Proof $I(X;X|Z)= H(X|Z)$
$begingroup$
$I(X;X|Z)= H(X|Z)$
Proof
$I(X;X|Z) = sum_{x,z}p(x,z) log frac{p(x,x|z)}{p(x|z)p(x|z)}$
since $p(x,x|z)= p(x|z)$
Then
$I(X;X|Z) = sum_{x,z} p(x,z) log frac{1}{p(x|z)}$
$= - sum_{x,z} p(x,z) log p(x|z)= H(X|Z)$
Can anyone please help to check if my answer is correct or not.
Thanks
probability probability-theory information-theory
asked Jan 13 at 7:14
Salwa MostafaSalwa Mostafa
274
$endgroup$
add a comment |
$begingroup$
$I(X;X|Z)= H(X|Z)$
Proof
$I(X;X|Z) = sum_{x,z}p(x,z) log frac{p(x,x|z)}{p(x|z)p(x|z)}$
since $p(x,x|z)= p(x|z)$
Then
$I(X;X|Z) = sum_{x,z} p(x,z) log frac{1}{p(x|z)}$
$= - sum_{x,z} p(x,z) log p(x|z)= H(X|Z)$
Can anyone please help to check if my answer is correct or not.
Thanks
probability probability-theory information-theory
asked Jan 13 at 7:14
Salwa MostafaSalwa Mostafa
274
$endgroup$
add a comment |
$begingroup$
$I(X;X|Z)= H(X|Z)$
Proof
$I(X;X|Z) = sum_{x,z}p(x,z) log frac{p(x,x|z)}{p(x|z)p(x|z)}$
since $p(x,x|z)= p(x|z)$
Then
$I(X;X|Z) = sum_{x,z} p(x,z) log frac{1}{p(x|z)}$
$= - sum_{x,z} p(x,z) log p(x|z)= H(X|Z)$
Can anyone please help to check if my answer is correct or not.
Thanks
probability probability-theory information-theory
asked Jan 13 at 7:14
Salwa MostafaSalwa Mostafa
274
$endgroup$
$I(X;X|Z)= H(X|Z)$
Proof
$I(X;X|Z) = sum_{x,z}p(x,z) log frac{p(x,x|z)}{p(x|z)p(x|z)}$
since $p(x,x|z)= p(x|z)$
Then
$I(X;X|Z) = sum_{x,z} p(x,z) log frac{1}{p(x|z)}$
$= - sum_{x,z} p(x,z) log p(x|z)= H(X|Z)$
Can anyone please help to check if my answer is correct or not.
Thanks
probability probability-theory information-theory
probability probability-theory information-theory
asked Jan 13 at 7:14
Salwa MostafaSalwa Mostafa
274
asked Jan 13 at 7:14
Salwa MostafaSalwa Mostafa
274
asked Jan 13 at 7:14
Salwa MostafaSalwa Mostafa
274
asked Jan 13 at 7:14
Salwa MostafaSalwa Mostafa
274
asked Jan 13 at 7:14
Salwa MostafaSalwa Mostafa
274
274
add a comment |
add a comment |
1 Answer
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$begingroup$
Looks good to me. An algebraic alternative proof :
The mutual information is a difference of entropy : $I(X;X|Z)=H(X|Z)-H(X|X,Z)$, the second term is $0$ since given $X$, $X$ is deterministic.
answered Jan 13 at 7:27
P. QuintonP. Quinton
1,9061213
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Looks good to me. An algebraic alternative proof :
The mutual information is a difference of entropy : $I(X;X|Z)=H(X|Z)-H(X|X,Z)$, the second term is $0$ since given $X$, $X$ is deterministic.
answered Jan 13 at 7:27
P. QuintonP. Quinton
1,9061213
$endgroup$
add a comment |
$begingroup$
Looks good to me. An algebraic alternative proof :
The mutual information is a difference of entropy : $I(X;X|Z)=H(X|Z)-H(X|X,Z)$, the second term is $0$ since given $X$, $X$ is deterministic.
answered Jan 13 at 7:27
P. QuintonP. Quinton
1,9061213
$endgroup$
add a comment |
$begingroup$
Looks good to me. An algebraic alternative proof :
The mutual information is a difference of entropy : $I(X;X|Z)=H(X|Z)-H(X|X,Z)$, the second term is $0$ since given $X$, $X$ is deterministic.
answered Jan 13 at 7:27
P. QuintonP. Quinton
1,9061213
$endgroup$
Looks good to me. An algebraic alternative proof :
The mutual information is a difference of entropy : $I(X;X|Z)=H(X|Z)-H(X|X,Z)$, the second term is $0$ since given $X$, $X$ is deterministic.
answered Jan 13 at 7:27
P. QuintonP. Quinton
1,9061213
answered Jan 13 at 7:27
P. QuintonP. Quinton
1,9061213
answered Jan 13 at 7:27
P. QuintonP. Quinton
1,9061213
answered Jan 13 at 7:27
P. QuintonP. Quinton
1,9061213
1,9061213
add a comment |
add a comment |
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