Permutation probability
$begingroup$
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
asked Dec 15 '18 at 14:48
PlanchePlanche
887212
$endgroup$
add a comment |
$begingroup$
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
asked Dec 15 '18 at 14:48
PlanchePlanche
887212
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Dec 15 '18 at 14:48
PlanchePlanche
887212
$endgroup$
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
asked Dec 15 '18 at 14:48
PlanchePlanche
887212
asked Dec 15 '18 at 14:48
PlanchePlanche
887212
edited Dec 15 '18 at 16:20
user593746
asked Dec 15 '18 at 14:48
PlanchePlanche
887212
asked Dec 15 '18 at 14:48
PlanchePlanche
887212
asked Dec 15 '18 at 14:48
PlanchePlanche
887212
887212
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
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,
$$sum_{tauinOmega_n}P_n^Phi(tau)=sum_{k=1}^nsum_{tauinOmega_k(n)}P_n^Phi(tau)=sum_{k=1}^nfrac{Phi(k)}{sum_{i=1}^nPhi(i)}=1.$$
$endgroup$
add a comment |
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1 Answer
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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,
$$sum_{tauinOmega_n}P_n^Phi(tau)=sum_{k=1}^nsum_{tauinOmega_k(n)}P_n^Phi(tau)=sum_{k=1}^nfrac{Phi(k)}{sum_{i=1}^nPhi(i)}=1.$$
$endgroup$
add a comment |
$begingroup$
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,
$$sum_{tauinOmega_n}P_n^Phi(tau)=sum_{k=1}^nsum_{tauinOmega_k(n)}P_n^Phi(tau)=sum_{k=1}^nfrac{Phi(k)}{sum_{i=1}^nPhi(i)}=1.$$
$endgroup$
add a comment |
$begingroup$
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,
$$sum_{tauinOmega_n}P_n^Phi(tau)=sum_{k=1}^nsum_{tauinOmega_k(n)}P_n^Phi(tau)=sum_{k=1}^nfrac{Phi(k)}{sum_{i=1}^nPhi(i)}=1.$$
$endgroup$
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,
$$sum_{tauinOmega_n}P_n^Phi(tau)=sum_{k=1}^nsum_{tauinOmega_k(n)}P_n^Phi(tau)=sum_{k=1}^nfrac{Phi(k)}{sum_{i=1}^nPhi(i)}=1.$$
edited Dec 15 '18 at 18:10
answered Dec 15 '18 at 16:48
user593746
add a comment |
add a comment |
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$begingroup$
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