Mixing
sequence of inequalities information theory
0
$begingroup$
can you please give a hint about how to solve this problem ?
probability-theory information-theory
asked Jan 25 at 9:11
Salwa MostafaSalwa Mostafa
274
$endgroup$
add a comment |
0
$begingroup$
can you please give a hint about how to solve this problem ?
probability-theory information-theory
asked Jan 25 at 9:11
Salwa MostafaSalwa Mostafa
274
$endgroup$
1
$begingroup$
To warm up: Some easy questions you should (might) already have asked yourself, to check we understand the problem . 1) how many terms have that sum ? 2) can we prove the inequalities if the variables are iid ?
$endgroup$
– leonbloy
Jan 25 at 15:31
$begingroup$
Assume that $n=4 , k = 3 then alpha = {1,3,4} and beta = {1,4} L= 3$ $H(X_{alpha}) = H(X_{beta} ) + H(X_L |X_{beta}) where |beta|=k-1, alpha = beta cup {L}$ Then, $H(X_1,X_3,X_4) = H(X_1,X_4) + H(X_3|X_1,X_4)$ we should use the theorem $H(X_1,X_2) + H(X_2,X_3) + H(X_1,X_3) ≥ 2H(X_1,X_2,X_3).$ and $H(X_1,X_2,X_3) = H(X_j, j= i) + H(X_i | X_j, j = i)$ for $i = 1, 2, 3$ to complete the proof but I stuck
$endgroup$
– Salwa Mostafa
Jan 26 at 2:33
2
$begingroup$
Hint: just try to prove $H_n le H_{n-1}.$ The natural proof you come up with can be generalised to give the following lemma: for any non-empty set $S subset [1:n],$ $$ H(X_S) le frac{1}{|S| -1} sum_{i in S} H(X_{S - {i}} ).$$ Now try to use this lemma to show $H_{k+1} le H_k$ for any $k$.
$endgroup$
– stochasticboy321
Jan 26 at 4:40
add a comment |
0
0
0
$begingroup$
can you please give a hint about how to solve this problem ?
probability-theory information-theory
asked Jan 25 at 9:11
Salwa MostafaSalwa Mostafa
274
$endgroup$
can you please give a hint about how to solve this problem ?
probability-theory information-theory
probability-theory information-theory
asked Jan 25 at 9:11
Salwa MostafaSalwa Mostafa
274
asked Jan 25 at 9:11
Salwa MostafaSalwa Mostafa
274
asked Jan 25 at 9:11
Salwa MostafaSalwa Mostafa
274
asked Jan 25 at 9:11
Salwa MostafaSalwa Mostafa
274
asked Jan 25 at 9:11
Salwa MostafaSalwa Mostafa
274
274
1
$begingroup$
To warm up: Some easy questions you should (might) already have asked yourself, to check we understand the problem . 1) how many terms have that sum ? 2) can we prove the inequalities if the variables are iid ?
$endgroup$
– leonbloy
Jan 25 at 15:31
$begingroup$
Assume that $n=4 , k = 3 then alpha = {1,3,4} and beta = {1,4} L= 3$ $H(X_{alpha}) = H(X_{beta} ) + H(X_L |X_{beta}) where |beta|=k-1, alpha = beta cup {L}$ Then, $H(X_1,X_3,X_4) = H(X_1,X_4) + H(X_3|X_1,X_4)$ we should use the theorem $H(X_1,X_2) + H(X_2,X_3) + H(X_1,X_3) ≥ 2H(X_1,X_2,X_3).$ and $H(X_1,X_2,X_3) = H(X_j, j= i) + H(X_i | X_j, j = i)$ for $i = 1, 2, 3$ to complete the proof but I stuck
$endgroup$
– Salwa Mostafa
Jan 26 at 2:33
2
$begingroup$
Hint: just try to prove $H_n le H_{n-1}.$ The natural proof you come up with can be generalised to give the following lemma: for any non-empty set $S subset [1:n],$ $$ H(X_S) le frac{1}{|S| -1} sum_{i in S} H(X_{S - {i}} ).$$ Now try to use this lemma to show $H_{k+1} le H_k$ for any $k$.
$endgroup$
– stochasticboy321
Jan 26 at 4:40
add a comment |
1
$begingroup$
To warm up: Some easy questions you should (might) already have asked yourself, to check we understand the problem . 1) how many terms have that sum ? 2) can we prove the inequalities if the variables are iid ?
$endgroup$
– leonbloy
Jan 25 at 15:31
$begingroup$
Assume that $n=4 , k = 3 then alpha = {1,3,4} and beta = {1,4} L= 3$ $H(X_{alpha}) = H(X_{beta} ) + H(X_L |X_{beta}) where |beta|=k-1, alpha = beta cup {L}$ Then, $H(X_1,X_3,X_4) = H(X_1,X_4) + H(X_3|X_1,X_4)$ we should use the theorem $H(X_1,X_2) + H(X_2,X_3) + H(X_1,X_3) ≥ 2H(X_1,X_2,X_3).$ and $H(X_1,X_2,X_3) = H(X_j, j= i) + H(X_i | X_j, j = i)$ for $i = 1, 2, 3$ to complete the proof but I stuck
$endgroup$
– Salwa Mostafa
Jan 26 at 2:33
2
$begingroup$
Hint: just try to prove $H_n le H_{n-1}.$ The natural proof you come up with can be generalised to give the following lemma: for any non-empty set $S subset [1:n],$ $$ H(X_S) le frac{1}{|S| -1} sum_{i in S} H(X_{S - {i}} ).$$ Now try to use this lemma to show $H_{k+1} le H_k$ for any $k$.
$endgroup$
– stochasticboy321
Jan 26 at 4:40
1
1
$begingroup$
To warm up: Some easy questions you should (might) already have asked yourself, to check we understand the problem . 1) how many terms have that sum ? 2) can we prove the inequalities if the variables are iid ?
$endgroup$
– leonbloy
Jan 25 at 15:31
$begingroup$
To warm up: Some easy questions you should (might) already have asked yourself, to check we understand the problem . 1) how many terms have that sum ? 2) can we prove the inequalities if the variables are iid ?
$endgroup$
– leonbloy
Jan 25 at 15:31
$begingroup$
Assume that $n=4 , k = 3 then alpha = {1,3,4} and beta = {1,4} L= 3$ $H(X_{alpha}) = H(X_{beta} ) + H(X_L |X_{beta}) where |beta|=k-1, alpha = beta cup {L}$ Then, $H(X_1,X_3,X_4) = H(X_1,X_4) + H(X_3|X_1,X_4)$ we should use the theorem $H(X_1,X_2) + H(X_2,X_3) + H(X_1,X_3) ≥ 2H(X_1,X_2,X_3).$ and $H(X_1,X_2,X_3) = H(X_j, j= i) + H(X_i | X_j, j = i)$ for $i = 1, 2, 3$ to complete the proof but I stuck
$endgroup$
– Salwa Mostafa
Jan 26 at 2:33
$begingroup$
Assume that $n=4 , k = 3 then alpha = {1,3,4} and beta = {1,4} L= 3$ $H(X_{alpha}) = H(X_{beta} ) + H(X_L |X_{beta}) where |beta|=k-1, alpha = beta cup {L}$ Then, $H(X_1,X_3,X_4) = H(X_1,X_4) + H(X_3|X_1,X_4)$ we should use the theorem $H(X_1,X_2) + H(X_2,X_3) + H(X_1,X_3) ≥ 2H(X_1,X_2,X_3).$ and $H(X_1,X_2,X_3) = H(X_j, j= i) + H(X_i | X_j, j = i)$ for $i = 1, 2, 3$ to complete the proof but I stuck
$endgroup$
– Salwa Mostafa
Jan 26 at 2:33
2
2
$begingroup$
Hint: just try to prove $H_n le H_{n-1}.$ The natural proof you come up with can be generalised to give the following lemma: for any non-empty set $S subset [1:n],$ $$ H(X_S) le frac{1}{|S| -1} sum_{i in S} H(X_{S - {i}} ).$$ Now try to use this lemma to show $H_{k+1} le H_k$ for any $k$.
$endgroup$
– stochasticboy321
Jan 26 at 4:40
$begingroup$
Hint: just try to prove $H_n le H_{n-1}.$ The natural proof you come up with can be generalised to give the following lemma: for any non-empty set $S subset [1:n],$ $$ H(X_S) le frac{1}{|S| -1} sum_{i in S} H(X_{S - {i}} ).$$ Now try to use this lemma to show $H_{k+1} le H_k$ for any $k$.
$endgroup$
– stochasticboy321
Jan 26 at 4:40
add a comment |
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1
$begingroup$
To warm up: Some easy questions you should (might) already have asked yourself, to check we understand the problem . 1) how many terms have that sum ? 2) can we prove the inequalities if the variables are iid ?
$endgroup$
– leonbloy
Jan 25 at 15:31
$begingroup$
Assume that $n=4 , k = 3 then alpha = {1,3,4} and beta = {1,4} L= 3$ $H(X_{alpha}) = H(X_{beta} ) + H(X_L |X_{beta}) where |beta|=k-1, alpha = beta cup {L}$ Then, $H(X_1,X_3,X_4) = H(X_1,X_4) + H(X_3|X_1,X_4)$ we should use the theorem $H(X_1,X_2) + H(X_2,X_3) + H(X_1,X_3) ≥ 2H(X_1,X_2,X_3).$ and $H(X_1,X_2,X_3) = H(X_j, j= i) + H(X_i | X_j, j = i)$ for $i = 1, 2, 3$ to complete the proof but I stuck
$endgroup$
– Salwa Mostafa
Jan 26 at 2:33
2
$begingroup$
Hint: just try to prove $H_n le H_{n-1}.$ The natural proof you come up with can be generalised to give the following lemma: for any non-empty set $S subset [1:n],$ $$ H(X_S) le frac{1}{|S| -1} sum_{i in S} H(X_{S - {i}} ).$$ Now try to use this lemma to show $H_{k+1} le H_k$ for any $k$.
$endgroup$
– stochasticboy321
Jan 26 at 4:40