Probability I
FS 2011
Mid-term exam
Exercise 1
Let X be a beta random variable of parameters 1/2 and 1/2, i.e. whose density f is given
1x∈(0,1) . Let Y be an exponential random variables with parameter 1, i.e.
by f (x) = √ 1
π
x(1−x)
whose density g is given by g(y) = e−y 1y>0 . Assume that X and Y are independent.
1. Prove that the application φ from (0, 1) × (0, ∞) to (0, ∞) × (0, ∞), and given by φ(x, y) =
(xy, (1 − x)y) is a bijection, and compute its inverse function φ−1 .
2. Show that φ−1 is differentiable and prove that the absolute value of the determinant of its
Jacobian matrix at the point (u, v) ∈ (0, ∞) × (0, ∞) is equal to 1/(u + v).
3. Compute the density of the law of (XY, (1 − X)Y ) and deduce that XY and (1 − X)Y
are independent.
Exercise 2
Let X, Y, Z be three independent random variables, uniform on [0, 1].
1. Compute the expectation of the following quantities:
• X + Y + Z,
• X 2 + Y 2 + Z 2,
• X 2Y 2Z 2,
• (X + Y + Z)2 .
2. Compute the probability of the following events:
• A = {X ≤ 1/2, Y ≤ 1/3, Z ≥ 1/5},
• B = {X ≤ Y ≤ Z},
• C = {X 2 + Y 2 + Z 2 ≤ 1}.
Exercise 3
Let X1 and X2 be two independent real-valued random variables with the same distribution,
and let f be the characteristic function of X1 (resp. X2 ).
1. Compute the characteristic function of −X1 , X1 + X2 , X1 − X2 .
2. Construct a random variable whose characteristic function is g1 , where g1 (u) = |f (u)|4 .
3. Let be a random variable, independent of X1 , and such that P ( = 0) = P ( = 1) = 1/2.
Prove that the characteristic function g2 of X1 is given by
g2 (u) =
1
1 + f (u)
.
2
Exercise 4
Let X be a real-valued random variable, whose distribution is symmetric, i.e. P (X ∈ B) =
P (X ∈ −B) for all B ∈ B(R), where −B := {−x| x ∈ B}.
1. Assume that X ∈ L3 , i.e. E[|X|3 ] < ∞. Show that Cov(X, X 2 ) = 0.
2. Assume that X is discrete (i.e. there exists a countable set C ⊂ R such that X ∈ C almost
surely). Show that X and X 2 are independent, if and only if X takes at most two values
with positive probability.
Let Z = (Z1 , Z2 ) be uniform on the unit circle K1 (0), i.e. P(Z ∈ A) =
A ∈ B(R2 ), m2 denoting the Lebesgue measure on R2 .
m2 (A∩K1 (0))
m2 (K1 (0))
for all
3. Does Z have a density? If yes, write it down. If no, why not?
4. Compute the distribution of Z1 . Is it symmetric?
5. Calculate Cov(Z1 , Z2 ). Are Z1 and Z2 independent?
Exercise 5
Let (Xn )n≥1 be a sequence
of independent Bernoulli random variables, the parameter of Xn
√
√
being equal to n − n − 1. Let (Qn )n≥1 be the deterministic sequence defined as follows:
Qn = 1 if n is a perfect square (n ∈ {1, 4, 9, 16, 25, . . . }), and Qn = 0 otherwise.
1. For all integers N ≥ 1, compute the expectation of X1 + X2 + · · · + XN and show that
this expectation is equivalent to Q1 + Q2 + · · · + QN when N tends to infinity.
2. For all√n ≥ 1, compute P[Xn = X2n = 1], and prove that this probability is equivalent to
1/(4n 2) when n goes to infinity.
3. Prove that the events {Xn = X2n = 1}, for n odd, are independent.
4. Prove that almost surely, there exists infinitely many n ≥ 1 such that Xn = X2n = 1.
5. Compare the situation with the sequence (Qn )n≥1 .
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Solution
Exercise 1
1. Let (u, v) ∈ (0, ∞) × (0, ∞). For (x, y) ∈ (0, 1) × (0, ∞), one has φ(x, y) = (u, v) if and
only if xy = u, (1 − x)y = v, i.e. xy = u, y = u + v, or x = u/(u + v), y = u + v. Since the
application ψ defined on (0, ∞) × (0, ∞) by ψ(u, v) = (u/(u + v), u + v) takes its values in
(0, 1) × (0, ∞), one deduces that φ is bijective and φ−1 = ψ.
2. One checks that the jacobian matrix of φ−1 is well-defined and equals
!
v
1
(u+v)2
u
− (u+v)
1
2
at any point (u, v) ∈ (0, ∞) × (0, ∞). Its determinant is then equal to
[v/(u + v)2 ] · 1 − [−u/(u + v)2 ] · 1 = 1/(u + v).
3. The density of (XY, (1 − X)Y ) at (u, v) ∈ (0, ∞) × (0, ∞) is 1/(u + v) times the density
of (X, Y ) at φ−1 (u, v) = (u/(u + v), u + v), i.e.
1
1
e−u e−v
p
e−(u+v) = √ √ .
u + v π (u/(u + v))(v/(u + v))
πu πv
The splitting of the density as the product of a function of u and a function of v implies
that XY and (1 − X)Y are independent.
Exercise 2
By linearity of the expectation and the fact that X, Y, Z have the same law:
Z 1
E(X + Y + Z) = E(X) + E(Y ) + E(Z) = 3E(X) = 3
x dx = 3/2,
0
1
Z
E(X 2 + Y 2 + Z 2 ) = 3E(X 2 ) = 3
x2 dx = 1.
0
By independence,
2
2
2
2
2
Z
2
E(X Y Z ) = E(X )E(Y )E(Z ) =
1
2
3
x dx
= 1/27,
0
E((X + Y + Z)2 ) = E(X 2 ) + E(Y 2 ) + E(Z 2 ) + 2E(XY ) + 2E(XZ) + 2E(Y Z)
= 3E(X 2 ) + 6E(XY ) = 3E(X 2 ) + 6E(X)E(Y )
= 3(1/3) + 6(1/2)2 = 5/2,
P (X ≤ 1/2, Y ≤ 1/3, Z ≥ 1/5) = P (X ≤ 1/2)P (Y ≤ 1/3)P (Z ≥ 1/5) = (1/2)(1/3)(4/5) = 2/15.
Moreover,
Z
P (X ≤ Y ≤ Z) =
1
Z
dx
0
x
Z
dy
0
y
Z
dz =
0
Z
dx
0
P (X 2 + Y 2 + Z 2
1
x
Z
y dy =
0
1
(x2 /2)dx = 1/6.
0
Now,
≤ 1) is equal to the volume of the part of the unit sphere consisting of the
points whose coordinates are all nonnegative. Hence, it is 1/8 of the volume of the unit sphere,
and then π/6.
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Exercise 3
1. One has for u ∈ R,
ϕ−X1 (u) = E(e−iuX1 ) = E(eiuX1 ) = ϕX1 (u) = f (u),
ϕX1 +X2 (u) = ϕX1 (u)ϕX2 (u) = (f (u))2 ,
ϕX1 −X2 (u) = ϕX1 (u)ϕ−X2 (u) = f (u)f (u) = |f (u)|2 .
2. One can take the sum of two independent variables, each of them having the law of X1 −X2 .
3. For u ∈ R, and by independence of and X1 :
Z
Z Z
1
1
iuxy
iuX1
e
P (dx)PX1 (dy) =
ϕX1 (u) = E(e
)=
(1+eiuy )PX1 (dy) = (1+ϕX1 (u)).
2
R 2
R R
Exercise 4
1. Since X is symmetric, X and −X are equal in law. Moreover, X ∈ L3 , and then E(X),
E(X 2 ) and E(X 3 ) are well-defined. One has
E(X) = E(−X) = −E(X)
and
E(X 3 ) = E((−X)3 ) = −E(X 3 ),
which implies that E(X) = E(X 3 ) = 0. Hence
Cov(X, X 2 ) = E(X 3 ) − E(X) · E(X 2 ) = 0 − 0E · (X 2 ) = 0.
2. If X takes only one value, it is equal to zero almost surely, which trivially implies that X
and X 2 are independent. Similarly, if X takes only two values, we must have P (X = x) =
P (X = −x) = 1/2 for some x > 0. But then P (X 2 = x2 ) = 1 almost surely, which implies
that X and X 2 are independent.
Now assume X takes more than two values with positive probability. Then there exists
x > 0 and p ∈ (0, 1/2) with P (X = x) = P (X = −x) = p and P (X 2 = x2 ) = 2p. But
then the events {X = x}, {X 2 = x2 } are not independent:
P (X = x, X 2 = x2 ) = P (X = x) = p 6= 2p2 = P (X = x) · P (X 2 = x2 ).
3. The density of Z is given by the formula:
fZ (z1 , z2 ) =
1
1 2 2 .
π {z1 +z2 ≤1}
4. The distribution function of Z1 is given by:
Z z1 √
1
2 1 − x2
dydx =
dx
√
π
−1 − 1−x2 π
−1
q
z1
xp
1
z1
1
1
1 − x2 + arcsin x =
1 − z12 + arcsin z1 + .
π
π
π
π
2
−1
Z
FZ1 (z1 ) =
=
z1
√
Z
1−x2
Since Z1 admits a density fZ1 such that
√
2 1 − z2
fZ1 (z) =
1(−1,1) (z) = fZ1 (−z),
π
PZ1 is symmetric.
4
5. Since Z1 and Z2 are symmetric, E(Z1 ) = E(Z2 ) = 0 by 1. Thus,
Z
1
Cov(Z1 , Z2 ) =
−1
√
Z
1−x2
√
− 1−x2
xy
dydx =
π
Z
1
0dx = 0.
−1
Since fZ1 (z1 ) · fZ2 (z2 ) 6= fZ (z1 , z2 ), Z1 and Z2 are not independent.
Exercise 5
1. One has
√
√
√
√
√
√
√
E(X1 + · · · + XN ) = ( 1 − 0) + ( 2 − 1) + · · · + ( N − N − 1) = N ,
Since Q√
1 + · · · + QN is the number of perfect squares between 1 and N , i.e. the integer
part of N , the two quantities computed are equivalent for N going to infinity.
2. The events {Xn = 1} and {X2n = 1} are independent, hence
√
√
√
√
P (Xn = X2n = 1) = P (Xn = 1)P (X2n = 1) = ( n − n − 1)( 2n − 2n − 1)
1
1
1
√
√
= √
∼
= √ .
√
√
√
n→∞
( n + n − 1)( 2n + 2n − 1)
(2 n)(2 2n)
4n 2
3. The indices involved in the different events {Xn = X2n = 1}, for n odd, do not overlap
(since n is always odd and 2n always even). Hence, these events are independent, since
(Xn )n≥1 are independent.
4. The sum of the probabilities of the events given in 3) is divergent by 2), and then by BorelCantelli’s lemma, there exists almost surely infinitely many odd n such that Xn = X2n = 1.
√
5. Since 2 is irrational, there does not exist a strictly positive perfect square which is twice
another perfect square. Hence, contrarily to the case of the random sequence (Xn )n≥1 ,
one cannot have n ≥ 1 such that Qn = Q2n = 1.
Comment: Because of 1), the sequence (Xn )n≥1 can be viewed as some random analog of
the deterministic sequence (Qn )n≥1 . The question 5) shows the limits of this point of view.
Note that if (Qn )n≥1 is replaced by (Q0n )n≥1 , where Q0n = 1 if n − 1 is a perfect square (n ∈
{1, 2, 5, 10, 17, 26, . . . }), and Q0n = 0 otherwise, then Q0n = Q02n = 1 for infinitely many values of
n, the first ones being n = 1, 5, 145, 4901.
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