Probability and Entropy

According to the Wikipedia article on conditional entropy, $\sum p(x,y)\log p(x)=\sum p(x)\log p(x)$. Can someone please explain how?

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$$\sum_{x,y} p(x,y)\log p(x)=\sum_x\sum_yp(x,y)\log p(x)=\sum_x\left(\sum_yp(x,y)\right)\log p(x)=\sum_x p(x)\log p(x)$$

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@joriki- I still dint get it. Can you please give a little more detail? Might be I'm missing something very basic here.. some concept of probability –  user23844 Jan 27 '12 at 19:00
@Buxme In the step $\sum \limits_{y} p(x,y) = p(x)$, you could think of the variable $y$ as being marginalised out. –  Srivatsan Jan 27 '12 at 19:02
@Buxme: Which of the three steps are you having trouble with? –  joriki Jan 27 '12 at 19:42
@joriki- How did y get away? The step Srivatsan mentioned in the second comment.. –  user23844 Jan 27 '12 at 19:49
@Buxme: That step is essentially the law of total probability. The wikipedia page contains a detailed explanation. Read it and get back to us if you still don't understand. –  Srivatsan Jan 27 '12 at 20:54

I find the notation irritating, whereby the same letter, $p$, is used to refer to two or more different functions.

I could write an algebraic proof, but I wonder if a concrete example might not shed more light. Suppose we have \begin{align} P(X=0\ \&\ Y=0) = 1/10 & & P(X=0\ \&\ Y=1) = 2/10 \\ \\ P(X=1\ \&\ Y=0) = 3/10 & & P(X=1\ \&\ Y=1) = 4/10 \end{align} Then $P(X=0)=4/10$ and $P(X=1)= 6/10$. So the first sum above is $$\underbrace{(1/10)\log (4/10) + (3/10)\log(4/10)} + \underbrace{(2/10)\log(6/10) + (4/10)\log(6/10) }.$$ The second sum above is $$(4/10)\log(4/10) + (6/10)\log(6/10).$$ So it's really just the distributive law.

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In the article $p(x,y)$ is summed (integrated) over all possible values of y for a fixed value of x.

$\sum_{y\in\mathcal Y} p(x,y)= p(x)$

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