Download Solution Series 6 - D-MATH

Probability and statistics
April 19, 2016
Lecturer: Prof. Dr. Mete SONER
Coordinator: Yilin WANG
Solution Series 6
Q1. Suppose that internet users access a particular Web site according to a Poisson process with
rate Λ per hour, but Λ is unknown. The Web site maintainer believes that Λ has a continuous
distribution with probability density function:
(
2e−2λ for λ > 0
f (λ) =
.
0
otherwise
Let X be the number of users who access the Web site during a one-hour period. If X = 1
is observed, find the conditional p.d.f. of Λ given X = 1.
Solution:
We compute first the CDF of λ given X = 1.
By hypothesis,
P(X = 1 | Λ) = E(1X=1 | Λ) = e−Λ Λ.
Hence
Z
∞
e−λ λf (λ)dλ
P(X = 1) = E(1X=1 ) = E[E(1X=1 | Λ)] =
0
Z ∞
=
2e−λ λe−2λ dλ
Z0 ∞
=
2λe−3λ dλ = 2/9.
0
By definition of the conditional expectation, and that for any x > 0, 1Λ≤λ is σ(Λ)-measurable,
hence
P(X = 1, Λ ≤ λ) = E[1X=1 1Λ≤λ ] = E[E(1X=1 | Λ)1Λ≤λ ] = E[e−Λ Λ1Λ≤λ ],
and the CDF of λ given X = 1 is
Rλ
2se−3s ds
P(X = 1, Λ ≤ λ)
0
=
.
P(Λ ≤ λ | X = 1) =
P(X = 1)
P (X = 1)
By differentiate the CDF w.r.t. λ, we obtain the PDF:
f (λ | X = 1) = 9λe−3λ .
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Probability and statistics
April 19, 2016
Q2. Let X be a real-valued random variable. We define the characteristic function of X by
ϕX : R → C
t 7→ ϕX (t) := E[e
itX
Z
]=
eitx µ(dx),
in which µ is the distribution of X on R. It represents an important analytic tool, that the
distribution of a random variable is uniquely determined (characterized) by the characteristic
function. Show the following features:
(a)
•
•
•
•
ϕX (0) = 1,
|ϕX (t)| ≤ 1,
ϕX is continuous, and
ϕaX+b (t) = eitb ϕX (at) for all a, b ∈ R.
(b) Show that if the n-th moment of X exists, i.e. E[|X|n ] < ∞, then ϕX is n times
differentiable, and
(k)
ϕX (t) = ik E[X k eitX ], for all k ≤ n,
(k)
(in particular ϕX (0) = ik E[X k ]).
Hint: one can use the inequality | e
iα −1
α
| ≤ 1, (α ∈ R).
(c) Compute the characteristic function for the standard normal distribution N(0, 1), then
for N(µ, σ 2 ).
(d) Let X and Y be two independent random variables, defined on the same probability
space. What is the characteristic function of X + Y ?
Solution
(a)
• ϕX (0) = E[1] = 1.
R
R
• ϕX (t) = eitx µ(dx) ≤ |eitx |µ(dx) = 1, since µ is a probability measure on R.
• This follows from the classical dominated convergence theorem and that eitx is
bounded.
• ϕaX+b (t) = E[eit(aX+b) ] = eitb E[eitaX ] = eitb ϕX (at).
(b) We can prove it by induction. The case k = 0 is by definition of ϕX . Assume that for
k < n the proposition is true,
(k)
(k)
eihX − 1
ϕX (t + h) − ϕX (t)
= ik E[X k eitX X
],
h
hX
ihX
since | e hX−1 | ≤ 1 for any X and goes to i as h → 0. By dominated convergence, using
the fact that X k+1 is integrable, the above expression tends to ik+1 E[X k+1 eitX ].
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Probability and statistics
April 19, 2016
(c) Let X ∼ N(0, 1),
Z
+∞
1
2
eitx √ e−x /2 dx
2π
−∞
Z +∞
1
2
√ e−(x −2itx)/2 dx
=
2π
−∞
Z +∞
1
2
2
√ e−(x−it) /2 e−t /2 dx
=
2π
−∞
−t2 /2
=e
.
ϕX (t) =
If Y ∼ N(µ, σ 2 ), then (Y − µ)/σ ∼ N(0, 1). Thus
ϕY (t) = eitµ ϕX (σt) = eitµ−σ
2 t2 /2
.
(d) As X and Y are independent, for every f, g : R → R, f (X) and g(Y ) are also independent random variables. Thus
ϕX+Y (t) = E[eit(X+Y ) ] = E[eitX eitY ] = E[eitX ]E[eitY ] = ϕX (t)ϕY (t).
Let X1 , X2 , · · · , Xn be random variables defined on the same probability space. (X1 , X2 , · · · , Xn )
is said to be a Gaussian vector if all R-linear combinations of Xi are centered Gaussian random
variable.
Q3. Let X and Y two independent standard normal random variables (N(0, 1)). Define the
random variable
X if Y ≥ 0,
Z :=
−X if Y < 0.
(a) Compute the distribution of Z.
(b) Compute the correlation between X and Z.
(c) Compute P(X + Z = 0).
(d) Does (X, Z) follow a multivariate normal distribution (in other words, is (X, Z) a Gaussian vector)?
Solution:
(a) We just have to compute:
P(Z ≥ t) = P({X ≥ t, Y > 0} ∪ {X ≤ −t, Y ≤ 0})
1
1
= P(X ≥ t) + P(X ≤ −t)
2
2
= P(X ≥ t).
So Z has the same law as X, thus Z ∼ N(0, 1).
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Probability and statistics
April 19, 2016
(b) Using the definition of covariance
Cov(X, Z) = E (XZ) − E(X)E(Z)
= E X 2 1{y≥0} + E −X 2 1{y<0} = 0.
(c) We have that
1
P(X + Z = 0) = P(Y < 0) + P(Y ≥ 0, 2X = 0) = .
2
(d) It’s not a multivariate normal, because the sum of them is not a normal.
Q4. Assume that X := (X1 , X2 , · · · , Xn ) is a Gaussian vector, KX the covariance matrix of X,
which is defined by
KX (i, j) = Cov(Xi , Xj ).
P
(a) Let α1 , · · · , αn be n real numbers, what is the law of ni=1 αi Xi in terms of KX ?
(b) What can you say about KX ?
(c) If KX (1, 2) = 0, show that X1 and X2 are independent. Is that true if we don’t assume
that (X1 , X2 ) is a Gaussian vector?
Hint: The characteristic function of the pair of random variables X := (X1 , X2 ), which
is defined as
ϕX : R2 → C
t = (a, b) 7→ ϕX (t) := E[eit·X ] = E[ei(aX1 +bX2 ) ],
characterizes also the joint law of (X1 , X2 ).
Solution:
(a) As X is Gaussian vector, Y :=
Pn
αi Xi is also Gaussian. E(Y ) = 0 and
!
n
n
X
X
V ar(Y ) = Cov
α i Xi ,
αj Xj .
i=1
i=1
j=1
Cov is bilinear, hence
V ar(Y ) =
n X
n
X
αi αj KX (i, j) = αKX t α,
i=1 j=1
where α = (α1 , α2 , · · · , αn ).
(b) KX is symmetric, and as V ar is always non-negative, hence KX is symmetric positive.
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Probability and statistics
April 19, 2016
(c) First, Z := (X1 , X2 ) is Gaussian vector. The characteristic function of Z is
ϕZ (a, b) = E[ei(aX1 +bX2 ) ]
0
a
1)
(
)/2
= exp −(a, b) V ar(X
b
0
V ar(X2 )
= exp −(a2 V ar(X1 ) + b2 V ar(X2 ))/2
= exp −a2 V ar(X1 )/2 exp −b2 V ar(X2 )/2 ,
which follows from Q2.c).
The above expression is also the characteristic function of the joint law of two independent centered random gaussian variables, with variance respectively V ar(X1 ) and
V ar(X2 ). Since the characteristic function determines the joint law, X1 and X2 are independent. If we omit the hypothesis that (X1 , X2 ) is Gaussian vector, then 0 covariance
does not imply the independence, as the example given in Q3.
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