Mixing
Probabilities maximizing products
Clash Royale CLAN TAG#URR8PPP
up vote
8
down vote
favorite
Given is an expression of the form $P=P_1times P_2timesdotstimes P_n$, where each $P_i$ is a sum of some distinct elements from $x_1,x_2,dots,x_k$. (For example, $P=x_1(x_1+x_2)(x_1+x_3)$.) We want to maximize this expression subject to the constraints $x_igeq 0$ for all $i$, and $sum_i=1^k x_i=1$. Let $A$ be the value of $P_1$ at the maximum. Let $B$ be the value of $P_1$ at the maximum if we instead maximize the expression $P'=P_2times P_3timesdotstimes P_n$, subject to the same constraints.
Is it true that $Ageq fracn-1nB+frac1n$?
probability inequality optimization
asked Jul 30 at 10:28
nan
456213
 |Â
show 9 more comments
up vote
8
down vote
favorite
Given is an expression of the form $P=P_1times P_2timesdotstimes P_n$, where each $P_i$ is a sum of some distinct elements from $x_1,x_2,dots,x_k$. (For example, $P=x_1(x_1+x_2)(x_1+x_3)$.) We want to maximize this expression subject to the constraints $x_igeq 0$ for all $i$, and $sum_i=1^k x_i=1$. Let $A$ be the value of $P_1$ at the maximum. Let $B$ be the value of $P_1$ at the maximum if we instead maximize the expression $P'=P_2times P_3timesdotstimes P_n$, subject to the same constraints.
Is it true that $Ageq fracn-1nB+frac1n$?
probability inequality optimization
asked Jul 30 at 10:28
nan
456213
not necessarily. Take $q_1 = 1, q_j = 0$ for $j=2,...,n$ and $p_2=1, p_j = 0$ for $j neq 2$, and $S_1 = 1 $
– Francisco
Jul 30 at 10:56
@Francisco Your choice of $p_i$'s doesn't maximize the given product - it gives value $0$, whereas if we take $p_1=1$ we would get value $1$.
– nan
Jul 30 at 11:52
oh.. I see... then $q_j$ must be zero for any $j in S_1$. That makes it easy, no? just need to prove that $sum_j in S_1p_j geq 1/n$ for $j in S_1$. Which does not seems so difficult to prove.
– Francisco
Jul 30 at 12:01
@Francisco No, it's not necessary that $q_j=0$ for $jin S_1$.
– nan
Jul 30 at 12:09
what do you mean by "sum of distinct elements"? Each $x_i$ is in at most one $P_j$?
– LinAlg
Aug 2 at 16:52
 |Â
show 9 more comments
up vote
8
down vote
favorite
up vote
8
down vote
favorite
Given is an expression of the form $P=P_1times P_2timesdotstimes P_n$, where each $P_i$ is a sum of some distinct elements from $x_1,x_2,dots,x_k$. (For example, $P=x_1(x_1+x_2)(x_1+x_3)$.) We want to maximize this expression subject to the constraints $x_igeq 0$ for all $i$, and $sum_i=1^k x_i=1$. Let $A$ be the value of $P_1$ at the maximum. Let $B$ be the value of $P_1$ at the maximum if we instead maximize the expression $P'=P_2times P_3timesdotstimes P_n$, subject to the same constraints.
Is it true that $Ageq fracn-1nB+frac1n$?
probability inequality optimization
asked Jul 30 at 10:28
nan
456213
Given is an expression of the form $P=P_1times P_2timesdotstimes P_n$, where each $P_i$ is a sum of some distinct elements from $x_1,x_2,dots,x_k$. (For example, $P=x_1(x_1+x_2)(x_1+x_3)$.) We want to maximize this expression subject to the constraints $x_igeq 0$ for all $i$, and $sum_i=1^k x_i=1$. Let $A$ be the value of $P_1$ at the maximum. Let $B$ be the value of $P_1$ at the maximum if we instead maximize the expression $P'=P_2times P_3timesdotstimes P_n$, subject to the same constraints.
Is it true that $Ageq fracn-1nB+frac1n$?
probability inequality optimization
asked Jul 30 at 10:28
nan
456213
edited 2 days ago
asked Jul 30 at 10:28
nan
456213
asked Jul 30 at 10:28
nan
456213
asked Jul 30 at 10:28
nan
456213
456213
not necessarily. Take $q_1 = 1, q_j = 0$ for $j=2,...,n$ and $p_2=1, p_j = 0$ for $j neq 2$, and $S_1 = 1 $
– Francisco
Jul 30 at 10:56
@Francisco Your choice of $p_i$'s doesn't maximize the given product - it gives value $0$, whereas if we take $p_1=1$ we would get value $1$.
– nan
Jul 30 at 11:52
oh.. I see... then $q_j$ must be zero for any $j in S_1$. That makes it easy, no? just need to prove that $sum_j in S_1p_j geq 1/n$ for $j in S_1$. Which does not seems so difficult to prove.
– Francisco
Jul 30 at 12:01
@Francisco No, it's not necessary that $q_j=0$ for $jin S_1$.
– nan
Jul 30 at 12:09
what do you mean by "sum of distinct elements"? Each $x_i$ is in at most one $P_j$?
– LinAlg
Aug 2 at 16:52
 |Â
show 9 more comments
not necessarily. Take $q_1 = 1, q_j = 0$ for $j=2,...,n$ and $p_2=1, p_j = 0$ for $j neq 2$, and $S_1 = 1 $
– Francisco
Jul 30 at 10:56
@Francisco Your choice of $p_i$'s doesn't maximize the given product - it gives value $0$, whereas if we take $p_1=1$ we would get value $1$.
– nan
Jul 30 at 11:52
oh.. I see... then $q_j$ must be zero for any $j in S_1$. That makes it easy, no? just need to prove that $sum_j in S_1p_j geq 1/n$ for $j in S_1$. Which does not seems so difficult to prove.
– Francisco
Jul 30 at 12:01
@Francisco No, it's not necessary that $q_j=0$ for $jin S_1$.
– nan
Jul 30 at 12:09
what do you mean by "sum of distinct elements"? Each $x_i$ is in at most one $P_j$?
– LinAlg
Aug 2 at 16:52
not necessarily. Take $q_1 = 1, q_j = 0$ for $j=2,...,n$ and $p_2=1, p_j = 0$ for $j neq 2$, and $S_1 = 1 $
– Francisco
Jul 30 at 10:56
not necessarily. Take $q_1 = 1, q_j = 0$ for $j=2,...,n$ and $p_2=1, p_j = 0$ for $j neq 2$, and $S_1 = 1 $
– Francisco
Jul 30 at 10:56
@Francisco Your choice of $p_i$'s doesn't maximize the given product - it gives value $0$, whereas if we take $p_1=1$ we would get value $1$.
– nan
Jul 30 at 11:52
@Francisco Your choice of $p_i$'s doesn't maximize the given product - it gives value $0$, whereas if we take $p_1=1$ we would get value $1$.
– nan
Jul 30 at 11:52
oh.. I see... then $q_j$ must be zero for any $j in S_1$. That makes it easy, no? just need to prove that $sum_j in S_1p_j geq 1/n$ for $j in S_1$. Which does not seems so difficult to prove.
– Francisco
Jul 30 at 12:01
oh.. I see... then $q_j$ must be zero for any $j in S_1$. That makes it easy, no? just need to prove that $sum_j in S_1p_j geq 1/n$ for $j in S_1$. Which does not seems so difficult to prove.
– Francisco
Jul 30 at 12:01
@Francisco No, it's not necessary that $q_j=0$ for $jin S_1$.
– nan
Jul 30 at 12:09
@Francisco No, it's not necessary that $q_j=0$ for $jin S_1$.
– nan
Jul 30 at 12:09
what do you mean by "sum of distinct elements"? Each $x_i$ is in at most one $P_j$?
– LinAlg
Aug 2 at 16:52
what do you mean by "sum of distinct elements"? Each $x_i$ is in at most one $P_j$?
– LinAlg
Aug 2 at 16:52
 |Â
show 9 more comments
1 Answer
1
active
oldest
votes
up vote
0
down vote
For what it's worth, I tried some Lagrange multipliers: Let $X=(x_1,...,x_k)$ and $Y_iin 0,1^k$ be such that $P_i=Y_icdot X$. Putting $$L(X)=sum_i ln(Y_icdot X)-lambda(textbf 1cdot X-1)$$ gives for each $j$, either $x_j=0$ or $(sum_i frac1 P_iY_i)_j=lambda=n$. A couple observations:
- $P_i=Y_icdot X$, so the above can be rewritten as $(sum_i X+Z_i)_j=n$ where $Z_icdot X=0$.
- $(sum_i frac1 P_iY_i)cdot X=n$
Hopefully this leads to something useful...
answered Aug 7 at 1:37
Akababa
2,557922
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
For what it's worth, I tried some Lagrange multipliers: Let $X=(x_1,...,x_k)$ and $Y_iin 0,1^k$ be such that $P_i=Y_icdot X$. Putting $$L(X)=sum_i ln(Y_icdot X)-lambda(textbf 1cdot X-1)$$ gives for each $j$, either $x_j=0$ or $(sum_i frac1 P_iY_i)_j=lambda=n$. A couple observations:
- $P_i=Y_icdot X$, so the above can be rewritten as $(sum_i X+Z_i)_j=n$ where $Z_icdot X=0$.
- $(sum_i frac1 P_iY_i)cdot X=n$
Hopefully this leads to something useful...
answered Aug 7 at 1:37
Akababa
2,557922
add a comment |Â
up vote
0
down vote
For what it's worth, I tried some Lagrange multipliers: Let $X=(x_1,...,x_k)$ and $Y_iin 0,1^k$ be such that $P_i=Y_icdot X$. Putting $$L(X)=sum_i ln(Y_icdot X)-lambda(textbf 1cdot X-1)$$ gives for each $j$, either $x_j=0$ or $(sum_i frac1 P_iY_i)_j=lambda=n$. A couple observations:
- $P_i=Y_icdot X$, so the above can be rewritten as $(sum_i X+Z_i)_j=n$ where $Z_icdot X=0$.
- $(sum_i frac1 P_iY_i)cdot X=n$
Hopefully this leads to something useful...
answered Aug 7 at 1:37
Akababa
2,557922
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For what it's worth, I tried some Lagrange multipliers: Let $X=(x_1,...,x_k)$ and $Y_iin 0,1^k$ be such that $P_i=Y_icdot X$. Putting $$L(X)=sum_i ln(Y_icdot X)-lambda(textbf 1cdot X-1)$$ gives for each $j$, either $x_j=0$ or $(sum_i frac1 P_iY_i)_j=lambda=n$. A couple observations:
- $P_i=Y_icdot X$, so the above can be rewritten as $(sum_i X+Z_i)_j=n$ where $Z_icdot X=0$.
- $(sum_i frac1 P_iY_i)cdot X=n$
Hopefully this leads to something useful...
answered Aug 7 at 1:37
Akababa
2,557922
For what it's worth, I tried some Lagrange multipliers: Let $X=(x_1,...,x_k)$ and $Y_iin 0,1^k$ be such that $P_i=Y_icdot X$. Putting $$L(X)=sum_i ln(Y_icdot X)-lambda(textbf 1cdot X-1)$$ gives for each $j$, either $x_j=0$ or $(sum_i frac1 P_iY_i)_j=lambda=n$. A couple observations:
- $P_i=Y_icdot X$, so the above can be rewritten as $(sum_i X+Z_i)_j=n$ where $Z_icdot X=0$.
- $(sum_i frac1 P_iY_i)cdot X=n$
Hopefully this leads to something useful...
answered Aug 7 at 1:37
Akababa
2,557922
answered Aug 7 at 1:37
Akababa
2,557922
answered Aug 7 at 1:37
Akababa
2,557922
answered Aug 7 at 1:37
Akababa
2,557922
2,557922
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2866872%2fprobabilities-maximizing-products%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
not necessarily. Take $q_1 = 1, q_j = 0$ for $j=2,...,n$ and $p_2=1, p_j = 0$ for $j neq 2$, and $S_1 = 1 $
– Francisco
Jul 30 at 10:56
@Francisco Your choice of $p_i$'s doesn't maximize the given product - it gives value $0$, whereas if we take $p_1=1$ we would get value $1$.
– nan
Jul 30 at 11:52
oh.. I see... then $q_j$ must be zero for any $j in S_1$. That makes it easy, no? just need to prove that $sum_j in S_1p_j geq 1/n$ for $j in S_1$. Which does not seems so difficult to prove.
– Francisco
Jul 30 at 12:01
@Francisco No, it's not necessary that $q_j=0$ for $jin S_1$.
– nan
Jul 30 at 12:09
what do you mean by "sum of distinct elements"? Each $x_i$ is in at most one $P_j$?
– LinAlg
Aug 2 at 16:52