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I have to prove the chain-rule for conditional entropy. I kept getting stuck on one step, so I looked up a proof and found this: \begin{align}H(Y\mid X)&= \sum_{x\in\mathcal X, y\in\mathcal Y}p(x,y)\log \frac {p(x)} {p(x,y)}\tag{1}\\&= -\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) \tag{2}\\ &= H(X,Y) + \sum_{x \in \mathcal X} p(x)\log\,p(x)\tag{3}\\ &= H(X,Y) - H(X)\tag{4}\end{align}

This is identical to the proof I had made on my own, but it makes a step that I didn't think was allowed. Specifically, can someone explain how you are able to move from the joint probability to the marginal probability between steps 2 and 3? It seems like I'm missing something basic and obvious, but I don't see it.

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2 Answers 2

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The question is phrased as whether $$ \sum_{x\in\mathcal X, y\in\mathcal Y}p(x,y)\log\,p(x) = \sum_{x \in \mathcal X} p(x)\log\,p(x) \quad \text{?} $$

I disapprove of using the same symbol, $p$, for two different functions. If one instead writes $p_X(x)$ with capital $X$ and lower-case $x$ in the appropriate places, one can then understand such things as the difference between $p_X(4)$ and $p_Y(4)$, and the meaning of $p_X(4)$, and things like $\Pr(X\le x)$.

We have \begin{align} & \sum_{\begin{smallmatrix} x\in\mathcal X \\ y\in\mathcal Y \end{smallmatrix}} \Pr(X=x\ \&\ Y=y) \log p(x) & & \text{where $p(x)$ is some function of $x$} \\[10pt] = {} & \sum_{x\in\mathcal X} \left( \sum_{y\in\mathcal Y} \Pr(X=x\ \&\ Y=y) \log p(x) \right) & & \text{where $p(x)$ is some function of $x$.} \end{align}

The factor $\log p(x)$ within the sum $\displaystyle\sum_{y\in\mathcal Y}$ does not depend on $y$, i.e. it does not change as $y$ runs through the list of all members of $\mathcal Y$. Therefore we can pull it out, getting this: $$ \sum_{x\in\mathcal X} \left( (\log p(x)) \sum_{y\in\mathcal Y} \Pr(X=x\ \&\ Y=y) \right) $$ Now all we need to do is show that $$ \sum_{y\in\mathcal Y} \Pr(X=x\ \&\ Y=y) = \Pr(X=x). $$ For example, if $\mathcal Y=\{y_1,y_2,y_3\}$, we would need to show that $$ \Pr(X=x\ \&\ Y=y_1) + \Pr(X=x\ \&\ Y=y_2) + \Pr(X=x\ \&\ Y=y_3) = \Pr(X=x). $$ Can you do that?

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Simply because $$ \sum_{y\in\mathcal Y}p(x,y)= p(x). $$ Really, you had done all difficult work already.

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  • $\begingroup$ I realize now that I did essentially that in another proof in the assignment. Just so I'm sure, there's no problem in saying: \begin{align}&=\sum_{j=1}^{m}q_j\big(-\sum_{i=1}^{n}\mathbb{P}(P=i)\log\mathbb{P}(P=i)\big)\tag{$P\independent Q$)}\\&=\big(\sum_{j=1}^{m}q_j\big)\cdot\big(-\sum_{i=1}^{n}\mathbb{P}(P=i)\log\mathbb{P}(P=i)\big)\tag{factoring}\\&=\big(1\big)\cdot\big(-\sum_{i=1}^{n}\mathbb{P}(P=i)\log\mathbb{P}(P=i)\big)\tag{By definition, all $q_j$'s sum to 1}\end{align} $\endgroup$
    – Alex
    Feb 6, 2016 at 17:28

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