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Actually I would disagree with Nick or Ami. Although I am not familiar with the chain rules of entropy, we can use a simple counter example. Suppose $X\sim N(0,\sigma^2)\in \mathbb{R}$. Then its entropy is given by (derivation given here) $$ H=\frac{1}{2}\log(2\pi)+\log(\sigma)+\frac{1}{2} $$ The entropy is dependent on $\sigma$, which can be modified by scaling. e.g. $Y=2X$ implies $\sigma_Y=2\sigma_X$, which yields $H(Y)-H(X)=\log 2\neq 0$

Actually I would disagree with Nick or Ami. Although I am not familiar with the chain rules of entropy, we can use a simple counter example. Suppose $X\sim N(0,\sigma^2)\in \mathbb{R}$. Then its entropy is given by (derivation given here) $$ H=\frac{1}{2}\log(2\pi)+\log(\sigma)+\frac{1}{2} $$ The entropy is dependent on $\sigma$, which can be modified by scaling. e.g. $Y=2X$ implies $\sigma_Y=2\sigma_X$, which yields $H(Y)-H(X)=\log 2\neq 0$

Edit: I realised that depending on whether X is discrete or continuous, the answer could be different. The first link by R Carnell illustrates this difference.

Actually I would disagree with Nick or Ami. Although I am not familiar with the chain rules of entropy, we can use a simple counter example. Suppose $X\sim N(0,\sigma^2)\in \mathbb{R}$. Then its entropy is given by (derivation: https://gregorygundersen.com/blog/2020/09/01/gaussian-entropy/#:~:text=Entropy%20of%20the%20univariate%20Gaussian.&text=Step%20%E2%8B%86%20holds%20because%20E,average%20amount%20of%20surprise%20increases) $$ H=\frac{1}{2}\log(2\pi)+\log(\sigma)+\frac{1}{2} $$ The entropy is dependent on $\sigma$, which can be modified by scaling. e.g. $Y=2X$ implies $\sigma_Y=2\sigma_X$, which yields $H(Y)-H(X)=\log 2\neq 0$

Actually I would disagree with Nick or Ami. Although I am not familiar with the chain rules of entropy, we can use a simple counter example. Suppose $X\sim N(0,\sigma^2)\in \mathbb{R}$. Then its entropy is given by (derivation given here) $$ H=\frac{1}{2}\log(2\pi)+\log(\sigma)+\frac{1}{2} $$ The entropy is dependent on $\sigma$, which can be modified by scaling. e.g. $Y=2X$ implies $\sigma_Y=2\sigma_X$, which yields $H(Y)-H(X)=\log 2\neq 0$

Actually I would disagree with Nick or Ami. Although I am not familiar with the chain rules of entropy, we can use a simple counter example. Suppose $X\sim N(0,\sigma^2)\in \mathbb{R}$. Then its entropy is given by (derivation: https://gregorygundersen.com/blog/2020/09/01/gaussian-entropy/#:~:text=Entropy%20of%20the%20univariate%20Gaussian.&text=Step%20%E2%8B%86%20holds%20because%20E,average%20amount%20of%20surprise%20increases) $$ H=\frac{1}{2}\log(2\pi)+\log(\sigma)+\frac{1}{2} $$ The entropy is dependent on $\sigma$, which can be modified by scaling. e.g. $Y=2X$ implies $\sigma_Y=2\sigma_X$, which yields $H(Y)-H(X)=\log 2$

Actually I would disagree with Nick or Ami. Although I am not familiar with the chain rules of entropy, we can use a simple counter example. Suppose $X\sim N(0,\sigma^2)\in \mathbb{R}$. Then its entropy is given by (derivation: https://gregorygundersen.com/blog/2020/09/01/gaussian-entropy/#:~:text=Entropy%20of%20the%20univariate%20Gaussian.&text=Step%20%E2%8B%86%20holds%20because%20E,average%20amount%20of%20surprise%20increases) $$ H=\frac{1}{2}\log(2\pi)+\log(\sigma)+\frac{1}{2} $$ The entropy is dependent on $\sigma$, which can be modified by scaling. e.g. $Y=2X$ implies $\sigma_Y=2\sigma_X$, which yields $H(Y)-H(X)=\log 2\neq 0$