4 - Probas Stats - TD 17 10

TD 4 - Probas Stats
Hakim ABDOUROIHAMANE
18 Octobre 2019
Exercice 1
Calculer les fonctions génératrices des lois usuelles : Bernouilli, binomiale,
géométrique, Poisson. En déduire leur moyenne et leur variance.
E(X) = G0x (1)
+∞
P
E(X) =
x × P (X = x)
x=0
+∞
P
E[f (X)] =
i × P [f (X) = i]
i=0
Exemple
P (X = 1) = P (X = 3) = 12
P (X = i) = 0 si i ∈
/ 1,3
f (1) = 6 et f (3) = 14
f (x) = 6 → X = 1
donc P (f (X) = 6) = 12
de même : P [f (X) = 14] = f (X = 3) =
E[f (x)] = 12 × 6 + 12 × 14
1
2
• X suit la loi de Bernouilli de paramètre p ↔P(X = 1) = petP(X = 0) =
(1-p)
+∞
P
Gx (s) = E(sX ) =
P (X = i) × si
x=0
Gx (s) = P (X = 0) × s0 + P (X = 1) × s1
Gx (s) = (1 − p) × 1 + p × s
Gx (s) = 1 + p(s − 1)
G0x (s) = p E(X) = G0 x(1) = p
V (X) = E(X 2 ) − E(X)2
V (X) = E[X(X − 1)] + E(X) − E(X)2
V (X) = G00X (1) + G0X (1) + G0X (1)2
V (X) = 0 + p − p2 = p(1 − p)
1
• X(n, p) ↔P(X
+∞
P
Gx (s) = E(sX ) =
P (X = i) × si
i=0
GX (s) =
+∞
P
i=0
Complément de formules
– Formule 1 : Si a ∈ R, a 6= 1 et n ∈ N
n
P
(
n+1
a2 = 1−a
1−a
i=0
– Formule 2 : Soit a ∈ R, ea =
+∞
P
=
i=0
ai
i!
– Corollaire 1 : Si a 6= 0 et |a| < 1
+∞
n
P i
P
1
a = limn→+∞
ai = 1−a
i=0
i=0
– Corollaire 2 :
n
P
ai =
i=k
ak −an+1
1−a
– Loi géométrique
Si X tilde G(p) alors X prend ses valeurs dans N*
P (X = i) = (1 − p)i−1 × p
+∞
P
GX (s) = E(sX ) =
P (X = i) × si
GX (s) = 0 × s0 +
i=0
+∞
P
P (X = i) × si
i=1
GX (s) =
+∞
P
+∞
P
i=1
i=1
(1 − p)i−1 × p × si = p ×
GX (s) = p × s
+∞
P
(1 − p)i−1 × si
[(1 − p)i−1 × si−1 ]
i=1
En posant j = i − 1, on voit que j est un indice de sommation allant
de 0 à +∞ donc :
+∞
P
GX (s) = ps ×
[(1 − p) × s]j
j=0
1
GX (s) = ps × 1−(1−p)s
ps
GX (s) = 1−(1−p)s
– Loi de Poisson : X tilde Poisson(theta)
i
P (X = i) = e−Θ × Θi!
+∞
P
GX (s) = E(S X ) =
P (X = i)si
i=0
GX (s) =
+∞
P
i=0
e−Θ ×
Θi
i!
× si
2
GX (s) = e−Θ ×
+∞
P
i=0
(Θ×s)i
i!
GX (s) = e−Θ × eΘ×s
GX (s) = eΘ×s−Θ
GX (s) = eΘ×(s−1)
G0X (1) = Θ
G00X (1) = Θ2 × eΘ×(s−1)
V (X) = Θ
Exercice 2
1. 9!
2. P (X = 0) =
3. P (X = 1) =
4. P (X = 2) =
5. P (X = 3) =
6. P (X = 4) =
7.
4
P
x=0
5×8!
70
9! = 126
4×5!
33
9! = 126
13
4×3×5×6!
= 126
9!
4×3×2×5×5!
5
= 126
9!
4!×5!
1
9! = 126
P (X = x) × x =
84
126
=
2
3
3