Can someone help me on this?
X = {$X_1,X_2,X_3,X_4$}
Y = {$Y_1,Y_2,Y_3,Y_4$}
Suppose p($X_i$) = p($Y_j$) = 1/4 (each X and each Y equally likely)
$1 \leq i, 4 \ge j$
and now suppose
$Y_1 : X_1 or X_2$
$Y_2 : X_3 or X_4$
$Y_3 : X_2 or X_3$
$Y_4 : X_1 or X_4$
what is $H(X|Y)$?
I tried to create a table, like $$ \left[ \begin{array}{ccc} & y1 & y2 & y3 & y4 \\ x1 & & 0 & 0 & \\ x2 & & 0 & & 0 \\ x3 & 0 & & & 0 \\ x4 & 0 & & 0 & \end{array} \right] $$ I know p(X) = 1/2, because only two choices. And p(Y)=1/4 is same.
Thank you
$\sum_{y}\sum_{x}p(y)p(x|y)log_2(p(x|y))$
$(1/4)p log_2 p + (1/4)((1/4)-p) log_2 ((1/4)-p)$
$4* [(1/4)p log_2 p + (1/4)((1/4)-p) log_2 ((1/4)-p)]$
$p log_2 p + ((1/4)-p) log_2 ((1/4)-p)$
Let $p\mathop{:=}\mathsf P(X_1,Y_1)$. Then since each column and each row must total $\tfrac 14$, we complete your table as follows:
$$\boxed{\begin{array}{|c|c:c:c|}\hline & Y_1 & Y_2 & Y_3 & Y_4 \\ \hline X_1 & p & 0 & 0 & \tfrac 14-p \\ \hdashline X_2 & \tfrac 14-p & 0 & p & 0 \\ \hdashline X_3 & 0 & p & \tfrac 14-p & 0 \\ \hdashline X_4 & 0 & \tfrac 14-p & 0 & p \\ \hline \end{array}}$$
The information you provided is insufficient to determine what $p$ actually is.
If the choice between each of the two $X_k$ for each $Y_h$ were unbiased, then $p=\tfrac 18$. However, nothing indicates this is so.
But since there are the same two conditional probabilities for the two $X_k$ any given $Y_h$ (although which event they match differ), we can calculate the conditional entropy as:
$$\begin{align}\mathsf H(X\mid Y_h) & = - \sum_{k=1}^4 \mathsf P(X_k\mid Y_h)\log_2(\mathsf P(X_k\mid Y_h)) & \big[h\in\{1,2,3,4\}\big] \\ & = -\big( 4p \log_2 (4p) + 4(1-p)\log_2(4(1-p)) + 0\log 0 + 0\log 0\big) \\ & = -4\big( p \log_2(p) + (1-p) \log_2(1-p)+2 ) \big) \end{align}$$