Lois de probabilités continues, cours, terminale S - MathsFG

I[a;b]a b
a < b [a; +∞[a
•RIf(t)dt =Rb
af(x)dt
•RIf(t)dt = limb→+∞Rx
af(t)dt
•I f
ZI
f(t)dt = 1
•X I
f I J
P(X∈J) = ZJ f(t)dt
I
f f(x) = 1
4[0; 4]
f[0; 4] R4
0f(t)dt =1
4×4=1
f X P (X≤x) = Rx
0f(t)dt =x
4
X
f I
•P(X∈I) = 1
•x P (X=x) = Rx
xf(t)dt = 0
•P(X∈[a;b]) = P(X∈]a;b]) = P(X∈[a;b[) = P(X∈]a;b[)
P(X≤a)) = P(X < a)a b a < b

P(a≤X≤b) = P(X∈[a;b])
X
f J E(X)
E(X) = ZJ tf(t)dt
X E(X) = R4
0
t
4dt =1 2×1×4 = 2
[a;b]a b [a;b]
f
f(x) = 1
b−a
x∈[a;b]
f≥0Rb
af(t)dt = 1
X[a;b]
E(X)X
E(X) = Zb
a
tf(t)dt =a+b
2
[0; 1]

λ
λ f
f(t) = λe−λt
t∈[0; +∞[
f≥0 limb→+∞Rb
0f(t)dt = 1
X
λ b
•P(X≤b) = P(X < b)=1−e−λb
•P(X≥b) = P(X > b) = e−λb
P(X≤b) = P(0 ≤X≤b) = Rb
0λe−λtdt =λ[−1
λe−λt]b
0=−λ
λ(e−λb −e0) = 1 −e−λb
X
λ t h
PX≥t(X≥t+h) = P(X≥h)
PX≥t(X≥t+h) = P((X≥t+h)∩(X≥t))
P(x≥t)=P(X≥t+h) P(X≥h)=1−P(X<t+h) 1−P(X<t)
PX≥t(X≥t+h) = 1−(1−e−λ(t+h))
1−(1−e−λt)=e−λ(t+h) e−λt =e−λte−λh e−λt =e−λh = 1 −P(X < h) = P(X≥h)
X
λ
E(X) = 1
λ

J
g g(t) = λte−λt t≥0
G g G(t)=(at +b)e−λt
G0(t) = ae−λt −(at +b)λe−λt = (a−λb −at)e−λt
G0(t) = g(t)a−λb = 0 −a=λ
a=−λ b =−1
λ
G(t) = (−t−1
λ)e−λt
Rb
0g(t)dt =G(b)−G(0)
G(0) = −1
λG(b) = −be−λb −1 λe−λb
limb→+∞be−λb = 0 limb→+∞e−λb = 0 limb→+∞G(b)−G(0) = 1
λ