Permutation probability












2












$begingroup$


Let $S = {1,2,3,4,5,...,n}$.



Let $Omega$ be set of permutation maps of $S$.



Let $Phi : mathbb{R} to mathbb{R}$ be strictly positive and strictly increasing map.



Consider positive function $P: Omega to mathbb{R}$ defined by



$$P(tau) = prod_{j=1}^{n} frac{Phi(tau(j))}{sum_{k=j}^n Phi(tau(k))}.$$



I want to show that
$P$ is probability function on $Omega$. For that , I should show that $$sum_{tau in Omega} P(tau)=1.$$



I tried to calculate
$$sum_{l=1}^nsum_{tau(l)=1} P(tau).$$



But it is difficult.



Is anyone want to help me?






probability combinatorics discrete-mathematics probability-distributions permutations







share|cite|improve this question











$endgroup$












  • $begingroup$
    I'm confused. $P$ doesn't take an argument in $S$. Did you want to say that $P$ is a probability measure on $color{red}Omega$?
    $endgroup$
    – user593746
    Dec 15 '18 at 16:09










  • $begingroup$
    @Zvi you are right
    $endgroup$
    – Planche
    Dec 15 '18 at 16:17
















2












$begingroup$


Let $S = {1,2,3,4,5,...,n}$.



Let $Omega$ be set of permutation maps of $S$.



Let $Phi : mathbb{R} to mathbb{R}$ be strictly positive and strictly increasing map.



Consider positive function $P: Omega to mathbb{R}$ defined by



$$P(tau) = prod_{j=1}^{n} frac{Phi(tau(j))}{sum_{k=j}^n Phi(tau(k))}.$$



I want to show that
$P$ is probability function on $Omega$. For that , I should show that $$sum_{tau in Omega} P(tau)=1.$$



I tried to calculate
$$sum_{l=1}^nsum_{tau(l)=1} P(tau).$$



But it is difficult.



Is anyone want to help me?






probability combinatorics discrete-mathematics probability-distributions permutations







share|cite|improve this question











$endgroup$












  • $begingroup$
    I'm confused. $P$ doesn't take an argument in $S$. Did you want to say that $P$ is a probability measure on $color{red}Omega$?
    $endgroup$
    – user593746
    Dec 15 '18 at 16:09










  • $begingroup$
    @Zvi you are right
    $endgroup$
    – Planche
    Dec 15 '18 at 16:17














2












2








2


1



$begingroup$


Let $S = {1,2,3,4,5,...,n}$.



Let $Omega$ be set of permutation maps of $S$.



Let $Phi : mathbb{R} to mathbb{R}$ be strictly positive and strictly increasing map.



Consider positive function $P: Omega to mathbb{R}$ defined by



$$P(tau) = prod_{j=1}^{n} frac{Phi(tau(j))}{sum_{k=j}^n Phi(tau(k))}.$$



I want to show that
$P$ is probability function on $Omega$. For that , I should show that $$sum_{tau in Omega} P(tau)=1.$$



I tried to calculate
$$sum_{l=1}^nsum_{tau(l)=1} P(tau).$$



But it is difficult.



Is anyone want to help me?






probability combinatorics discrete-mathematics probability-distributions permutations







share|cite|improve this question











$endgroup$




Let $S = {1,2,3,4,5,...,n}$.



Let $Omega$ be set of permutation maps of $S$.



Let $Phi : mathbb{R} to mathbb{R}$ be strictly positive and strictly increasing map.



Consider positive function $P: Omega to mathbb{R}$ defined by



$$P(tau) = prod_{j=1}^{n} frac{Phi(tau(j))}{sum_{k=j}^n Phi(tau(k))}.$$



I want to show that
$P$ is probability function on $Omega$. For that , I should show that $$sum_{tau in Omega} P(tau)=1.$$



I tried to calculate
$$sum_{l=1}^nsum_{tau(l)=1} P(tau).$$



But it is difficult.



Is anyone want to help me?






probability combinatorics discrete-mathematics probability-distributions permutations




probability combinatorics discrete-mathematics probability-distributions permutations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 16:20







user593746

















asked Dec 15 '18 at 14:48









PlanchePlanche

885212




885212












  • $begingroup$
    I'm confused. $P$ doesn't take an argument in $S$. Did you want to say that $P$ is a probability measure on $color{red}Omega$?
    $endgroup$
    – user593746
    Dec 15 '18 at 16:09










  • $begingroup$
    @Zvi you are right
    $endgroup$
    – Planche
    Dec 15 '18 at 16:17


















  • $begingroup$
    I'm confused. $P$ doesn't take an argument in $S$. Did you want to say that $P$ is a probability measure on $color{red}Omega$?
    $endgroup$
    – user593746
    Dec 15 '18 at 16:09










  • $begingroup$
    @Zvi you are right
    $endgroup$
    – Planche
    Dec 15 '18 at 16:17
















$begingroup$
I'm confused. $P$ doesn't take an argument in $S$. Did you want to say that $P$ is a probability measure on $color{red}Omega$?
$endgroup$
– user593746
Dec 15 '18 at 16:09




$begingroup$
I'm confused. $P$ doesn't take an argument in $S$. Did you want to say that $P$ is a probability measure on $color{red}Omega$?
$endgroup$
– user593746
Dec 15 '18 at 16:09












$begingroup$
@Zvi you are right
$endgroup$
– Planche
Dec 15 '18 at 16:17




$begingroup$
@Zvi you are right
$endgroup$
– Planche
Dec 15 '18 at 16:17










1 Answer
1






active

oldest

votes


















2












$begingroup$

I do not need the monotonicity of $Phi$, only positivity. And since we only need the values of $Phi$ on $Bbb N_1$, I will assume $Phi:Bbb{N}_1to (0,infty)$.



We prove this by induction on $n$. And since dealing with many values of $n$ at the same time, things can get confusing. So, I denote $S_n$ and $Omega_n$ for $S$ and $Omega$ corresponding to $n$. Since $P$ depends on both $n$ and $Phi$, we write $P^{Phi}_n$ for $P$ corresponding to a given pair $(n,Phi)$.



The cases $n=1$ and $n=2$ are trivial. Suppose that $ngeq 3$ and we know the claim holds for $n-1$. For $kin {1,2,ldots,n}$, let $Omega_n(k)$ denote the subset of $Omega_n$ consisting of $tauin Omega_n$ such that $tau(1)=k$. Define $Phi_{k}:Bbb{N}_1to(0,infty)$ by
$$Phi_k(m)=begin{cases}Phi(m)&text{if} m<k,\ Phi(m+1)&text{if} mgeq k.end{cases}$$
Define $s_k:Bbb{N}_1toBbb{N}_1$ to be
$$s_k(m)=begin{cases}m&text{if }mleq k,\m-1&text{if }m>k.end{cases}$$ Let $Gamma_{n,k}:Omega_n(k)toOmega_{n-1}$ be the bijective map sending
$$tau=begin{pmatrix}1&2&cdots&n\tau(1)&tau(2)&cdots&tau(n)end{pmatrix}mapsto begin{pmatrix} 1 & 2 & cdots &n-1\ s_kbig(tau(2)big)&s_kbig(tau(3)big)&cdots&s_kbig(tau(n)big)end{pmatrix}=Gamma_{n,k}tau.$$
It follows that
$$P^Phi_n(tau)=frac{Phi(k)}{sum_{i=1}^nPhi(i)}P^{Phi_k}_{n-1}left(Gamma_{n,k}tauright)$$
for every $tauinOmega_n(k)$. By induction,
$$sum_{tauinOmega_n(k)}P^{Phi_k}_{n-1}(Gamma_{n,k}tau)=sum_{sigmain Omega_{n-1}}P^{Phi_k}_{n-1}(sigma)=1.$$
That is,
$$sum_{tauinOmega_n(k)}P_n^Phi(tau)=frac{Phi(k)}{sum_{i=1}^nPhi(i)}sum_{tauinOmega_n(k)}P^{Phi_k}_{n-1}(Gamma_{n,k}tau)=frac{Phi(k)}{sum_{i=1}^nPhi(i)}.$$
Consequently,
$$sum_{tauinOmega_n}P_n^Phi(tau)=sum_{k=1}^nsum_{tauinOmega_k(n)}P_n^Phi(tau)=sum_{k=1}^nfrac{Phi(k)}{sum_{i=1}^nPhi(i)}=1.$$






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    $begingroup$

    I do not need the monotonicity of $Phi$, only positivity. And since we only need the values of $Phi$ on $Bbb N_1$, I will assume $Phi:Bbb{N}_1to (0,infty)$.



    We prove this by induction on $n$. And since dealing with many values of $n$ at the same time, things can get confusing. So, I denote $S_n$ and $Omega_n$ for $S$ and $Omega$ corresponding to $n$. Since $P$ depends on both $n$ and $Phi$, we write $P^{Phi}_n$ for $P$ corresponding to a given pair $(n,Phi)$.



    The cases $n=1$ and $n=2$ are trivial. Suppose that $ngeq 3$ and we know the claim holds for $n-1$. For $kin {1,2,ldots,n}$, let $Omega_n(k)$ denote the subset of $Omega_n$ consisting of $tauin Omega_n$ such that $tau(1)=k$. Define $Phi_{k}:Bbb{N}_1to(0,infty)$ by
    $$Phi_k(m)=begin{cases}Phi(m)&text{if} m<k,\ Phi(m+1)&text{if} mgeq k.end{cases}$$
    Define $s_k:Bbb{N}_1toBbb{N}_1$ to be
    $$s_k(m)=begin{cases}m&text{if }mleq k,\m-1&text{if }m>k.end{cases}$$ Let $Gamma_{n,k}:Omega_n(k)toOmega_{n-1}$ be the bijective map sending
    $$tau=begin{pmatrix}1&2&cdots&n\tau(1)&tau(2)&cdots&tau(n)end{pmatrix}mapsto begin{pmatrix} 1 & 2 & cdots &n-1\ s_kbig(tau(2)big)&s_kbig(tau(3)big)&cdots&s_kbig(tau(n)big)end{pmatrix}=Gamma_{n,k}tau.$$
    It follows that
    $$P^Phi_n(tau)=frac{Phi(k)}{sum_{i=1}^nPhi(i)}P^{Phi_k}_{n-1}left(Gamma_{n,k}tauright)$$
    for every $tauinOmega_n(k)$. By induction,
    $$sum_{tauinOmega_n(k)}P^{Phi_k}_{n-1}(Gamma_{n,k}tau)=sum_{sigmain Omega_{n-1}}P^{Phi_k}_{n-1}(sigma)=1.$$
    That is,
    $$sum_{tauinOmega_n(k)}P_n^Phi(tau)=frac{Phi(k)}{sum_{i=1}^nPhi(i)}sum_{tauinOmega_n(k)}P^{Phi_k}_{n-1}(Gamma_{n,k}tau)=frac{Phi(k)}{sum_{i=1}^nPhi(i)}.$$
    Consequently,
    $$sum_{tauinOmega_n}P_n^Phi(tau)=sum_{k=1}^nsum_{tauinOmega_k(n)}P_n^Phi(tau)=sum_{k=1}^nfrac{Phi(k)}{sum_{i=1}^nPhi(i)}=1.$$






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      I do not need the monotonicity of $Phi$, only positivity. And since we only need the values of $Phi$ on $Bbb N_1$, I will assume $Phi:Bbb{N}_1to (0,infty)$.



      We prove this by induction on $n$. And since dealing with many values of $n$ at the same time, things can get confusing. So, I denote $S_n$ and $Omega_n$ for $S$ and $Omega$ corresponding to $n$. Since $P$ depends on both $n$ and $Phi$, we write $P^{Phi}_n$ for $P$ corresponding to a given pair $(n,Phi)$.



      The cases $n=1$ and $n=2$ are trivial. Suppose that $ngeq 3$ and we know the claim holds for $n-1$. For $kin {1,2,ldots,n}$, let $Omega_n(k)$ denote the subset of $Omega_n$ consisting of $tauin Omega_n$ such that $tau(1)=k$. Define $Phi_{k}:Bbb{N}_1to(0,infty)$ by
      $$Phi_k(m)=begin{cases}Phi(m)&text{if} m<k,\ Phi(m+1)&text{if} mgeq k.end{cases}$$
      Define $s_k:Bbb{N}_1toBbb{N}_1$ to be
      $$s_k(m)=begin{cases}m&text{if }mleq k,\m-1&text{if }m>k.end{cases}$$ Let $Gamma_{n,k}:Omega_n(k)toOmega_{n-1}$ be the bijective map sending
      $$tau=begin{pmatrix}1&2&cdots&n\tau(1)&tau(2)&cdots&tau(n)end{pmatrix}mapsto begin{pmatrix} 1 & 2 & cdots &n-1\ s_kbig(tau(2)big)&s_kbig(tau(3)big)&cdots&s_kbig(tau(n)big)end{pmatrix}=Gamma_{n,k}tau.$$
      It follows that
      $$P^Phi_n(tau)=frac{Phi(k)}{sum_{i=1}^nPhi(i)}P^{Phi_k}_{n-1}left(Gamma_{n,k}tauright)$$
      for every $tauinOmega_n(k)$. By induction,
      $$sum_{tauinOmega_n(k)}P^{Phi_k}_{n-1}(Gamma_{n,k}tau)=sum_{sigmain Omega_{n-1}}P^{Phi_k}_{n-1}(sigma)=1.$$
      That is,
      $$sum_{tauinOmega_n(k)}P_n^Phi(tau)=frac{Phi(k)}{sum_{i=1}^nPhi(i)}sum_{tauinOmega_n(k)}P^{Phi_k}_{n-1}(Gamma_{n,k}tau)=frac{Phi(k)}{sum_{i=1}^nPhi(i)}.$$
      Consequently,
      $$sum_{tauinOmega_n}P_n^Phi(tau)=sum_{k=1}^nsum_{tauinOmega_k(n)}P_n^Phi(tau)=sum_{k=1}^nfrac{Phi(k)}{sum_{i=1}^nPhi(i)}=1.$$






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        I do not need the monotonicity of $Phi$, only positivity. And since we only need the values of $Phi$ on $Bbb N_1$, I will assume $Phi:Bbb{N}_1to (0,infty)$.



        We prove this by induction on $n$. And since dealing with many values of $n$ at the same time, things can get confusing. So, I denote $S_n$ and $Omega_n$ for $S$ and $Omega$ corresponding to $n$. Since $P$ depends on both $n$ and $Phi$, we write $P^{Phi}_n$ for $P$ corresponding to a given pair $(n,Phi)$.



        The cases $n=1$ and $n=2$ are trivial. Suppose that $ngeq 3$ and we know the claim holds for $n-1$. For $kin {1,2,ldots,n}$, let $Omega_n(k)$ denote the subset of $Omega_n$ consisting of $tauin Omega_n$ such that $tau(1)=k$. Define $Phi_{k}:Bbb{N}_1to(0,infty)$ by
        $$Phi_k(m)=begin{cases}Phi(m)&text{if} m<k,\ Phi(m+1)&text{if} mgeq k.end{cases}$$
        Define $s_k:Bbb{N}_1toBbb{N}_1$ to be
        $$s_k(m)=begin{cases}m&text{if }mleq k,\m-1&text{if }m>k.end{cases}$$ Let $Gamma_{n,k}:Omega_n(k)toOmega_{n-1}$ be the bijective map sending
        $$tau=begin{pmatrix}1&2&cdots&n\tau(1)&tau(2)&cdots&tau(n)end{pmatrix}mapsto begin{pmatrix} 1 & 2 & cdots &n-1\ s_kbig(tau(2)big)&s_kbig(tau(3)big)&cdots&s_kbig(tau(n)big)end{pmatrix}=Gamma_{n,k}tau.$$
        It follows that
        $$P^Phi_n(tau)=frac{Phi(k)}{sum_{i=1}^nPhi(i)}P^{Phi_k}_{n-1}left(Gamma_{n,k}tauright)$$
        for every $tauinOmega_n(k)$. By induction,
        $$sum_{tauinOmega_n(k)}P^{Phi_k}_{n-1}(Gamma_{n,k}tau)=sum_{sigmain Omega_{n-1}}P^{Phi_k}_{n-1}(sigma)=1.$$
        That is,
        $$sum_{tauinOmega_n(k)}P_n^Phi(tau)=frac{Phi(k)}{sum_{i=1}^nPhi(i)}sum_{tauinOmega_n(k)}P^{Phi_k}_{n-1}(Gamma_{n,k}tau)=frac{Phi(k)}{sum_{i=1}^nPhi(i)}.$$
        Consequently,
        $$sum_{tauinOmega_n}P_n^Phi(tau)=sum_{k=1}^nsum_{tauinOmega_k(n)}P_n^Phi(tau)=sum_{k=1}^nfrac{Phi(k)}{sum_{i=1}^nPhi(i)}=1.$$






        share|cite|improve this answer











        $endgroup$



        I do not need the monotonicity of $Phi$, only positivity. And since we only need the values of $Phi$ on $Bbb N_1$, I will assume $Phi:Bbb{N}_1to (0,infty)$.



        We prove this by induction on $n$. And since dealing with many values of $n$ at the same time, things can get confusing. So, I denote $S_n$ and $Omega_n$ for $S$ and $Omega$ corresponding to $n$. Since $P$ depends on both $n$ and $Phi$, we write $P^{Phi}_n$ for $P$ corresponding to a given pair $(n,Phi)$.



        The cases $n=1$ and $n=2$ are trivial. Suppose that $ngeq 3$ and we know the claim holds for $n-1$. For $kin {1,2,ldots,n}$, let $Omega_n(k)$ denote the subset of $Omega_n$ consisting of $tauin Omega_n$ such that $tau(1)=k$. Define $Phi_{k}:Bbb{N}_1to(0,infty)$ by
        $$Phi_k(m)=begin{cases}Phi(m)&text{if} m<k,\ Phi(m+1)&text{if} mgeq k.end{cases}$$
        Define $s_k:Bbb{N}_1toBbb{N}_1$ to be
        $$s_k(m)=begin{cases}m&text{if }mleq k,\m-1&text{if }m>k.end{cases}$$ Let $Gamma_{n,k}:Omega_n(k)toOmega_{n-1}$ be the bijective map sending
        $$tau=begin{pmatrix}1&2&cdots&n\tau(1)&tau(2)&cdots&tau(n)end{pmatrix}mapsto begin{pmatrix} 1 & 2 & cdots &n-1\ s_kbig(tau(2)big)&s_kbig(tau(3)big)&cdots&s_kbig(tau(n)big)end{pmatrix}=Gamma_{n,k}tau.$$
        It follows that
        $$P^Phi_n(tau)=frac{Phi(k)}{sum_{i=1}^nPhi(i)}P^{Phi_k}_{n-1}left(Gamma_{n,k}tauright)$$
        for every $tauinOmega_n(k)$. By induction,
        $$sum_{tauinOmega_n(k)}P^{Phi_k}_{n-1}(Gamma_{n,k}tau)=sum_{sigmain Omega_{n-1}}P^{Phi_k}_{n-1}(sigma)=1.$$
        That is,
        $$sum_{tauinOmega_n(k)}P_n^Phi(tau)=frac{Phi(k)}{sum_{i=1}^nPhi(i)}sum_{tauinOmega_n(k)}P^{Phi_k}_{n-1}(Gamma_{n,k}tau)=frac{Phi(k)}{sum_{i=1}^nPhi(i)}.$$
        Consequently,
        $$sum_{tauinOmega_n}P_n^Phi(tau)=sum_{k=1}^nsum_{tauinOmega_k(n)}P_n^Phi(tau)=sum_{k=1}^nfrac{Phi(k)}{sum_{i=1}^nPhi(i)}=1.$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 15 '18 at 18:10

























        answered Dec 15 '18 at 16:48







        user593746





































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