Rapport du TP1 Problème Inverse Algorithme de Viterbi

−→
X= (X1, X2, ..., XT)
XtS= (s1, s2, ..., sN)
Xt−1
P(Xt=sk|X1, X2, X3, . . . , Xt−1) = P(Xt=st|Xt−1)
=P(X2=st|X1),
skP
Xtt
−→
X Yt
Xt
−→
X
−→
X= (X1, X2, ..., Xn)−→ Y= (Y1, Y2, ..., Yn)
−→
X−→ Y
b
X=ArgMax
−→
X
P(−→ X|−→ Y).

b
X
T= 100 N= 3 2100
P(−→ X|−→ Y) = P(−→
X , −→ Y)
P(−→ Y),
P(−→ Y)−→ X
b
X=ArgMax
−→
XhP(−→
Y|−→ X)P(−→ X)i
| {z }
PY(
−→
X)
−→
Y
−→
X
P(−→ Y|−→ X) =
n
Y
t=1
P(Yt=yt|Xt=xt),
PY(−→ X) = (
n
Y
t=1
P(yt|xt))(
n
Y
t=2
P(xt|xt−1))P(x1),
PY(−→ X) = P(y1|x1)(P(x1)(
n
Y
t=2
(P(yt|xtP(xt|xt−1),
PY(−→ X) =
x1
x1, x2
qx2, x3
qx3, x4
...
xn−1, xn
PY(−→ X) = f1(x1)f2(x1, x2)f3(x2, x3)f4(x3, x4)...fn(xn−1, xn).
log PY(−→ X)
b
X

L(−→ X) = log PY(−→ X)
= log(P(y1|x1)(P(x1)) +
n
X
t=2
log(P(yt|xtP(xt|xt−1).
L(−→ X) = L1(x1) +
n
X
t=2
Lt(xt, xt−1).
L(−→ X
b
X=Arg Max
−→
X=(x1,x2,x3)
L(x1, x2, x3)