I'm trying to figure out the correct method to work through an entropy problem. I'm given the following probability distribution table below, and asked to find different a few different values, for example $H(X)$, $H(X|Y)$, etc.
$$\begin{array}{|c|l|l|}\hline X|Y & \mathsf{0} & \mathsf{1} \\ \hline \mathsf{0} & \frac{1}{3} & \frac{1}{3}\\ \mathsf{1} & 0 & \frac{1}{3} \\ \hline \end{array}$$ just to be clear, the probability of $X=1$, and $Y=0$ ($prob(1,0)$) is zero ($X$ is the left side, $Y$ is the top)
To find $H(X)$, I believe I find the marginal of $X$, which should just be adding up the different probabilities for each $Y$ value. which should be
$prob(X|Y=0) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$
and
$prob(X|Y=1) = 0 + \frac{1}{3} = \frac{1}{3}$.
From there I think I find $H(X)$ with
$H(X) = \sum_{i=0}^1 prob(X=i)\cdot log_2(\frac{1}{prob(X=i)}) = \frac{2}{3}\cdot log_2(\frac{3}{2})+\frac{1}{3}\cdot log_2(3)$, is that correct?
I'm a little more confused on the $H(X|Y)$. I know the definition of this should be $H(X|Y)=-\sum_{y}\sum_{x} prob(x,y)\cdot log_2(prob(y|x))$ but in another example I found it looks like this should also equal $H(X|Y)=\sum_{i} prob(Y=i)H(X|Y=i)$ which looks like it's a lot easier to me. using this method, is the following correct?
$prob(Y=0) = \frac{1}{3} + 0 = \frac{1}{3}$, and $prob(Y=1) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$,
$H(X|Y=0) = H(1) + H(0) $ where $H(1)$ comes from if $Y=0$, then $X=0$ 100% of the time. if that's true, then the entropy of 1 is 0 ($1\cdot log_2(1) = 0$)
$H(X|Y=1) = H(.5) + H(.5) $ where $H(.5)$ comes from if $Y=1$, then $X=0$ and $X=1$ are equalily likely, and happen 50% of the time. if that's true, then the entropy of .5 is .5 ($.5\cdot log_2(1/.5) = .5$) and so $H(X|Y=1) = 1$
Is that the correct way of thinking about these? is there a most systematic approach to this? Thanks for any help with understanding this!