De la loi géométrique au problème du collectionneur

FF
1 2
2 3

6
7 4 5
2
EΩ
P X
XΩRX
N∗={1,2,3, . . .}X
P(X=k)k
lim
n→+∞
n
X
k=0
P(X=k) = lim
n→+∞P(X∈ {1, . . . , n}) = P(X∈N∗)=1.
T
A B
P(A B) = P(A∩B) = P(A)×P(B).
p= 1/2
G
G
BY:
$
\
C

ne
G= 1
P(G= 1)
G= 2
P(G= 2)
P(G=n)n≥1
G
p∈]0,1[
G
G p = 1/2
p
G
n≥1G=n
n
n≥1
P(G=n) = (1 −p)n−1p.
G
∀n≥1, P (G=n) = (1 −p)n−1×p,
p∈]0,1[ p
(P(G=n))n≥1
X n
E[X] =
n
X
k=0
k P (X=k) = P(X= 1) + 2P(X= 2) + · · · +nP (X=n).
e e
BY:
$
\
C

XN
E[X] = lim
n→+∞
n
X
k=0
k P (X=k) = P(X= 1) + 2P(X= 2) + · · · +nP (X=n) + . . .
+∞
X1X2
E[X1+X2] = E[X1] + E[X2].
G
p
E[G] = 1
p.(∗)
E[G] = 1
p
1/p
p
E[G] = 1
p
N
T
T=T0+T1+· · · +TN−1,
Tkk∈ {0,...N −1}
k
T0
T1
Tkk∈
{0, . . . , N −1}
BY:
$
\
C

E[T] = N1 + 1
2+1 3+· · · +1 N=N
N
X
k=1
1
k.
N≥2
E(T) = N1 + 1
2+... +1 N=N
N
X
k=1
1
k
E(T)
N
E(T)
8
N12
c150
c21
c2> c1
BY:
$
\
C