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I only know the answer for the first one. If you see the proof of data processing inequality in Cover page 34, it is easy to find out data processing inequality turns out to be equality only when the "double Markovity" is satisfies.

Let $X, Y$ and $Z$ are three random variables representing the input, output of the first channel and output of the second, respectively. Hence, we have the Markov chain $X\to Y\to Z$ and due to data processing inequality we have $I(X;Y)\geq I(X;Z)$. The equality occurs if and only if $I(X;Y|Z)=0$ which implies the Markov chain $X\to Z\to Y$. In this case $I(X;Y)=I(X;Z)$. This is why $Z$ is called sufficient statistics of $Y$ with respect to $X$. (see this post11)

I only know the answer for the first one. If you see the proof of data processing inequality in Cover page 34, it is easy to find out data processing inequality turns out to be equality only when the "double Markovity" is satisfies.

Let $X, Y$ and $Z$ are three random variables representing the input, output of the first channel and output of the second, respectively. Hence, we have the Markov chain $X\to Y\to Z$ and due to data processing inequality we have $I(X;Y)\geq I(X;Z)$. The equality occurs if and only if $I(X;Y|Z)=0$ which implies the Markov chain $X\to Z\to Y$. In this case $I(X;Y)=I(X;Z)$. This is why $Z$ is called sufficient statistics of $Y$ with respect to $X$. (see this post1)

I only know the answer for the first one. If you see the proof of data processing inequality in Cover page 34, it is easy to find out data processing inequality turns out to be equality only when the "double Markovity" is satisfies.

Let $X, Y$ and $Z$ are three random variables representing the input, output of the first channel and output of the second, respectively. Hence, we have the Markov chain $X\to Y\to Z$ and due to data processing inequality we have $I(X;Y)\geq I(X;Z)$. The equality occurs if and only if $I(X;Y|Z)=0$ which implies the Markov chain $X\to Z\to Y$. In this case $I(X;Y)=I(X;Z)$. This is why $Z$ is called sufficient statistics of $Y$ with respect to $X$. (see this post1)

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math-Student
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I only know the answer for the first one. If you see the proof of data processing inequality in Cover page 34, it is easy to find out data processing inequality turns out to be equality only when the "double Markovity" is satisfies.

Let $X, Y$ and $Z$ are three random variables representing the input, output of the first channel and output of the second, respectively. Hence, we have the Markov chain $X\to Y\to Z$ and due to data processing inequality we have $I(X;Y)\geq I(X;Z)$. The equality occurs if and only if $I(X;Y|Z)=0$ which implies the Markov chain $X\to Z\to Y$. In this case $I(X;Y)=I(X;Z)$. This is why $Z$ is called sufficient statistics of $Y$ with respect to $X$. (see this postenter link description here1)

I only know the answer for the first one. If you see the proof of data processing inequality in Cover page 34, it is easy to find out data processing inequality turns out to be equality only when the "double Markovity" is satisfies.

Let $X, Y$ and $Z$ are three random variables representing the input, output of the first channel and output of the second, respectively. Hence, we have the Markov chain $X\to Y\to Z$ and due to data processing inequality we have $I(X;Y)\geq I(X;Z)$. The equality occurs if and only if $I(X;Y|Z)=0$ which implies the Markov chain $X\to Z\to Y$. In this case $I(X;Y)=I(X;Z)$. This is why $Z$ is called sufficient statistics of $Y$ with respect to $X$. (see this postenter link description here)

I only know the answer for the first one. If you see the proof of data processing inequality in Cover page 34, it is easy to find out data processing inequality turns out to be equality only when the "double Markovity" is satisfies.

Let $X, Y$ and $Z$ are three random variables representing the input, output of the first channel and output of the second, respectively. Hence, we have the Markov chain $X\to Y\to Z$ and due to data processing inequality we have $I(X;Y)\geq I(X;Z)$. The equality occurs if and only if $I(X;Y|Z)=0$ which implies the Markov chain $X\to Z\to Y$. In this case $I(X;Y)=I(X;Z)$. This is why $Z$ is called sufficient statistics of $Y$ with respect to $X$. (see this post1)

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math-Student
math-Student
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I only know the answer for the first one. If you see the proof of data processing inequality in Cover page 34, it is easy to find out data processing inequality turns out to be equality only when the "double Markovity" is satisfies.

Let $X, Y$ and $Z$ are three random variables representing the input, output of the first channel and output of the second, respectively. Hence, we have the Markov chain $X\to Y\to Z$ and due to data processing inequality we have $I(X;Y)\geq I(X;Z)$. The equality occurs if and only if $I(X;Y|Z)=0$ which implies the Markov chain $X\to Z\to Y$. In this case $I(X;Y)=I(X;Z)$. This is why $Z$ is called sufficient statistics of $Y$ with respect to $X$. (see this postenter link description here)