lois binomiales, continues - Institut de Mathématiques de Toulouse

a≥4f(a)x2+bx+
c= 0 b c
[−a, a]
lima→+∞f(a)
lima→0+f(a)f(a)a≤4
n p =P( )
pn
X X
pn
(x+y)n(x−y)n
P0≤k≡0(2)≤nn
kxkyn−kpn
p0= 1 n≥1pn=p(1 −pn−1) + (1 −p)pn−1
pn

p(x) = x2+bx +c b2−4c≥0
Obc c=b2/4 (b, c)
[−a, a]×[−a, a]
a
b=a(±2√a, a)
2√a≤a a ≤4
0< a ≤4
2a2+ 2a(a−2√a)+2Za
0
t2
4dt = 2a2+ 2a(a−2√a)+2a3 12dt =
f(a) = 4a2−23a3/2/6
4a2= 1 −23
24√a.
2Za
0
2√tdt =8a3/2
3.
lima→+∞f(a)=1
p(x) = x2+bx +c= 0
1
a≤4a
q(b) = b2/4 0 a2/4q
0 1/2
f(a)2a2+ 2 Ra
0t2dt/4
4a2=1 2+a 24.

n
X
n B(n, p)
pn=P(X= 0) + P(x= 2) + ··· +P(X= 2E(n/2)) =
E(n/2)
X
k=0 2k
np2k(1 −p)n−2k.
x, y ∈R
(x+y)n+ (−x+y)n=
n
X
l=0 l
nxlyn−l+
n
X
l=0 l
n(−1)lxlyn−l= 2
E(n/2)
X
k=0 2k
nx2kyn−2k.
x=p y = 1 −p
1 + (1 −2p)n= 2
E(n/2)
X
k=0 2k
np2k(1 −p)n−2k= 2pn⇒pn=1 + (1 −2p)n
2.
n≥2
pn=p(1 −pn−1) + (1 −p)pn−1=p+ (1 −2p)pn−1.
pn
Y Y