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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.
Apr 8, 2018 · This is what I have solved so far, by using theory, I can simply say that $H(X,Y)=H(X)+H(Y\mid X)$ and therefore, $H(Y\mid X) \le H(Y)$. This is true and when $X$ and $Y$ are independent, then $H(Y\mid X) = H(Y)$.
6 days ago · 三生三世十里桃花. 天地不容-胡鴻鈞 高音牧童笛. 從未知道你最好. 寵愛-TFBOYS 高音牧童笛. マクロスF ライオン (小節) Truth Is Too Adventure 真心話太冒險. Mt heart will go on (鐵達尼號) Side Angle Side. 牧童笛指法.
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.
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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.
