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I'm new to information theory and I am struggling to understand this problem. Let $p(x,y)$ given by:

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How can we calculate $H(X|Y)$? I know $H(X|Y)=H(X|Y=0)+H(X|Y=1)$ but then I don't know how to go further.

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$${\displaystyle \mathrm {H} (Y|X)\ =-\sum _{x\in {\mathcal {X}},y\in {\mathcal {Y}}}p(x,y)\log {\frac {p(x,y)}{p(x)}}}$$

Based on chain rule, we have:

\begin{aligned}\mathrm {H} (Y|X)&=\sum _{x\in {\mathcal {X}},y\in {\mathcal {Y}}}p(x,y)\log \left({\frac {p(x)}{p(x,y)}}\right)\\[4pt]&=-\sum _{x\in {\mathcal {X}},y\in {\mathcal {Y}}}p(x,y)\log(p(x,y))+\sum _{x\in {\mathcal {X}},y\in {\mathcal {Y}}}{p(x,y)\log(p(x))}\\[4pt]&=\mathrm {H} (X,Y)+\sum _{x\in {\mathcal {X}}}p(x)\log(p(x))\\[4pt]&=\mathrm {H} (X,Y)-\mathrm {H} (X).\end{aligned}

So for $H(X|Y)$ in your sample, we have: $$ H(X|Y) = -( 0.25 \times \log(\frac{1}{2}) ) - ( 0.25 \times \log(\frac{1}{2}) ) -( 0 \times \log(\frac{0}{0.5}) ) -( 0.5 \times \log(\frac{0.5}{0.5}) ) $$

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  • $\begingroup$ I know the formula for conditional entropy but I wanted to calculate it the other way. Or let's say we want to calculate H(X|Y=0). For this, what would be probability p(x=0|y=0)? $\endgroup$ – Niousha Oct 26 '19 at 11:18
  • $\begingroup$ p(X=0 | Y=0) = P(X=0,Y=0) / P(Y=0) = (0.25)/(0.25) = 1 $\endgroup$ – aminrd Oct 26 '19 at 19:47

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