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Apr 6, 2018 · If the claim $$ H(X+Y)=H(X) + H(Y\mid X) $$ were true, it would follow that $H(X)=0$, which is impossible when $X$ is non-constant.
Jul 31, 2018 · Does anyone know how to prove the $H(Y|X)=H(Y)$ when X and Y are independent? I know the proof of $H(Y|X)=\sum p(x,y)\log_2p(y|x)$,but I found that I can't prove $H(Y|X)=H(Y)$ when $X$ and $Y$ are independent from this formula below.
In information theory, the conditional entropy quantifies the amount of information needed to describe the outcome of a random variable given that the value of another random variable is known. Here, information is measured in shannons, nats, or hartleys. The entropy of conditioned on is written as .
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I'm trying to show that $y[n]=x[n]*h[n]$ turns into $Y(z) = X(z)H(z)$ in Z-domain by first applying convolution then by taking the inverse Z-transform of the $Y(z)$, stating that it's the same sequence after all.
h(x;y) = h(x) + h(y) (2.1) However, if X and Y are not independent, observing one might give some information about the other, so simply adding the informations is over counting.
Entropy of a sum. Let X and Y be random variables that take on values x1; x2; : : : ; xr and y1; y2; : : : ; ys. Let Z = X + Y . (a) Show that H(ZjX) = H(Y jX). Argue that if X; Y are independent then H(Y ) H(Z) and H(X) H(Z). Thus the addition of independent random variables adds uncertainty.
