Probability inequality of multiple continuous random variables

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Let $X_0, ldots , X_n-1$ be i.i.d positive random variables with a continuous probability function $f_X$. I'm trying to prove that
$$
mathbbP[X_0 ge max(X_1, ldots , X_n-1)] = int_0^infty (F_X_0)^n-1(a)f_X_0(a) ,da
$$
Where $F$ is the CDF and $f$ is the PDF



I guess that $mathbbP[X_0 ge max(X_1, ldots , X_n-1)]= (n-1) mathbbP[X_0 ge X_1]$ because they're i.d.d, but how do I translate it to the CDF?






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    Let $X_0, ldots , X_n-1$ be i.i.d positive random variables with a continuous probability function $f_X$. I'm trying to prove that
    $$
    mathbbP[X_0 ge max(X_1, ldots , X_n-1)] = int_0^infty (F_X_0)^n-1(a)f_X_0(a) ,da
    $$
    Where $F$ is the CDF and $f$ is the PDF



    I guess that $mathbbP[X_0 ge max(X_1, ldots , X_n-1)]= (n-1) mathbbP[X_0 ge X_1]$ because they're i.d.d, but how do I translate it to the CDF?






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      Let $X_0, ldots , X_n-1$ be i.i.d positive random variables with a continuous probability function $f_X$. I'm trying to prove that
      $$
      mathbbP[X_0 ge max(X_1, ldots , X_n-1)] = int_0^infty (F_X_0)^n-1(a)f_X_0(a) ,da
      $$
      Where $F$ is the CDF and $f$ is the PDF



      I guess that $mathbbP[X_0 ge max(X_1, ldots , X_n-1)]= (n-1) mathbbP[X_0 ge X_1]$ because they're i.d.d, but how do I translate it to the CDF?






      probability







      share|cite|improve this question













      Let $X_0, ldots , X_n-1$ be i.i.d positive random variables with a continuous probability function $f_X$. I'm trying to prove that
      $$
      mathbbP[X_0 ge max(X_1, ldots , X_n-1)] = int_0^infty (F_X_0)^n-1(a)f_X_0(a) ,da
      $$
      Where $F$ is the CDF and $f$ is the PDF



      I guess that $mathbbP[X_0 ge max(X_1, ldots , X_n-1)]= (n-1) mathbbP[X_0 ge X_1]$ because they're i.d.d, but how do I translate it to the CDF?






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      probability






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      asked 6 hours ago









      galah92

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          3 Answers
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          up vote
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          accepted










          Your guess is wong.



          Realize that $mathbb P(X_0geq X_1)=frac12$ on base of:



          • $P(X_0geq X_1)+P(X_1geq X_0)-P(X_0=X_1)$


          • $P(X_0geq X_1)=P(X_1geq X_0)$ on base of symmetry.


          • $P(X_0=X_1)=0$ on base of continuity.

          Likewise it can be verified that: $$Pleft(X_0geqmax(X_1,dots,X_n-1)right)=frac1n$$



          edit:



          Also $$Pleft(X_0geqmax(X_1,dots,X_n-1)right)=int Pleft(X_0geqmax(X_1,dots,X_n-1)mid X_0=aright)f_X(a)da=$$$$int P(max(X_1,dots,X_n-1)leq a)f_X(a)da=int F_X^n-1(a)f_X(a)da$$



          where the second equality rests on independence.






          share|cite|improve this answer






















          • I do understand that. I'm trying to get to the explicit general expression stated above.
            – galah92
            5 hours ago










          • I have added a justification of the equality.
            – drhab
            5 hours ago

















          up vote
          2
          down vote













          Disclaimer: this is just a remark on drhab's answer.



          Note that
          beginalign
          (F_X_0)^n-1(a)f_X_0(a) = fracdda frac1n (F_X_0)^n(a).
          endalign
          The fundamental theorem of calculus implies that the integral is equal to
          beginalign
          lim_a to infty frac1n (F_X_0)^n(a) - frac1n (F_X_0)^n(0) = frac1n,
          endalign
          where we used that $X_0$ is a positive random variable.






          share|cite|improve this answer



























            up vote
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            down vote













            Since the variables are i.i.d. it is
            beginalign*
            mathbbP(X_0geq max(X_1,ldots,X_n-1)|X_0=x) &= mathbbP(xgeq max(X_1,ldots,X_n-1)|X_0=x) \
            & = mathbbP(xgeq max(X_1,ldots,X_n-1)) \
            & = mathbbP(X_1leq x,ldots,X_n-1leq x) \
            & = mathbbP(X_1leq x)^n-1 \
            & = F_X_0(x)^n-1.
            endalign*
            Hence:
            beginalign*
            mathbbP(X_0geq max(X_1,ldots,X_n-1)) &=int_0^+inftymathbbP(X_0geq max(X_1,ldots,X_n-1)|X_0=x)f_X_0(x)dx \
            &=int_0^+infty F_X_0(x)^n-1f_X_0(x)dx
            endalign*
            Now you can use the fact that $F_X_0(x)^n-1f_X_0=frac1nfracddxF_X_0(x)^n$ to show that $mathbbP(X_0geq max(X_1,ldots,X_n-1))=frac1n$.



            Another way to reach this last result is the following. Let $X_(n)=max(X_0,X_1,ldots,X_n-1)$. The event $X_0geq max(X_1,ldots,X_n-1)$ is equivalent to the event $X_0=X_(n)$. Since the variables are i.i.d., $mathbbP(X_0=X_(n))=mathbbP(X_i=X_(n))=mathbbP(X_i textrm is the largest among X_0,ldots,X_n-1)$ for all $i=0,ldots,n-1$. Additionally, since ties have probability zero (as the distributions are continuous) it is
            $$1=sum_i=0^n-1 mathbbP(X_i=X_(n)) = nmathbbP(X_0=X_(n))=nmathbbP(X_0geq max(X_1,ldots,X_n-1))$$






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              3 Answers
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              active

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              3 Answers
              3






              active

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              active

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              active

              oldest

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              up vote
              3
              down vote



              accepted










              Your guess is wong.



              Realize that $mathbb P(X_0geq X_1)=frac12$ on base of:



              • $P(X_0geq X_1)+P(X_1geq X_0)-P(X_0=X_1)$


              • $P(X_0geq X_1)=P(X_1geq X_0)$ on base of symmetry.


              • $P(X_0=X_1)=0$ on base of continuity.

              Likewise it can be verified that: $$Pleft(X_0geqmax(X_1,dots,X_n-1)right)=frac1n$$



              edit:



              Also $$Pleft(X_0geqmax(X_1,dots,X_n-1)right)=int Pleft(X_0geqmax(X_1,dots,X_n-1)mid X_0=aright)f_X(a)da=$$$$int P(max(X_1,dots,X_n-1)leq a)f_X(a)da=int F_X^n-1(a)f_X(a)da$$



              where the second equality rests on independence.






              share|cite|improve this answer






















              • I do understand that. I'm trying to get to the explicit general expression stated above.
                – galah92
                5 hours ago










              • I have added a justification of the equality.
                – drhab
                5 hours ago














              up vote
              3
              down vote



              accepted










              Your guess is wong.



              Realize that $mathbb P(X_0geq X_1)=frac12$ on base of:



              • $P(X_0geq X_1)+P(X_1geq X_0)-P(X_0=X_1)$


              • $P(X_0geq X_1)=P(X_1geq X_0)$ on base of symmetry.


              • $P(X_0=X_1)=0$ on base of continuity.

              Likewise it can be verified that: $$Pleft(X_0geqmax(X_1,dots,X_n-1)right)=frac1n$$



              edit:



              Also $$Pleft(X_0geqmax(X_1,dots,X_n-1)right)=int Pleft(X_0geqmax(X_1,dots,X_n-1)mid X_0=aright)f_X(a)da=$$$$int P(max(X_1,dots,X_n-1)leq a)f_X(a)da=int F_X^n-1(a)f_X(a)da$$



              where the second equality rests on independence.






              share|cite|improve this answer






















              • I do understand that. I'm trying to get to the explicit general expression stated above.
                – galah92
                5 hours ago










              • I have added a justification of the equality.
                – drhab
                5 hours ago












              up vote
              3
              down vote



              accepted







              up vote
              3
              down vote



              accepted






              Your guess is wong.



              Realize that $mathbb P(X_0geq X_1)=frac12$ on base of:



              • $P(X_0geq X_1)+P(X_1geq X_0)-P(X_0=X_1)$


              • $P(X_0geq X_1)=P(X_1geq X_0)$ on base of symmetry.


              • $P(X_0=X_1)=0$ on base of continuity.

              Likewise it can be verified that: $$Pleft(X_0geqmax(X_1,dots,X_n-1)right)=frac1n$$



              edit:



              Also $$Pleft(X_0geqmax(X_1,dots,X_n-1)right)=int Pleft(X_0geqmax(X_1,dots,X_n-1)mid X_0=aright)f_X(a)da=$$$$int P(max(X_1,dots,X_n-1)leq a)f_X(a)da=int F_X^n-1(a)f_X(a)da$$



              where the second equality rests on independence.






              share|cite|improve this answer














              Your guess is wong.



              Realize that $mathbb P(X_0geq X_1)=frac12$ on base of:



              • $P(X_0geq X_1)+P(X_1geq X_0)-P(X_0=X_1)$


              • $P(X_0geq X_1)=P(X_1geq X_0)$ on base of symmetry.


              • $P(X_0=X_1)=0$ on base of continuity.

              Likewise it can be verified that: $$Pleft(X_0geqmax(X_1,dots,X_n-1)right)=frac1n$$



              edit:



              Also $$Pleft(X_0geqmax(X_1,dots,X_n-1)right)=int Pleft(X_0geqmax(X_1,dots,X_n-1)mid X_0=aright)f_X(a)da=$$$$int P(max(X_1,dots,X_n-1)leq a)f_X(a)da=int F_X^n-1(a)f_X(a)da$$



              where the second equality rests on independence.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 5 hours ago

























              answered 5 hours ago









              drhab

              91.3k542124




              91.3k542124











              • I do understand that. I'm trying to get to the explicit general expression stated above.
                – galah92
                5 hours ago










              • I have added a justification of the equality.
                – drhab
                5 hours ago
















              • I do understand that. I'm trying to get to the explicit general expression stated above.
                – galah92
                5 hours ago










              • I have added a justification of the equality.
                – drhab
                5 hours ago















              I do understand that. I'm trying to get to the explicit general expression stated above.
              – galah92
              5 hours ago




              I do understand that. I'm trying to get to the explicit general expression stated above.
              – galah92
              5 hours ago












              I have added a justification of the equality.
              – drhab
              5 hours ago




              I have added a justification of the equality.
              – drhab
              5 hours ago










              up vote
              2
              down vote













              Disclaimer: this is just a remark on drhab's answer.



              Note that
              beginalign
              (F_X_0)^n-1(a)f_X_0(a) = fracdda frac1n (F_X_0)^n(a).
              endalign
              The fundamental theorem of calculus implies that the integral is equal to
              beginalign
              lim_a to infty frac1n (F_X_0)^n(a) - frac1n (F_X_0)^n(0) = frac1n,
              endalign
              where we used that $X_0$ is a positive random variable.






              share|cite|improve this answer
























                up vote
                2
                down vote













                Disclaimer: this is just a remark on drhab's answer.



                Note that
                beginalign
                (F_X_0)^n-1(a)f_X_0(a) = fracdda frac1n (F_X_0)^n(a).
                endalign
                The fundamental theorem of calculus implies that the integral is equal to
                beginalign
                lim_a to infty frac1n (F_X_0)^n(a) - frac1n (F_X_0)^n(0) = frac1n,
                endalign
                where we used that $X_0$ is a positive random variable.






                share|cite|improve this answer






















                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  Disclaimer: this is just a remark on drhab's answer.



                  Note that
                  beginalign
                  (F_X_0)^n-1(a)f_X_0(a) = fracdda frac1n (F_X_0)^n(a).
                  endalign
                  The fundamental theorem of calculus implies that the integral is equal to
                  beginalign
                  lim_a to infty frac1n (F_X_0)^n(a) - frac1n (F_X_0)^n(0) = frac1n,
                  endalign
                  where we used that $X_0$ is a positive random variable.






                  share|cite|improve this answer












                  Disclaimer: this is just a remark on drhab's answer.



                  Note that
                  beginalign
                  (F_X_0)^n-1(a)f_X_0(a) = fracdda frac1n (F_X_0)^n(a).
                  endalign
                  The fundamental theorem of calculus implies that the integral is equal to
                  beginalign
                  lim_a to infty frac1n (F_X_0)^n(a) - frac1n (F_X_0)^n(0) = frac1n,
                  endalign
                  where we used that $X_0$ is a positive random variable.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 5 hours ago









                  Stockfish

                  1414




                  1414




















                      up vote
                      1
                      down vote













                      Since the variables are i.i.d. it is
                      beginalign*
                      mathbbP(X_0geq max(X_1,ldots,X_n-1)|X_0=x) &= mathbbP(xgeq max(X_1,ldots,X_n-1)|X_0=x) \
                      & = mathbbP(xgeq max(X_1,ldots,X_n-1)) \
                      & = mathbbP(X_1leq x,ldots,X_n-1leq x) \
                      & = mathbbP(X_1leq x)^n-1 \
                      & = F_X_0(x)^n-1.
                      endalign*
                      Hence:
                      beginalign*
                      mathbbP(X_0geq max(X_1,ldots,X_n-1)) &=int_0^+inftymathbbP(X_0geq max(X_1,ldots,X_n-1)|X_0=x)f_X_0(x)dx \
                      &=int_0^+infty F_X_0(x)^n-1f_X_0(x)dx
                      endalign*
                      Now you can use the fact that $F_X_0(x)^n-1f_X_0=frac1nfracddxF_X_0(x)^n$ to show that $mathbbP(X_0geq max(X_1,ldots,X_n-1))=frac1n$.



                      Another way to reach this last result is the following. Let $X_(n)=max(X_0,X_1,ldots,X_n-1)$. The event $X_0geq max(X_1,ldots,X_n-1)$ is equivalent to the event $X_0=X_(n)$. Since the variables are i.i.d., $mathbbP(X_0=X_(n))=mathbbP(X_i=X_(n))=mathbbP(X_i textrm is the largest among X_0,ldots,X_n-1)$ for all $i=0,ldots,n-1$. Additionally, since ties have probability zero (as the distributions are continuous) it is
                      $$1=sum_i=0^n-1 mathbbP(X_i=X_(n)) = nmathbbP(X_0=X_(n))=nmathbbP(X_0geq max(X_1,ldots,X_n-1))$$






                      share|cite|improve this answer
























                        up vote
                        1
                        down vote













                        Since the variables are i.i.d. it is
                        beginalign*
                        mathbbP(X_0geq max(X_1,ldots,X_n-1)|X_0=x) &= mathbbP(xgeq max(X_1,ldots,X_n-1)|X_0=x) \
                        & = mathbbP(xgeq max(X_1,ldots,X_n-1)) \
                        & = mathbbP(X_1leq x,ldots,X_n-1leq x) \
                        & = mathbbP(X_1leq x)^n-1 \
                        & = F_X_0(x)^n-1.
                        endalign*
                        Hence:
                        beginalign*
                        mathbbP(X_0geq max(X_1,ldots,X_n-1)) &=int_0^+inftymathbbP(X_0geq max(X_1,ldots,X_n-1)|X_0=x)f_X_0(x)dx \
                        &=int_0^+infty F_X_0(x)^n-1f_X_0(x)dx
                        endalign*
                        Now you can use the fact that $F_X_0(x)^n-1f_X_0=frac1nfracddxF_X_0(x)^n$ to show that $mathbbP(X_0geq max(X_1,ldots,X_n-1))=frac1n$.



                        Another way to reach this last result is the following. Let $X_(n)=max(X_0,X_1,ldots,X_n-1)$. The event $X_0geq max(X_1,ldots,X_n-1)$ is equivalent to the event $X_0=X_(n)$. Since the variables are i.i.d., $mathbbP(X_0=X_(n))=mathbbP(X_i=X_(n))=mathbbP(X_i textrm is the largest among X_0,ldots,X_n-1)$ for all $i=0,ldots,n-1$. Additionally, since ties have probability zero (as the distributions are continuous) it is
                        $$1=sum_i=0^n-1 mathbbP(X_i=X_(n)) = nmathbbP(X_0=X_(n))=nmathbbP(X_0geq max(X_1,ldots,X_n-1))$$






                        share|cite|improve this answer






















                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          Since the variables are i.i.d. it is
                          beginalign*
                          mathbbP(X_0geq max(X_1,ldots,X_n-1)|X_0=x) &= mathbbP(xgeq max(X_1,ldots,X_n-1)|X_0=x) \
                          & = mathbbP(xgeq max(X_1,ldots,X_n-1)) \
                          & = mathbbP(X_1leq x,ldots,X_n-1leq x) \
                          & = mathbbP(X_1leq x)^n-1 \
                          & = F_X_0(x)^n-1.
                          endalign*
                          Hence:
                          beginalign*
                          mathbbP(X_0geq max(X_1,ldots,X_n-1)) &=int_0^+inftymathbbP(X_0geq max(X_1,ldots,X_n-1)|X_0=x)f_X_0(x)dx \
                          &=int_0^+infty F_X_0(x)^n-1f_X_0(x)dx
                          endalign*
                          Now you can use the fact that $F_X_0(x)^n-1f_X_0=frac1nfracddxF_X_0(x)^n$ to show that $mathbbP(X_0geq max(X_1,ldots,X_n-1))=frac1n$.



                          Another way to reach this last result is the following. Let $X_(n)=max(X_0,X_1,ldots,X_n-1)$. The event $X_0geq max(X_1,ldots,X_n-1)$ is equivalent to the event $X_0=X_(n)$. Since the variables are i.i.d., $mathbbP(X_0=X_(n))=mathbbP(X_i=X_(n))=mathbbP(X_i textrm is the largest among X_0,ldots,X_n-1)$ for all $i=0,ldots,n-1$. Additionally, since ties have probability zero (as the distributions are continuous) it is
                          $$1=sum_i=0^n-1 mathbbP(X_i=X_(n)) = nmathbbP(X_0=X_(n))=nmathbbP(X_0geq max(X_1,ldots,X_n-1))$$






                          share|cite|improve this answer












                          Since the variables are i.i.d. it is
                          beginalign*
                          mathbbP(X_0geq max(X_1,ldots,X_n-1)|X_0=x) &= mathbbP(xgeq max(X_1,ldots,X_n-1)|X_0=x) \
                          & = mathbbP(xgeq max(X_1,ldots,X_n-1)) \
                          & = mathbbP(X_1leq x,ldots,X_n-1leq x) \
                          & = mathbbP(X_1leq x)^n-1 \
                          & = F_X_0(x)^n-1.
                          endalign*
                          Hence:
                          beginalign*
                          mathbbP(X_0geq max(X_1,ldots,X_n-1)) &=int_0^+inftymathbbP(X_0geq max(X_1,ldots,X_n-1)|X_0=x)f_X_0(x)dx \
                          &=int_0^+infty F_X_0(x)^n-1f_X_0(x)dx
                          endalign*
                          Now you can use the fact that $F_X_0(x)^n-1f_X_0=frac1nfracddxF_X_0(x)^n$ to show that $mathbbP(X_0geq max(X_1,ldots,X_n-1))=frac1n$.



                          Another way to reach this last result is the following. Let $X_(n)=max(X_0,X_1,ldots,X_n-1)$. The event $X_0geq max(X_1,ldots,X_n-1)$ is equivalent to the event $X_0=X_(n)$. Since the variables are i.i.d., $mathbbP(X_0=X_(n))=mathbbP(X_i=X_(n))=mathbbP(X_i textrm is the largest among X_0,ldots,X_n-1)$ for all $i=0,ldots,n-1$. Additionally, since ties have probability zero (as the distributions are continuous) it is
                          $$1=sum_i=0^n-1 mathbbP(X_i=X_(n)) = nmathbbP(X_0=X_(n))=nmathbbP(X_0geq max(X_1,ldots,X_n-1))$$







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                          answered 5 hours ago









                          a.arfe

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                              What does second last employer means? [closed]

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                              Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote -3 down vote favorite Say, for example I have worked in the following companies, Company A => 2010 - 2011 Company B => 2011 - 2012 Company C => 2012 - Till Date Company D => Have offer Now assume Company D is asking to submit documents from my second last employer. So from the list of companies given, which company would be qualified as second last employer? software-industry employer share | improve this question asked Aug 10 '14 at 5:16 Rajaprabhu Aravindasamy 559 1 6 13 closed as off-topic by Jim G., jmort253 ♦ Aug 11 '14 at 1:58 This question appears to be off-topic. The users who voted to close gave this specific reason: "Questions seeking advice on company-specific regulations, agreements, or policies should be directed to your manager or HR department. Questions that address...
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                              Installing NextGIS Connect into QGIS 3?

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                              Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugins nextgis share | improve this question edited 3 hours ago PolyGeo ♦ 51.9k 17 77 233 asked 4 hours ago IanGS 16 4 New contributor IanGS is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugin...
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                              One-line joke

                              A one-liner is a joke that is delivered in a single line. A good one-liner is said to be pithy - concise and meaningful. [1] Comedians and actors use this comedic method as part of their act, e.g. Jimmy Carr, Tommy Cooper, Rodney Dangerfield, Norm Macdonald, Ken Dodd, Stewart Francis, Zach Galifianakis, Mitch Hedberg, Anthony Jeselnik, Milton Jones, Shappi Khorsandi, Jay London, Mark Linn-Baker, Demetri Martin, Groucho Marx, Dan Mintz, Emo Philips, Tim Vine, Steven Wright and Henny Youngman and Arnold Schwarzenegger. [ citation needed ] Many fictional characters are also known to deliver one-liners, including James Bond, who usually includes pithy and laconic quips after disposing of a villain. [ citation needed ] Examples "A baby seal walks into a club." "A dyslexic man walks into a bra." (Anonymous) "There are three types of people, those who can count and those who can't." "Venison's dear, isn't it?" (Jimmy Carr) "An escala...
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