $Y=max(b_1X_1,..,b_nX_n)$ for some constants $a_1,...,a_n$, $b_1,...,b_n$.(All real, positive numbers)



      What is the joint pdf (or cdf) of X and Y?



      My idea: We have $f_{X,Y}(x,y)=f_{Xmid Y}(x)f_Y(y)$, Now the latter term can be calculated very easily because we have $F_Y(x)=prod_{i=1}^n F_{b_iX_i}(x)$. So now we need to calculate the conditional probability;



      However, I do not know how to continue here;



      Any idea?







      probability probability-distributions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 29 '18 at 9:24







      J.Doe

















      asked Nov 29 '18 at 9:16









      J.DoeJ.Doe

      19910




      19910






















          1 Answer
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          $begingroup$

          We have $$Pr(X<x,Y<y){=Pr(max(a_1X_1,..,a_nX_n)<x,max(b_1X_1,..,b_nX_n)<y)\=Pr(a_1X_1,..,a_nX_n<x,b_1X_1,..,b_nX_n<y)\=Pr(X_1<min{{xover a_1},{yover b_1}},cdots , X_n<min{{xover a_n},{yover b_n}})\=U(min{{xover a_1},{yover b_1}})cdots U(min{{xover a_n},{yover b_n}})}$$where $U(x)$ is the cmf of uniform distribution on $[0,1]$. It is not generally easy to find the pmf from the above cmf by differentiation. It is all I got for now!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much; Ill try to work with it - maybe I can work only with the cdf instead of the pdf;
            $endgroup$
            – J.Doe
            Nov 29 '18 at 10:03










          • $begingroup$
            But if you have an idea how to get to the pdf let me know
            $endgroup$
            – J.Doe
            Nov 29 '18 at 10:04










          • $begingroup$
            You're welcome. If you felt more details i'll try to make some calculations but the results are easy to be derived only for some particular cases :)
            $endgroup$
            – Mostafa Ayaz
            Nov 29 '18 at 10:04










          • $begingroup$
            Actually you must divide each $min{{xover a_i},{yover b_i}}$ to the sections where $0<{xover a_i}<1$ or $0<{yover b_i}<1$. Also notice that the cdf is nonzero only if $0<min{{xover a_i},{yover b_i}}<1$ for all $i$
            $endgroup$
            – Mostafa Ayaz
            Nov 29 '18 at 10:06











          Your Answer





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          $begingroup$

          We have $$Pr(X<x,Y<y){=Pr(max(a_1X_1,..,a_nX_n)<x,max(b_1X_1,..,b_nX_n)<y)\=Pr(a_1X_1,..,a_nX_n<x,b_1X_1,..,b_nX_n<y)\=Pr(X_1<min{{xover a_1},{yover b_1}},cdots , X_n<min{{xover a_n},{yover b_n}})\=U(min{{xover a_1},{yover b_1}})cdots U(min{{xover a_n},{yover b_n}})}$$where $U(x)$ is the cmf of uniform distribution on $[0,1]$. It is not generally easy to find the pmf from the above cmf by differentiation. It is all I got for now!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much; Ill try to work with it - maybe I can work only with the cdf instead of the pdf;
            $endgroup$
            – J.Doe
            Nov 29 '18 at 10:03










          • $begingroup$
            But if you have an idea how to get to the pdf let me know
            $endgroup$
            – J.Doe
            Nov 29 '18 at 10:04










          • $begingroup$
            You're welcome. If you felt more details i'll try to make some calculations but the results are easy to be derived only for some particular cases :)
            $endgroup$
            – Mostafa Ayaz
            Nov 29 '18 at 10:04










          • $begingroup$
            Actually you must divide each $min{{xover a_i},{yover b_i}}$ to the sections where $0<{xover a_i}<1$ or $0<{yover b_i}<1$. Also notice that the cdf is nonzero only if $0<min{{xover a_i},{yover b_i}}<1$ for all $i$
            $endgroup$
            – Mostafa Ayaz
            Nov 29 '18 at 10:06
















          2












          $begingroup$

          We have $$Pr(X<x,Y<y){=Pr(max(a_1X_1,..,a_nX_n)<x,max(b_1X_1,..,b_nX_n)<y)\=Pr(a_1X_1,..,a_nX_n<x,b_1X_1,..,b_nX_n<y)\=Pr(X_1<min{{xover a_1},{yover b_1}},cdots , X_n<min{{xover a_n},{yover b_n}})\=U(min{{xover a_1},{yover b_1}})cdots U(min{{xover a_n},{yover b_n}})}$$where $U(x)$ is the cmf of uniform distribution on $[0,1]$. It is not generally easy to find the pmf from the above cmf by differentiation. It is all I got for now!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much; Ill try to work with it - maybe I can work only with the cdf instead of the pdf;
            $endgroup$
            – J.Doe
            Nov 29 '18 at 10:03










          • $begingroup$
            But if you have an idea how to get to the pdf let me know
            $endgroup$
            – J.Doe
            Nov 29 '18 at 10:04










          • $begingroup$
            You're welcome. If you felt more details i'll try to make some calculations but the results are easy to be derived only for some particular cases :)
            $endgroup$
            – Mostafa Ayaz
            Nov 29 '18 at 10:04










          • $begingroup$
            Actually you must divide each $min{{xover a_i},{yover b_i}}$ to the sections where $0<{xover a_i}<1$ or $0<{yover b_i}<1$. Also notice that the cdf is nonzero only if $0<min{{xover a_i},{yover b_i}}<1$ for all $i$
            $endgroup$
            – Mostafa Ayaz
            Nov 29 '18 at 10:06














          2












          2








          2





          $begingroup$

          We have $$Pr(X<x,Y<y){=Pr(max(a_1X_1,..,a_nX_n)<x,max(b_1X_1,..,b_nX_n)<y)\=Pr(a_1X_1,..,a_nX_n<x,b_1X_1,..,b_nX_n<y)\=Pr(X_1<min{{xover a_1},{yover b_1}},cdots , X_n<min{{xover a_n},{yover b_n}})\=U(min{{xover a_1},{yover b_1}})cdots U(min{{xover a_n},{yover b_n}})}$$where $U(x)$ is the cmf of uniform distribution on $[0,1]$. It is not generally easy to find the pmf from the above cmf by differentiation. It is all I got for now!






          share|cite|improve this answer









          $endgroup$



          We have $$Pr(X<x,Y<y){=Pr(max(a_1X_1,..,a_nX_n)<x,max(b_1X_1,..,b_nX_n)<y)\=Pr(a_1X_1,..,a_nX_n<x,b_1X_1,..,b_nX_n<y)\=Pr(X_1<min{{xover a_1},{yover b_1}},cdots , X_n<min{{xover a_n},{yover b_n}})\=U(min{{xover a_1},{yover b_1}})cdots U(min{{xover a_n},{yover b_n}})}$$where $U(x)$ is the cmf of uniform distribution on $[0,1]$. It is not generally easy to find the pmf from the above cmf by differentiation. It is all I got for now!







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 '18 at 9:47









          Mostafa AyazMostafa Ayaz

          15.5k3939




          15.5k3939












          • $begingroup$
            Thank you very much; Ill try to work with it - maybe I can work only with the cdf instead of the pdf;
            $endgroup$
            – J.Doe
            Nov 29 '18 at 10:03










          • $begingroup$
            But if you have an idea how to get to the pdf let me know
            $endgroup$
            – J.Doe
            Nov 29 '18 at 10:04










          • $begingroup$
            You're welcome. If you felt more details i'll try to make some calculations but the results are easy to be derived only for some particular cases :)
            $endgroup$
            – Mostafa Ayaz
            Nov 29 '18 at 10:04










          • $begingroup$
            Actually you must divide each $min{{xover a_i},{yover b_i}}$ to the sections where $0<{xover a_i}<1$ or $0<{yover b_i}<1$. Also notice that the cdf is nonzero only if $0<min{{xover a_i},{yover b_i}}<1$ for all $i$
            $endgroup$
            – Mostafa Ayaz
            Nov 29 '18 at 10:06


















          • $begingroup$
            Thank you very much; Ill try to work with it - maybe I can work only with the cdf instead of the pdf;
            $endgroup$
            – J.Doe
            Nov 29 '18 at 10:03










          • $begingroup$
            But if you have an idea how to get to the pdf let me know
            $endgroup$
            – J.Doe
            Nov 29 '18 at 10:04










          • $begingroup$
            You're welcome. If you felt more details i'll try to make some calculations but the results are easy to be derived only for some particular cases :)
            $endgroup$
            – Mostafa Ayaz
            Nov 29 '18 at 10:04










          • $begingroup$
            Actually you must divide each $min{{xover a_i},{yover b_i}}$ to the sections where $0<{xover a_i}<1$ or $0<{yover b_i}<1$. Also notice that the cdf is nonzero only if $0<min{{xover a_i},{yover b_i}}<1$ for all $i$
            $endgroup$
            – Mostafa Ayaz
            Nov 29 '18 at 10:06
















          $begingroup$
          Thank you very much; Ill try to work with it - maybe I can work only with the cdf instead of the pdf;
          $endgroup$
          – J.Doe
          Nov 29 '18 at 10:03




          $begingroup$
          Thank you very much; Ill try to work with it - maybe I can work only with the cdf instead of the pdf;
          $endgroup$
          – J.Doe
          Nov 29 '18 at 10:03












          $begingroup$
          But if you have an idea how to get to the pdf let me know
          $endgroup$
          – J.Doe
          Nov 29 '18 at 10:04




          $begingroup$
          But if you have an idea how to get to the pdf let me know
          $endgroup$
          – J.Doe
          Nov 29 '18 at 10:04












          $begingroup$
          You're welcome. If you felt more details i'll try to make some calculations but the results are easy to be derived only for some particular cases :)
          $endgroup$
          – Mostafa Ayaz
          Nov 29 '18 at 10:04




          $begingroup$
          You're welcome. If you felt more details i'll try to make some calculations but the results are easy to be derived only for some particular cases :)
          $endgroup$
          – Mostafa Ayaz
          Nov 29 '18 at 10:04












          $begingroup$
          Actually you must divide each $min{{xover a_i},{yover b_i}}$ to the sections where $0<{xover a_i}<1$ or $0<{yover b_i}<1$. Also notice that the cdf is nonzero only if $0<min{{xover a_i},{yover b_i}}<1$ for all $i$
          $endgroup$
          – Mostafa Ayaz
          Nov 29 '18 at 10:06




          $begingroup$
          Actually you must divide each $min{{xover a_i},{yover b_i}}$ to the sections where $0<{xover a_i}<1$ or $0<{yover b_i}<1$. Also notice that the cdf is nonzero only if $0<min{{xover a_i},{yover b_i}}<1$ for all $i$
          $endgroup$
          – Mostafa Ayaz
          Nov 29 '18 at 10:06


















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