Independence of probability of $a_k$ being the largest element among the first $k$ elements in the...












3












$begingroup$


The question is:



Let $n ge 2$ be an integer and consider a uniformly random permutation
($a_1$, $a_2$, . . . , $a_n$) of the set (1, 2, . . . , n).



For each $k$ with $1 le k le n$, define the event



$A_k$ = "$a_k$ is the largest element among the first $k$ elements
in the permutation".



Let $k$ and $l$ be two integers with $1 le k lt l le n$. Prove that the events $A_k$ and $A_l$ are independent.



I know that we should define Pr($A_k$), Pr($A_k$), Pr($A_k cap A_l$) and than check relation between those but I keep getting into a dead end where those events are dependent.










share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    The question is:



    Let $n ge 2$ be an integer and consider a uniformly random permutation
    ($a_1$, $a_2$, . . . , $a_n$) of the set (1, 2, . . . , n).



    For each $k$ with $1 le k le n$, define the event



    $A_k$ = "$a_k$ is the largest element among the first $k$ elements
    in the permutation".



    Let $k$ and $l$ be two integers with $1 le k lt l le n$. Prove that the events $A_k$ and $A_l$ are independent.



    I know that we should define Pr($A_k$), Pr($A_k$), Pr($A_k cap A_l$) and than check relation between those but I keep getting into a dead end where those events are dependent.










    share|cite|improve this question









    $endgroup$















      3












      3








      3


      1



      $begingroup$


      The question is:



      Let $n ge 2$ be an integer and consider a uniformly random permutation
      ($a_1$, $a_2$, . . . , $a_n$) of the set (1, 2, . . . , n).



      For each $k$ with $1 le k le n$, define the event



      $A_k$ = "$a_k$ is the largest element among the first $k$ elements
      in the permutation".



      Let $k$ and $l$ be two integers with $1 le k lt l le n$. Prove that the events $A_k$ and $A_l$ are independent.



      I know that we should define Pr($A_k$), Pr($A_k$), Pr($A_k cap A_l$) and than check relation between those but I keep getting into a dead end where those events are dependent.










      share|cite|improve this question









      $endgroup$




      The question is:



      Let $n ge 2$ be an integer and consider a uniformly random permutation
      ($a_1$, $a_2$, . . . , $a_n$) of the set (1, 2, . . . , n).



      For each $k$ with $1 le k le n$, define the event



      $A_k$ = "$a_k$ is the largest element among the first $k$ elements
      in the permutation".



      Let $k$ and $l$ be two integers with $1 le k lt l le n$. Prove that the events $A_k$ and $A_l$ are independent.



      I know that we should define Pr($A_k$), Pr($A_k$), Pr($A_k cap A_l$) and than check relation between those but I keep getting into a dead end where those events are dependent.







      probability combinatorics permutations independence






      share|cite|improve this question













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      share|cite|improve this question










      asked Dec 24 '18 at 14:12









      MrKadek750MrKadek750

      253




      253






















          2 Answers
          2






          active

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          1












          $begingroup$

          To specify a permutation where $a_k$ is the largest among ${a_1,dots,a_k}$, you choose which $k$ numbers form the set ${a_1,dots,a_k}$ in $binom{n}k$ ways, you place the largest at the $k^{th}$ spot in $1$ way, you order the other elements in $(k-1)!$ ways, then you order the other $n-k$ elements in $(n-k)!$ ways. Therefore,
          $$
          P(A_k)=frac{text{# of valid permutations}}{text{total number of permutations}}=frac{binom{n}k(k-1)!(n-k)!}{n!}=frac1k
          $$

          Similarly, letting $k<ell$, in order to specify a permutation where $a_ell$ is the largest among $a_1,dots a_ell$ and $a_k$ among $a_1,dots,a_k$, you




          • Choose the set ${a_1,dots,a_ell}$ in $binom{n}ell$ ways.

          • Place the largest element in the $ell^{th}$ spot.

          • From the remaining $ell-1$ numbers, choose ${a_1,dots,a_k}$ in $binom{ell-1}k$ ways.

          • Place the largest of those $k$ numbers in the $k^{th}$ spot.

          • Choose an order for $a_{k+1},a_{k+2},dots,a_{ell-1}$ in $(ell-1-k)!$ ways.

          • Choose an order for $a_1,dots,a_{k-1}$ in $(k-1)!$ ways.

          • Choose an order for$a_{ell+1},dots,a_n$ in $(n-ell)!$ ways.


          If you multiply all hose numbers out and divide by $n!$, wou will get $frac{1}{kell}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Beautiful, thank you!
            $endgroup$
            – MrKadek750
            Dec 25 '18 at 15:04



















          1












          $begingroup$

          For each $k$ let $[n]_k$ be the family of all subsets of the set ${1,dots,n}$ of size $k$. Then



          $$p_{k,n}=P(A_k)=sum_{Sin [n]_k} P(A_k|{a_1,dots,a_k}=S) P({a_1,dots,a_k}=S)=$$ $$sum_{Sin [n]_k} frac 1kP({a_1,dots,a_k}=S)=frac 1ksum_{Sin [n]_k} P({a_1,dots,a_k}=S)=frac 1k.$$



          $$P(A_kcap A_l)=sum_{Sin [n]_l} P(A_kcap A_l|{a_1,dots,a_l}=S) P({a_1,dots,a_l}=S)=$$
          $$sum_{Sin [n]_l} frac 1l P(A_k|{a_1,dots,a_{l-1}}=Ssetminusmax S) P({a_1,dots,a_l}=S)=$$
          $$sum_{Sin [n]_l} frac 1l p_{k,l-1}P({a_1,dots,a_l}=S)=sum_{Sin [n]_l} frac 1lfrac 1{k} P({a_1,dots,a_l}=S)=$$ $$ frac 1lfrac 1{k}sum_{Sin [n]_l} P({a_1,dots,a_l}=S)= frac 1lfrac 1{k}=P(A_l)P(A_k).$$






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            2 Answers
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            active

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            1












            $begingroup$

            To specify a permutation where $a_k$ is the largest among ${a_1,dots,a_k}$, you choose which $k$ numbers form the set ${a_1,dots,a_k}$ in $binom{n}k$ ways, you place the largest at the $k^{th}$ spot in $1$ way, you order the other elements in $(k-1)!$ ways, then you order the other $n-k$ elements in $(n-k)!$ ways. Therefore,
            $$
            P(A_k)=frac{text{# of valid permutations}}{text{total number of permutations}}=frac{binom{n}k(k-1)!(n-k)!}{n!}=frac1k
            $$

            Similarly, letting $k<ell$, in order to specify a permutation where $a_ell$ is the largest among $a_1,dots a_ell$ and $a_k$ among $a_1,dots,a_k$, you




            • Choose the set ${a_1,dots,a_ell}$ in $binom{n}ell$ ways.

            • Place the largest element in the $ell^{th}$ spot.

            • From the remaining $ell-1$ numbers, choose ${a_1,dots,a_k}$ in $binom{ell-1}k$ ways.

            • Place the largest of those $k$ numbers in the $k^{th}$ spot.

            • Choose an order for $a_{k+1},a_{k+2},dots,a_{ell-1}$ in $(ell-1-k)!$ ways.

            • Choose an order for $a_1,dots,a_{k-1}$ in $(k-1)!$ ways.

            • Choose an order for$a_{ell+1},dots,a_n$ in $(n-ell)!$ ways.


            If you multiply all hose numbers out and divide by $n!$, wou will get $frac{1}{kell}$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Beautiful, thank you!
              $endgroup$
              – MrKadek750
              Dec 25 '18 at 15:04
















            1












            $begingroup$

            To specify a permutation where $a_k$ is the largest among ${a_1,dots,a_k}$, you choose which $k$ numbers form the set ${a_1,dots,a_k}$ in $binom{n}k$ ways, you place the largest at the $k^{th}$ spot in $1$ way, you order the other elements in $(k-1)!$ ways, then you order the other $n-k$ elements in $(n-k)!$ ways. Therefore,
            $$
            P(A_k)=frac{text{# of valid permutations}}{text{total number of permutations}}=frac{binom{n}k(k-1)!(n-k)!}{n!}=frac1k
            $$

            Similarly, letting $k<ell$, in order to specify a permutation where $a_ell$ is the largest among $a_1,dots a_ell$ and $a_k$ among $a_1,dots,a_k$, you




            • Choose the set ${a_1,dots,a_ell}$ in $binom{n}ell$ ways.

            • Place the largest element in the $ell^{th}$ spot.

            • From the remaining $ell-1$ numbers, choose ${a_1,dots,a_k}$ in $binom{ell-1}k$ ways.

            • Place the largest of those $k$ numbers in the $k^{th}$ spot.

            • Choose an order for $a_{k+1},a_{k+2},dots,a_{ell-1}$ in $(ell-1-k)!$ ways.

            • Choose an order for $a_1,dots,a_{k-1}$ in $(k-1)!$ ways.

            • Choose an order for$a_{ell+1},dots,a_n$ in $(n-ell)!$ ways.


            If you multiply all hose numbers out and divide by $n!$, wou will get $frac{1}{kell}$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Beautiful, thank you!
              $endgroup$
              – MrKadek750
              Dec 25 '18 at 15:04














            1












            1








            1





            $begingroup$

            To specify a permutation where $a_k$ is the largest among ${a_1,dots,a_k}$, you choose which $k$ numbers form the set ${a_1,dots,a_k}$ in $binom{n}k$ ways, you place the largest at the $k^{th}$ spot in $1$ way, you order the other elements in $(k-1)!$ ways, then you order the other $n-k$ elements in $(n-k)!$ ways. Therefore,
            $$
            P(A_k)=frac{text{# of valid permutations}}{text{total number of permutations}}=frac{binom{n}k(k-1)!(n-k)!}{n!}=frac1k
            $$

            Similarly, letting $k<ell$, in order to specify a permutation where $a_ell$ is the largest among $a_1,dots a_ell$ and $a_k$ among $a_1,dots,a_k$, you




            • Choose the set ${a_1,dots,a_ell}$ in $binom{n}ell$ ways.

            • Place the largest element in the $ell^{th}$ spot.

            • From the remaining $ell-1$ numbers, choose ${a_1,dots,a_k}$ in $binom{ell-1}k$ ways.

            • Place the largest of those $k$ numbers in the $k^{th}$ spot.

            • Choose an order for $a_{k+1},a_{k+2},dots,a_{ell-1}$ in $(ell-1-k)!$ ways.

            • Choose an order for $a_1,dots,a_{k-1}$ in $(k-1)!$ ways.

            • Choose an order for$a_{ell+1},dots,a_n$ in $(n-ell)!$ ways.


            If you multiply all hose numbers out and divide by $n!$, wou will get $frac{1}{kell}$.






            share|cite|improve this answer









            $endgroup$



            To specify a permutation where $a_k$ is the largest among ${a_1,dots,a_k}$, you choose which $k$ numbers form the set ${a_1,dots,a_k}$ in $binom{n}k$ ways, you place the largest at the $k^{th}$ spot in $1$ way, you order the other elements in $(k-1)!$ ways, then you order the other $n-k$ elements in $(n-k)!$ ways. Therefore,
            $$
            P(A_k)=frac{text{# of valid permutations}}{text{total number of permutations}}=frac{binom{n}k(k-1)!(n-k)!}{n!}=frac1k
            $$

            Similarly, letting $k<ell$, in order to specify a permutation where $a_ell$ is the largest among $a_1,dots a_ell$ and $a_k$ among $a_1,dots,a_k$, you




            • Choose the set ${a_1,dots,a_ell}$ in $binom{n}ell$ ways.

            • Place the largest element in the $ell^{th}$ spot.

            • From the remaining $ell-1$ numbers, choose ${a_1,dots,a_k}$ in $binom{ell-1}k$ ways.

            • Place the largest of those $k$ numbers in the $k^{th}$ spot.

            • Choose an order for $a_{k+1},a_{k+2},dots,a_{ell-1}$ in $(ell-1-k)!$ ways.

            • Choose an order for $a_1,dots,a_{k-1}$ in $(k-1)!$ ways.

            • Choose an order for$a_{ell+1},dots,a_n$ in $(n-ell)!$ ways.


            If you multiply all hose numbers out and divide by $n!$, wou will get $frac{1}{kell}$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 24 '18 at 16:46









            Mike EarnestMike Earnest

            24.3k22151




            24.3k22151












            • $begingroup$
              Beautiful, thank you!
              $endgroup$
              – MrKadek750
              Dec 25 '18 at 15:04


















            • $begingroup$
              Beautiful, thank you!
              $endgroup$
              – MrKadek750
              Dec 25 '18 at 15:04
















            $begingroup$
            Beautiful, thank you!
            $endgroup$
            – MrKadek750
            Dec 25 '18 at 15:04




            $begingroup$
            Beautiful, thank you!
            $endgroup$
            – MrKadek750
            Dec 25 '18 at 15:04











            1












            $begingroup$

            For each $k$ let $[n]_k$ be the family of all subsets of the set ${1,dots,n}$ of size $k$. Then



            $$p_{k,n}=P(A_k)=sum_{Sin [n]_k} P(A_k|{a_1,dots,a_k}=S) P({a_1,dots,a_k}=S)=$$ $$sum_{Sin [n]_k} frac 1kP({a_1,dots,a_k}=S)=frac 1ksum_{Sin [n]_k} P({a_1,dots,a_k}=S)=frac 1k.$$



            $$P(A_kcap A_l)=sum_{Sin [n]_l} P(A_kcap A_l|{a_1,dots,a_l}=S) P({a_1,dots,a_l}=S)=$$
            $$sum_{Sin [n]_l} frac 1l P(A_k|{a_1,dots,a_{l-1}}=Ssetminusmax S) P({a_1,dots,a_l}=S)=$$
            $$sum_{Sin [n]_l} frac 1l p_{k,l-1}P({a_1,dots,a_l}=S)=sum_{Sin [n]_l} frac 1lfrac 1{k} P({a_1,dots,a_l}=S)=$$ $$ frac 1lfrac 1{k}sum_{Sin [n]_l} P({a_1,dots,a_l}=S)= frac 1lfrac 1{k}=P(A_l)P(A_k).$$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              For each $k$ let $[n]_k$ be the family of all subsets of the set ${1,dots,n}$ of size $k$. Then



              $$p_{k,n}=P(A_k)=sum_{Sin [n]_k} P(A_k|{a_1,dots,a_k}=S) P({a_1,dots,a_k}=S)=$$ $$sum_{Sin [n]_k} frac 1kP({a_1,dots,a_k}=S)=frac 1ksum_{Sin [n]_k} P({a_1,dots,a_k}=S)=frac 1k.$$



              $$P(A_kcap A_l)=sum_{Sin [n]_l} P(A_kcap A_l|{a_1,dots,a_l}=S) P({a_1,dots,a_l}=S)=$$
              $$sum_{Sin [n]_l} frac 1l P(A_k|{a_1,dots,a_{l-1}}=Ssetminusmax S) P({a_1,dots,a_l}=S)=$$
              $$sum_{Sin [n]_l} frac 1l p_{k,l-1}P({a_1,dots,a_l}=S)=sum_{Sin [n]_l} frac 1lfrac 1{k} P({a_1,dots,a_l}=S)=$$ $$ frac 1lfrac 1{k}sum_{Sin [n]_l} P({a_1,dots,a_l}=S)= frac 1lfrac 1{k}=P(A_l)P(A_k).$$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                For each $k$ let $[n]_k$ be the family of all subsets of the set ${1,dots,n}$ of size $k$. Then



                $$p_{k,n}=P(A_k)=sum_{Sin [n]_k} P(A_k|{a_1,dots,a_k}=S) P({a_1,dots,a_k}=S)=$$ $$sum_{Sin [n]_k} frac 1kP({a_1,dots,a_k}=S)=frac 1ksum_{Sin [n]_k} P({a_1,dots,a_k}=S)=frac 1k.$$



                $$P(A_kcap A_l)=sum_{Sin [n]_l} P(A_kcap A_l|{a_1,dots,a_l}=S) P({a_1,dots,a_l}=S)=$$
                $$sum_{Sin [n]_l} frac 1l P(A_k|{a_1,dots,a_{l-1}}=Ssetminusmax S) P({a_1,dots,a_l}=S)=$$
                $$sum_{Sin [n]_l} frac 1l p_{k,l-1}P({a_1,dots,a_l}=S)=sum_{Sin [n]_l} frac 1lfrac 1{k} P({a_1,dots,a_l}=S)=$$ $$ frac 1lfrac 1{k}sum_{Sin [n]_l} P({a_1,dots,a_l}=S)= frac 1lfrac 1{k}=P(A_l)P(A_k).$$






                share|cite|improve this answer









                $endgroup$



                For each $k$ let $[n]_k$ be the family of all subsets of the set ${1,dots,n}$ of size $k$. Then



                $$p_{k,n}=P(A_k)=sum_{Sin [n]_k} P(A_k|{a_1,dots,a_k}=S) P({a_1,dots,a_k}=S)=$$ $$sum_{Sin [n]_k} frac 1kP({a_1,dots,a_k}=S)=frac 1ksum_{Sin [n]_k} P({a_1,dots,a_k}=S)=frac 1k.$$



                $$P(A_kcap A_l)=sum_{Sin [n]_l} P(A_kcap A_l|{a_1,dots,a_l}=S) P({a_1,dots,a_l}=S)=$$
                $$sum_{Sin [n]_l} frac 1l P(A_k|{a_1,dots,a_{l-1}}=Ssetminusmax S) P({a_1,dots,a_l}=S)=$$
                $$sum_{Sin [n]_l} frac 1l p_{k,l-1}P({a_1,dots,a_l}=S)=sum_{Sin [n]_l} frac 1lfrac 1{k} P({a_1,dots,a_l}=S)=$$ $$ frac 1lfrac 1{k}sum_{Sin [n]_l} P({a_1,dots,a_l}=S)= frac 1lfrac 1{k}=P(A_l)P(A_k).$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 24 '18 at 16:22









                Alex RavskyAlex Ravsky

                42.4k32383




                42.4k32383






























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