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An Entropy Inequalityentropy inequality

Let $X,Y$ be probability measures on $\{1,2,\dots,n\}$, and set $K=\sum_i\sqrt{X(i)Y(i)}$ so that $Z:=\frac{1}{K}\sqrt{XY}$ is also a probability measure on $\{1,2,\dots,n\}$. How can we prove the inequality $$H(X)+H(Y)\geq 2K^2 H(Z),$$ where $H(X)=-\sum_{i=1}^n X(i)\log X(i)$ is the Entropyentropy function.

The problem originates from this math stack exchange post, and cardinal's rewording of it in the comments. Despite having being asked over two years ago, with numerous bounties posted, the problem was never solved, and for that reason I am posting it here.

I checked the inequality numerically on matlab for millions of choices of $X$ and $Y$, with $n$ up to size $100$, and it always held, which suggests that finding a counter example is unlikely.

Remark: By Cauchy Schwarz, $1\geq K^2,$ so the above inequality would be implied by $H(X)+H(Y)\geq 2H(Z).$ However it is worth noting that this inequality does not hold, so the factor of $K^2$ is important.

An Entropy Inequality

Let $X,Y$ be probability measures on $\{1,2,\dots,n\}$, and set $K=\sum_i\sqrt{X(i)Y(i)}$ so that $Z:=\frac{1}{K}\sqrt{XY}$ is also a probability measure on $\{1,2,\dots,n\}$. How can we prove the inequality $$H(X)+H(Y)\geq 2K^2 H(Z),$$ where $H(X)=-\sum_{i=1}^n X(i)\log X(i)$ is the Entropy function.

The problem originates from this math stack exchange post, and cardinal's rewording of it in the comments. Despite having being asked over two years ago, with numerous bounties posted, the problem was never solved, and for that reason I am posting it here.

I checked the inequality numerically on matlab for millions of choices of $X$ and $Y$, with $n$ up to size $100$, and it always held, which suggests that finding a counter example is unlikely.

Remark: By Cauchy Schwarz, $1\geq K^2,$ so the above inequality would be implied by $H(X)+H(Y)\geq 2H(Z).$ However it is worth noting that this inequality does not hold, so the factor of $K^2$ is important.

An entropy inequality

Let $X,Y$ be probability measures on $\{1,2,\dots,n\}$, and set $K=\sum_i\sqrt{X(i)Y(i)}$ so that $Z:=\frac{1}{K}\sqrt{XY}$ is also a probability measure on $\{1,2,\dots,n\}$. How can we prove the inequality $$H(X)+H(Y)\geq 2K^2 H(Z),$$ where $H(X)=-\sum_{i=1}^n X(i)\log X(i)$ is the entropy function.

The problem originates from this math stack exchange post, and cardinal's rewording of it in the comments. Despite having being asked over two years ago, with numerous bounties posted, the problem was never solved, and for that reason I am posting it here.

I checked the inequality numerically on matlab for millions of choices of $X$ and $Y$, with $n$ up to size $100$, and it always held, which suggests that finding a counter example is unlikely.

Remark: By Cauchy Schwarz, $1\geq K^2,$ so the above inequality would be implied by $H(X)+H(Y)\geq 2H(Z).$ However it is worth noting that this inequality does not hold, so the factor of $K^2$ is important.

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Let $X,Y$ be probability measures on $\{1,2,\dots,n\}$, and set $K=\sum_i\sqrt{X(i)Y(i)}$ so that $Z:=\frac{1}{K}\sqrt{XY}$ is also a probability measure on $\{1,2,\dots,n\}$. How can we prove the inequality $$H(X)+H(Y)\geq 2K^2 H(Z),$$ where $H(X)=-\sum_{i=1}^n X(i)\log X(i)$ is the Entropy function.

The problem originates from this math stack exchange postmath stack exchange post, and cardinal's rewordingcardinal's rewording of it in the comments. Despite having being asked over two years ago, with numerous bounties posted, the problem was never solved, and for that reason I am posting it here.

I checked the inequality numerically on matlab for millions of choices of $X$ and $Y$, with $n$ up to size $100$, and it always held, which suggests that finding a counter example is unlikely.

Remark: By Cauchy Schwarz, $1\geq K^2,$ so the above inequality would be implied by $H(X)+H(Y)\geq 2H(Z).$ However it is worth noting that this inequality does not hold, so the factor of $K^2$ is important.

Let $X,Y$ be probability measures on $\{1,2,\dots,n\}$, and set $K=\sum_i\sqrt{X(i)Y(i)}$ so that $Z:=\frac{1}{K}\sqrt{XY}$ is also a probability measure on $\{1,2,\dots,n\}$. How can we prove the inequality $$H(X)+H(Y)\geq 2K^2 H(Z),$$ where $H(X)=-\sum_{i=1}^n X(i)\log X(i)$ is the Entropy function.

The problem originates from this math stack exchange post, and cardinal's rewording of it in the comments. Despite having being asked over two years ago, with numerous bounties posted, the problem was never solved, and for that reason I am posting it here.

I checked the inequality numerically on matlab for millions of choices of $X$ and $Y$, with $n$ up to size $100$, and it always held, which suggests that finding a counter example is unlikely.

Remark: By Cauchy Schwarz, $1\geq K^2,$ so the above inequality would be implied by $H(X)+H(Y)\geq 2H(Z).$ However it is worth noting that this inequality does not hold, so the factor of $K^2$ is important.

Let $X,Y$ be probability measures on $\{1,2,\dots,n\}$, and set $K=\sum_i\sqrt{X(i)Y(i)}$ so that $Z:=\frac{1}{K}\sqrt{XY}$ is also a probability measure on $\{1,2,\dots,n\}$. How can we prove the inequality $$H(X)+H(Y)\geq 2K^2 H(Z),$$ where $H(X)=-\sum_{i=1}^n X(i)\log X(i)$ is the Entropy function.

The problem originates from this math stack exchange post, and cardinal's rewording of it in the comments. Despite having being asked over two years ago, with numerous bounties posted, the problem was never solved, and for that reason I am posting it here.

I checked the inequality numerically on matlab for millions of choices of $X$ and $Y$, with $n$ up to size $100$, and it always held, which suggests that finding a counter example is unlikely.

Remark: By Cauchy Schwarz, $1\geq K^2,$ so the above inequality would be implied by $H(X)+H(Y)\geq 2H(Z).$ However it is worth noting that this inequality does not hold, so the factor of $K^2$ is important.

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Eric Naslund
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Let $X,Y$ be probability measures on $\{1,2,\dots,n\}$, and set $K=\sum_i\sqrt{X(i)Y(i)}$ so that $Z:=\frac{1}{K}\sqrt{XY}$ is also a probability measure on $\{1,2,\dots,n\}$. How can we prove the inequality $$H(X)+H(Y)\geq 2K^2 H(Z),$$ where $H(X)=-\sum_{i=1}^n X(i)\log X(i)$ is the Entropy function.

The problem originates from this math stack exchange post, and cardinal's rewording of it in the comments. Despite having being asked over two years ago, with numerous bounties posted, the problem was never solved, and for that reason I am posting it here.

I checked the inequality numerically on matlab for millions of choices of $X$ and $Y$, with $n$ up to size $100$, and it always held, which suggests that finding a counter example is unlikely.

RemarksRemark: By Cauchy Schwarz, $1\geq K^2,$ so the above inequality would be implied by $H(X)+H(Y)\geq 2H(Z).$ However, it is worth noting that this inequality is falsedoes not hold, so the factor of $K^2$ is important.

Let $X,Y$ be probability measures on $\{1,2,\dots,n\}$, and set $K=\sum_i\sqrt{X(i)Y(i)}$ so that $Z:=\frac{1}{K}\sqrt{XY}$ is also a probability measure on $\{1,2,\dots,n\}$. How can we prove the inequality $$H(X)+H(Y)\geq 2K^2 H(Z),$$ where $H(X)=-\sum_{i=1}^n X(i)\log X(i)$ is the Entropy function.

The problem originates from this math stack exchange post, and cardinal's rewording of it in the comments. Despite having being asked over two years ago, with numerous bounties posted, the problem was never solved, and for that reason I am posting it here.

I checked the inequality numerically on matlab for millions of choices, with $n$ up to size $100$, and it always held, which suggests that finding a counter example is unlikely.

Remarks: By Cauchy Schwarz, $1\geq K^2,$ so the above inequality would be implied by $H(X)+H(Y)\geq 2H(Z).$ However, this inequality is false, so the factor of $K^2$ is important.

Let $X,Y$ be probability measures on $\{1,2,\dots,n\}$, and set $K=\sum_i\sqrt{X(i)Y(i)}$ so that $Z:=\frac{1}{K}\sqrt{XY}$ is also a probability measure on $\{1,2,\dots,n\}$. How can we prove the inequality $$H(X)+H(Y)\geq 2K^2 H(Z),$$ where $H(X)=-\sum_{i=1}^n X(i)\log X(i)$ is the Entropy function.

The problem originates from this math stack exchange post, and cardinal's rewording of it in the comments. Despite having being asked over two years ago, with numerous bounties posted, the problem was never solved, and for that reason I am posting it here.

I checked the inequality numerically on matlab for millions of choices of $X$ and $Y$, with $n$ up to size $100$, and it always held, which suggests that finding a counter example is unlikely.

Remark: By Cauchy Schwarz, $1\geq K^2,$ so the above inequality would be implied by $H(X)+H(Y)\geq 2H(Z).$ However it is worth noting that this inequality does not hold, so the factor of $K^2$ is important.

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Eric Naslund
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Eric Naslund
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  • 106
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