Distribution of max and min












3














If $X_1,X_2,ldots,X_n,ldots$ are iid uniform random variables on $[-1,1]$. What's the distribution of $X_{max,n} = max_{1 leq i leq n} X_i$ and $X_{min,n} = min_{1 leq i leq n} X_i$? My understanding is
begin{align*}
mathbb {P}(X_{max,n}<a)&=mathbb {P}(X_1<a,X_2<a,...,X_n<a)\
&=mathbb {P}(X_1<a)mathbb {P}(X_2<a)...mathbb {P}(X_n<a)\
&=int_{-1}^{a} frac {1}{2}, dx_1 int_{-1}^{a} frac {1}{2}, dx_2ldotsint_{-1}^{a} frac {1}{2}, dx_n\
&=int_{-1}^{a} (frac {1}{2})^n, dx
end{align*}
Therefore,
begin{align*}
f(X_{max,n})=frac {1}{2^n},quadtext{$X_{max,n}in[-1,1]$}
end{align*}






probability uniform-distribution







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  • 1




    In the fouth line, the power $n$ should be outside the integral. You can see that the density you obtained does not integrate to $1$. If $M$ is the maximum, then $P(M le a) = (a+1)^n/2^n$. The density is then $n (a+1)^{n-1}/2^n$.
    – Fnacool
    Nov 13 '18 at 2:31
















3














If $X_1,X_2,ldots,X_n,ldots$ are iid uniform random variables on $[-1,1]$. What's the distribution of $X_{max,n} = max_{1 leq i leq n} X_i$ and $X_{min,n} = min_{1 leq i leq n} X_i$? My understanding is
begin{align*}
mathbb {P}(X_{max,n}<a)&=mathbb {P}(X_1<a,X_2<a,...,X_n<a)\
&=mathbb {P}(X_1<a)mathbb {P}(X_2<a)...mathbb {P}(X_n<a)\
&=int_{-1}^{a} frac {1}{2}, dx_1 int_{-1}^{a} frac {1}{2}, dx_2ldotsint_{-1}^{a} frac {1}{2}, dx_n\
&=int_{-1}^{a} (frac {1}{2})^n, dx
end{align*}
Therefore,
begin{align*}
f(X_{max,n})=frac {1}{2^n},quadtext{$X_{max,n}in[-1,1]$}
end{align*}






probability uniform-distribution







share|cite|improve this question


















  • 1




    In the fouth line, the power $n$ should be outside the integral. You can see that the density you obtained does not integrate to $1$. If $M$ is the maximum, then $P(M le a) = (a+1)^n/2^n$. The density is then $n (a+1)^{n-1}/2^n$.
    – Fnacool
    Nov 13 '18 at 2:31














3












3








3







If $X_1,X_2,ldots,X_n,ldots$ are iid uniform random variables on $[-1,1]$. What's the distribution of $X_{max,n} = max_{1 leq i leq n} X_i$ and $X_{min,n} = min_{1 leq i leq n} X_i$? My understanding is
begin{align*}
mathbb {P}(X_{max,n}<a)&=mathbb {P}(X_1<a,X_2<a,...,X_n<a)\
&=mathbb {P}(X_1<a)mathbb {P}(X_2<a)...mathbb {P}(X_n<a)\
&=int_{-1}^{a} frac {1}{2}, dx_1 int_{-1}^{a} frac {1}{2}, dx_2ldotsint_{-1}^{a} frac {1}{2}, dx_n\
&=int_{-1}^{a} (frac {1}{2})^n, dx
end{align*}
Therefore,
begin{align*}
f(X_{max,n})=frac {1}{2^n},quadtext{$X_{max,n}in[-1,1]$}
end{align*}






probability uniform-distribution







share|cite|improve this question













If $X_1,X_2,ldots,X_n,ldots$ are iid uniform random variables on $[-1,1]$. What's the distribution of $X_{max,n} = max_{1 leq i leq n} X_i$ and $X_{min,n} = min_{1 leq i leq n} X_i$? My understanding is
begin{align*}
mathbb {P}(X_{max,n}<a)&=mathbb {P}(X_1<a,X_2<a,...,X_n<a)\
&=mathbb {P}(X_1<a)mathbb {P}(X_2<a)...mathbb {P}(X_n<a)\
&=int_{-1}^{a} frac {1}{2}, dx_1 int_{-1}^{a} frac {1}{2}, dx_2ldotsint_{-1}^{a} frac {1}{2}, dx_n\
&=int_{-1}^{a} (frac {1}{2})^n, dx
end{align*}
Therefore,
begin{align*}
f(X_{max,n})=frac {1}{2^n},quadtext{$X_{max,n}in[-1,1]$}
end{align*}






probability uniform-distribution




probability uniform-distribution






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asked Nov 13 '18 at 2:21









Jiexiong687691

705




705








  • 1




    In the fouth line, the power $n$ should be outside the integral. You can see that the density you obtained does not integrate to $1$. If $M$ is the maximum, then $P(M le a) = (a+1)^n/2^n$. The density is then $n (a+1)^{n-1}/2^n$.
    – Fnacool
    Nov 13 '18 at 2:31














  • 1




    In the fouth line, the power $n$ should be outside the integral. You can see that the density you obtained does not integrate to $1$. If $M$ is the maximum, then $P(M le a) = (a+1)^n/2^n$. The density is then $n (a+1)^{n-1}/2^n$.
    – Fnacool
    Nov 13 '18 at 2:31








1




1




In the fouth line, the power $n$ should be outside the integral. You can see that the density you obtained does not integrate to $1$. If $M$ is the maximum, then $P(M le a) = (a+1)^n/2^n$. The density is then $n (a+1)^{n-1}/2^n$.
– Fnacool
Nov 13 '18 at 2:31




In the fouth line, the power $n$ should be outside the integral. You can see that the density you obtained does not integrate to $1$. If $M$ is the maximum, then $P(M le a) = (a+1)^n/2^n$. The density is then $n (a+1)^{n-1}/2^n$.
– Fnacool
Nov 13 '18 at 2:31










3 Answers
3






active

oldest

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2














For the cdf of the sample maximum we have



$$begin{align*}
F_{X_{(n)}}(x)
&=mathsf P(text{max}{{X_1,...,X_n}}leq x)\\
&=mathsf P(X_1leq x,...,X_nleq x)\\
&=mathsf P(Xleq x)^n\\
&=F_X(x)^n
end{align*}$$



where



$$ F_{X}(x)=
begin{cases}
0 & x lt -1 \
frac{x+1}{2} & -1leq xleq1 \
1 & x gt 1
end{cases} $$



Hence taking the derivative we get



$$f_{X_{(n)}} = frac{n(x+1)^{n-1}}{2^n} I_{[-1,1]}(x)$$



Similarly for the sample minimum we have



$$begin{align*}
F_{X_{(1)}}(x)
&=mathsf P(text{min}{{X_1,...,X_n}}leq x)\\
&=1-mathsf P(text{min}{{X_1,...,X_n}}gt x)\\
&=1-left(1-F_X(x)right)^n\\
&=1-left(1-frac{x+1}{2}right)^n
end{align*}$$



Hence taking the derivative we get



$$f_{X_{(1)}} = frac{nleft(-x+1right)^{n-1}}{2^n} I_{[-1,1]}(x)$$






share|cite|improve this answer































    1














    Your error is in the step $$int_{-1}^{a} frac {1}{2}, dx_1 int_{-1}^{a} frac {1}{2}, dx_2ldotsint_{-1}^{a} frac {1}{2}, dx_n =int_{-1}^{a}frac{1}{2^n}, dx$$ which is not right. You can combine them into a multiple integral, not a single integral. Really, the easiest thing to do is to just compute $$ int_{-1}^a frac{1}{2}dx_1 = frac{1}{2}(a+1),$$ and then realize it is just a product of $n$ of these, so $$P(X_{max,n} le a) = frac{1}{2^n}(a+1)^n$$ and then differentiate with respect to $a$ to get the PDF.






    share|cite|improve this answer





























      0














      Let $Y = X_{max}$ and $Z = X_{min}$



      Then the CDF: $F_{Y}(a) = P(Yleq a) = P(X_{1}, ..., X_{n} leq a) = prod_{i=1}^{n}P(X_{i} leq a) = left(frac{a +1}{2} right)^{n}$.



      Similarly, $F_{Z}(a) = P(Z leq a) = 1 - P(Z > a)$
      $ = 1 - prod_{i=1}^{n}P(X_{i} > a) = 1 - prod_{i=1}^{n} left[1 - P(X_{i=1} leq a) right ] = 1 - left(1 - frac{a +1}{2} right)^{n} = 1 - left(frac{1-a}{2} right)^{n}$.



      You can now differentiate the CDFs to get your pdfs.






      share|cite|improve this answer





















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        3 Answers
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        3 Answers
        3






        active

        oldest

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        active

        oldest

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        active

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        2














        For the cdf of the sample maximum we have



        $$begin{align*}
        F_{X_{(n)}}(x)
        &=mathsf P(text{max}{{X_1,...,X_n}}leq x)\\
        &=mathsf P(X_1leq x,...,X_nleq x)\\
        &=mathsf P(Xleq x)^n\\
        &=F_X(x)^n
        end{align*}$$



        where



        $$ F_{X}(x)=
        begin{cases}
        0 & x lt -1 \
        frac{x+1}{2} & -1leq xleq1 \
        1 & x gt 1
        end{cases} $$



        Hence taking the derivative we get



        $$f_{X_{(n)}} = frac{n(x+1)^{n-1}}{2^n} I_{[-1,1]}(x)$$



        Similarly for the sample minimum we have



        $$begin{align*}
        F_{X_{(1)}}(x)
        &=mathsf P(text{min}{{X_1,...,X_n}}leq x)\\
        &=1-mathsf P(text{min}{{X_1,...,X_n}}gt x)\\
        &=1-left(1-F_X(x)right)^n\\
        &=1-left(1-frac{x+1}{2}right)^n
        end{align*}$$



        Hence taking the derivative we get



        $$f_{X_{(1)}} = frac{nleft(-x+1right)^{n-1}}{2^n} I_{[-1,1]}(x)$$






        share|cite|improve this answer




























          2














          For the cdf of the sample maximum we have



          $$begin{align*}
          F_{X_{(n)}}(x)
          &=mathsf P(text{max}{{X_1,...,X_n}}leq x)\\
          &=mathsf P(X_1leq x,...,X_nleq x)\\
          &=mathsf P(Xleq x)^n\\
          &=F_X(x)^n
          end{align*}$$



          where



          $$ F_{X}(x)=
          begin{cases}
          0 & x lt -1 \
          frac{x+1}{2} & -1leq xleq1 \
          1 & x gt 1
          end{cases} $$



          Hence taking the derivative we get



          $$f_{X_{(n)}} = frac{n(x+1)^{n-1}}{2^n} I_{[-1,1]}(x)$$



          Similarly for the sample minimum we have



          $$begin{align*}
          F_{X_{(1)}}(x)
          &=mathsf P(text{min}{{X_1,...,X_n}}leq x)\\
          &=1-mathsf P(text{min}{{X_1,...,X_n}}gt x)\\
          &=1-left(1-F_X(x)right)^n\\
          &=1-left(1-frac{x+1}{2}right)^n
          end{align*}$$



          Hence taking the derivative we get



          $$f_{X_{(1)}} = frac{nleft(-x+1right)^{n-1}}{2^n} I_{[-1,1]}(x)$$






          share|cite|improve this answer


























            2












            2








            2






            For the cdf of the sample maximum we have



            $$begin{align*}
            F_{X_{(n)}}(x)
            &=mathsf P(text{max}{{X_1,...,X_n}}leq x)\\
            &=mathsf P(X_1leq x,...,X_nleq x)\\
            &=mathsf P(Xleq x)^n\\
            &=F_X(x)^n
            end{align*}$$



            where



            $$ F_{X}(x)=
            begin{cases}
            0 & x lt -1 \
            frac{x+1}{2} & -1leq xleq1 \
            1 & x gt 1
            end{cases} $$



            Hence taking the derivative we get



            $$f_{X_{(n)}} = frac{n(x+1)^{n-1}}{2^n} I_{[-1,1]}(x)$$



            Similarly for the sample minimum we have



            $$begin{align*}
            F_{X_{(1)}}(x)
            &=mathsf P(text{min}{{X_1,...,X_n}}leq x)\\
            &=1-mathsf P(text{min}{{X_1,...,X_n}}gt x)\\
            &=1-left(1-F_X(x)right)^n\\
            &=1-left(1-frac{x+1}{2}right)^n
            end{align*}$$



            Hence taking the derivative we get



            $$f_{X_{(1)}} = frac{nleft(-x+1right)^{n-1}}{2^n} I_{[-1,1]}(x)$$






            share|cite|improve this answer














            For the cdf of the sample maximum we have



            $$begin{align*}
            F_{X_{(n)}}(x)
            &=mathsf P(text{max}{{X_1,...,X_n}}leq x)\\
            &=mathsf P(X_1leq x,...,X_nleq x)\\
            &=mathsf P(Xleq x)^n\\
            &=F_X(x)^n
            end{align*}$$



            where



            $$ F_{X}(x)=
            begin{cases}
            0 & x lt -1 \
            frac{x+1}{2} & -1leq xleq1 \
            1 & x gt 1
            end{cases} $$



            Hence taking the derivative we get



            $$f_{X_{(n)}} = frac{n(x+1)^{n-1}}{2^n} I_{[-1,1]}(x)$$



            Similarly for the sample minimum we have



            $$begin{align*}
            F_{X_{(1)}}(x)
            &=mathsf P(text{min}{{X_1,...,X_n}}leq x)\\
            &=1-mathsf P(text{min}{{X_1,...,X_n}}gt x)\\
            &=1-left(1-F_X(x)right)^n\\
            &=1-left(1-frac{x+1}{2}right)^n
            end{align*}$$



            Hence taking the derivative we get



            $$f_{X_{(1)}} = frac{nleft(-x+1right)^{n-1}}{2^n} I_{[-1,1]}(x)$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 13 '18 at 9:01

























            answered Nov 13 '18 at 2:46









            Remy

            6,369721




            6,369721























                1














                Your error is in the step $$int_{-1}^{a} frac {1}{2}, dx_1 int_{-1}^{a} frac {1}{2}, dx_2ldotsint_{-1}^{a} frac {1}{2}, dx_n =int_{-1}^{a}frac{1}{2^n}, dx$$ which is not right. You can combine them into a multiple integral, not a single integral. Really, the easiest thing to do is to just compute $$ int_{-1}^a frac{1}{2}dx_1 = frac{1}{2}(a+1),$$ and then realize it is just a product of $n$ of these, so $$P(X_{max,n} le a) = frac{1}{2^n}(a+1)^n$$ and then differentiate with respect to $a$ to get the PDF.






                share|cite|improve this answer


























                  1














                  Your error is in the step $$int_{-1}^{a} frac {1}{2}, dx_1 int_{-1}^{a} frac {1}{2}, dx_2ldotsint_{-1}^{a} frac {1}{2}, dx_n =int_{-1}^{a}frac{1}{2^n}, dx$$ which is not right. You can combine them into a multiple integral, not a single integral. Really, the easiest thing to do is to just compute $$ int_{-1}^a frac{1}{2}dx_1 = frac{1}{2}(a+1),$$ and then realize it is just a product of $n$ of these, so $$P(X_{max,n} le a) = frac{1}{2^n}(a+1)^n$$ and then differentiate with respect to $a$ to get the PDF.






                  share|cite|improve this answer
























                    1












                    1








                    1






                    Your error is in the step $$int_{-1}^{a} frac {1}{2}, dx_1 int_{-1}^{a} frac {1}{2}, dx_2ldotsint_{-1}^{a} frac {1}{2}, dx_n =int_{-1}^{a}frac{1}{2^n}, dx$$ which is not right. You can combine them into a multiple integral, not a single integral. Really, the easiest thing to do is to just compute $$ int_{-1}^a frac{1}{2}dx_1 = frac{1}{2}(a+1),$$ and then realize it is just a product of $n$ of these, so $$P(X_{max,n} le a) = frac{1}{2^n}(a+1)^n$$ and then differentiate with respect to $a$ to get the PDF.






                    share|cite|improve this answer












                    Your error is in the step $$int_{-1}^{a} frac {1}{2}, dx_1 int_{-1}^{a} frac {1}{2}, dx_2ldotsint_{-1}^{a} frac {1}{2}, dx_n =int_{-1}^{a}frac{1}{2^n}, dx$$ which is not right. You can combine them into a multiple integral, not a single integral. Really, the easiest thing to do is to just compute $$ int_{-1}^a frac{1}{2}dx_1 = frac{1}{2}(a+1),$$ and then realize it is just a product of $n$ of these, so $$P(X_{max,n} le a) = frac{1}{2^n}(a+1)^n$$ and then differentiate with respect to $a$ to get the PDF.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 13 '18 at 2:35









                    spaceisdarkgreen

                    32.5k21753




                    32.5k21753























                        0














                        Let $Y = X_{max}$ and $Z = X_{min}$



                        Then the CDF: $F_{Y}(a) = P(Yleq a) = P(X_{1}, ..., X_{n} leq a) = prod_{i=1}^{n}P(X_{i} leq a) = left(frac{a +1}{2} right)^{n}$.



                        Similarly, $F_{Z}(a) = P(Z leq a) = 1 - P(Z > a)$
                        $ = 1 - prod_{i=1}^{n}P(X_{i} > a) = 1 - prod_{i=1}^{n} left[1 - P(X_{i=1} leq a) right ] = 1 - left(1 - frac{a +1}{2} right)^{n} = 1 - left(frac{1-a}{2} right)^{n}$.



                        You can now differentiate the CDFs to get your pdfs.






                        share|cite|improve this answer


























                          0














                          Let $Y = X_{max}$ and $Z = X_{min}$



                          Then the CDF: $F_{Y}(a) = P(Yleq a) = P(X_{1}, ..., X_{n} leq a) = prod_{i=1}^{n}P(X_{i} leq a) = left(frac{a +1}{2} right)^{n}$.



                          Similarly, $F_{Z}(a) = P(Z leq a) = 1 - P(Z > a)$
                          $ = 1 - prod_{i=1}^{n}P(X_{i} > a) = 1 - prod_{i=1}^{n} left[1 - P(X_{i=1} leq a) right ] = 1 - left(1 - frac{a +1}{2} right)^{n} = 1 - left(frac{1-a}{2} right)^{n}$.



                          You can now differentiate the CDFs to get your pdfs.






                          share|cite|improve this answer
























                            0












                            0








                            0






                            Let $Y = X_{max}$ and $Z = X_{min}$



                            Then the CDF: $F_{Y}(a) = P(Yleq a) = P(X_{1}, ..., X_{n} leq a) = prod_{i=1}^{n}P(X_{i} leq a) = left(frac{a +1}{2} right)^{n}$.



                            Similarly, $F_{Z}(a) = P(Z leq a) = 1 - P(Z > a)$
                            $ = 1 - prod_{i=1}^{n}P(X_{i} > a) = 1 - prod_{i=1}^{n} left[1 - P(X_{i=1} leq a) right ] = 1 - left(1 - frac{a +1}{2} right)^{n} = 1 - left(frac{1-a}{2} right)^{n}$.



                            You can now differentiate the CDFs to get your pdfs.






                            share|cite|improve this answer












                            Let $Y = X_{max}$ and $Z = X_{min}$



                            Then the CDF: $F_{Y}(a) = P(Yleq a) = P(X_{1}, ..., X_{n} leq a) = prod_{i=1}^{n}P(X_{i} leq a) = left(frac{a +1}{2} right)^{n}$.



                            Similarly, $F_{Z}(a) = P(Z leq a) = 1 - P(Z > a)$
                            $ = 1 - prod_{i=1}^{n}P(X_{i} > a) = 1 - prod_{i=1}^{n} left[1 - P(X_{i=1} leq a) right ] = 1 - left(1 - frac{a +1}{2} right)^{n} = 1 - left(frac{1-a}{2} right)^{n}$.



                            You can now differentiate the CDFs to get your pdfs.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 13 '18 at 2:44









                            akech

                            2,642618




                            2,642618






























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                                Thiostrepton

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                                Thiostrepton [1] Names Other names Alaninamide, Bryamycin, Thiactin Identifiers CAS Number 1393-48-2   Y 3D model (JSmol) Interactive image ChEMBL ChEMBL468719   N ECHA InfoCard 100.014.304 PubChem CID 16130278 InChI InChI=1S/C72H85N19O18S5/c1-14-26(3)47-63(105)78-30(7)57(99)75-28(5)56(98)76-31(8)58(100)91-72-19-18-40(66-85-43(22-111-66)59(101)77-29(6)55(97)74-27(4)54(73)96)81-52(72)42-21-112-67(83-42)49(34(11)109-69(107)41-20-37(32(9)92)36-16-17-39(79-47)51(95)50(36)80-41)89-60(102)44-24-113-68(86-44)53(71(13,108)35(12)94)90-62(104)45-23-110-65(84-45)38(15-2)82-64(106)48(33(10)93)88-61(103)46-25-114-70(72)87-46/h15-17,20-22,24-26,30-35,39,45,47-49,51-53,79,92-95,108H,4-6,14,18-19,23H2,1-3,7-13H3,(H2,73,96)(H,74,97)(H,75,99)(H,76,98)(H,77,101)(H,78,105)(H,82,106)(H,88,103)(H,89,102)(H,90,104)(H,91,100)/b38-15-/t26-,30-,31-,32-,33+,34+,35+,39+,45+,47-,48-,49-,51-,52+,53+,71+,72+/m0/s1 Key: NSFFHOGKXHRQEW-AIHSUZKVSA-N ...
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                                Caerphilly

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                                This article is about the town of Caerphilly. For the cheese of the same name, see Caerphilly cheese. For other uses, see Caerphilly (disambiguation). Caerphilly Welsh: Caerffili Caerphilly Castle Caerphilly Location within Caerphilly Population 30,388  OS grid reference ST1586 Principal area Caerphilly Ceremonial county Gwent Country Wales Sovereign state United Kingdom Post town CAERPHILLY Postcode district CF83 Dialling code 029 Police Gwent Fire South Wales Ambulance Welsh EU Parliament Wales UK Parliament Caerphilly Welsh Assembly Caerphilly List of places UK Wales Caerphilly 51°34′41″N 3°13′05″W  /  51.578°N 3.218°W  / 51.578; -3.218 Coordinates: 51°34′41″N 3°13′05″W  /  51.578°N 3.218°W  / 51.578; -3.218 Caerphilly ( / k aɪər ˈ f ɪ l i / ; Welsh: Caerffili , Welsh pronunciation:  [ˌkɑːɨrˈfɪlɪ] ) is a town and community in South Wales, at the southern end of the Rhymney Valley....
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