Mettre exo 0 sur fiche car c'est mieux

Exercice 1
1/
  conditionnelle, de P(Y|X)
2/
P(Y|X)

x\y|1  |0  |-1 |
---+---+---+---+
0  |0.2|0.7|0.1|
---+---+---+---+
1  |0.6|0.3|0.1|
---+---+---+---+

(0.7 0.2 0.1
 0.3 0.6 0.1)


3/
P(X,Y)

x\y|1  |0  |-1 |P_x|
---+---+---+---+---+
0  | .1|.35|.05| .5|
---+---+---+---+---+
1  | .3|.15|.05| .5|
---+---+---+---+---+
P_y| .4| .5| .1|
---+---+---+---+

P(X|Y)

x\y|1  |0  |-1 |
---+---+---+---+
0  |.25|0.7|0.5|
---+---+---+---+
1  |.75|0.3|0.5|
---+---+---+---+

distribution marginale P_x et P_y (voir P(X,Y) (dernière ligne, derniere colonne))

4/

H(X)= -\sigma_(i=1)^(n=2) p(x_i)log2(p(x_i))=log2(2)= 1 bit/symbole
H(Y)= -\sigma_(i=1)^(n=3) p(y_i)log2(p(y_i))=-0.5(log2(0.5)-0.4log2(0.4)-0.log2(0.1)) =~ 1.36 bits/symbole

H(X,Y)= -\sigma_(i=1)^(n=2)\sigma_(i=1)^(n=3) p(xi,yj)log2(p(xi,yj)) =~ 2.22 bits/symbole

H(X|Y)= -\sigma_(i=1)^(n=2)\sigma_(i=1)^(n=3) p(xi,yj)log2[p(xi,yj)/p(yj)]  =~ 1.22bits/symbole
                                                          \_____________/
                                                              p(xi|yj)

5/ I(X,Y)= H(Y)-H(Y|X)
         = H(X)-H(X|Y)
         =~ 0.14 bits/symbole

EXERCICE 2
1/ canal pas symétrique car :
  les lignes de la matrice de transition p(X,Y) sont toutes équivalentes par permutation mais pas les colones

2/
x\y|1  |0  |-1 |
---+---+---+---+
0  |  0|0.2|0.8|
---+---+---+---+
1  |0.7|  0|0.2|
---+---+---+---+


x\y|1  |0  |-1 |P_x|
---+---+---+---+---+
0  |0  |   |   |   |
---+---+---+---+---+
1  |   |   |   |   |
---+---+---+---+---+
P_y|   |   |   |
---+---+---+---+
matrice pour simplifier

(
0         0.8p   0.2p       p
0.8(1-p)  0      0.2(1-p)  (1-p)
0.8(1-p)  .8p    0.2
)

/3

Hx(P)= -\sigma_(i=1)^2 p(xi)log2[p(xi)] = -p*log2(p)-(1-p)log(1-p)

4/
....