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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                              Long meetings (6-7 hours a day): Being “babysat” by supervisor

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                              Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote 31 down vote favorite 4 I hold a PhD in mechanical engineering with a 7 year background in self-guided, self-paced research including 2 years as a graduate level course instructor and a subject matter expert for various engineering projects in a university setting in the United States. For the last 10-11 months, I am working in a research industry in France where my task is to debug C++ spaghetti code written over a period of several years by my current supervisor. It is a team of 2: just I and my supervisor. Diurnally, I am posed with a "bug of the day" to get out of this C++ code. My supervisor generally gives me about 15-30 minutes to solve it and then accosts me about whether or not the task is completed and then rinses and repeats this until the end of the day. Off-late, my supervisor has been holding 6-7 hour long "meetin
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                              Is the Concept of Multiple Fantasy Races Scientifically Flawed? [closed]

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                              Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite Many fantasy worlds have multiple humanoid species (elves, orcs, humans, etc) living in the same world, in addition to that, they often have the ability to interbreed. And I can't help but question their believability... fantasy-races reproduction interspecies-relations share | improve this question edited Aug 9 at 14:46 Michael Kjörling ♦ 21.1k 10 85 161 asked Aug 9 at 13:13 Chlodio 118 6 closed as unclear what you're asking by MichaelK, Gryphon, John, Vincent, Frostfyre Aug 9 at 15:25 Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question. 3 Stack Exchange exist to
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                              Confectionery

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                              Prepared foods made with a lot of sugar or other cabohydrates "Sweetmeat" redirects here. For the racehorse, see Sweetmeat (horse). Not to be confused with Sweetbread. This Kransekake is a traditional Scandinavian baker's confection. Confectionery is the art of making confections , which are food items that are rich in sugar and carbohydrates. Exact definitions are difficult. [1] In general, though, confectionery is divided into two broad and somewhat overlapping categories, bakers' confections and sugar confections. [2] Bakers' confectionery, also called flour confections , includes principally sweet pastries, cakes, and similar baked goods. Sugar confectionery includes candies ( sweets in British English), candied nuts, chocolates, chewing gum, bubble gum, pastillage, and other confections that are made primarily of sugar. In some cases, chocolate confections (confections made of chocolate) are treated as a separate category, as are sugar-free versions of
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