Probability Theory

"
Sn − nµ
s − nµ
√
≤ √
nσ 2"
nσ 2
!
s − nµ
CLT
& Φ √
.
nσ 2
In this case, we have n = 196, Xi ∼ Exponential(2.5), µ =
Department of Mathematics, IST — Probability and Statistics Unit
P (Sn ≤ s)
Probability Theory
2nd. Test
1st. Semester — 2013/14
Duration: 1h30m
2014/01/25 — 9:45 AM, Room P12
P
therefore:
• Please justify your answers.
with rate 2.5 per year, and that an individual dies when exactly 196 such mistakes have
and σ 2 =
1
,
2.52
1 − Φ(2.07)
table
A certain model supposes that mistakes in cell division occur according to a Poisson process
1
2.5

1
90 − 196 × 2.5

1 − Φ %
196 × 2.51 2
&
3.0 points
!

&
P (S196 > 90)
• This test has two pages and three groups. The total of points is 20.0.
Group I — Independence, Poisson processes and expectation
=
=
1 − 0.9808
=
0.0192.
• [Exact value and another approximate value
occurred. Assuming this model is valid:
(a) find the mean and variance of the lifetime of an individual;
P (S196 > 90)
(1.5)
=
M athematica
=
• Stochastic process
f orm.
P (S196 > 90)
{N (t) : t ≥ 0} ∼ P P (λ)
=
N (t) = number of mistakes in cell division up to time t (in years)
M athematica
λ = 2.5 mistakes per year
f orm.
=
P (S196 > 90)
• R.v.
S196 = lifetime of an individual
=
FP oisson(225) (195)
!
"
195 − 225
√
Φ
225
Φ(−2)
=
=
According to Prop. 3.107,
table
S196 ∼ Gamma(196, 2.5).
0.022707;
FP oisson(2.5×90) (196 − 1)
&
• Distribution of S196
1 − Fχ22×196 (2 × 2.5 × 90)
=
CLT
= time of the occurrence of the 196th mistake in cell division
1 − FGamma(196,2.5) (90)
0.022707;
1 − Φ(2)
=
1 − 0.9772
=
0.0228.]
• Requested mean and variance
f orm. 196
2.5
f orm. 196
= 2.52
E(S196 ) =
= 78.4
V (S196 )
= 31.36
Group II — Expectation
(b) obtain an approximate value for the probability that an individual dies after her/his (1.5)
9.0 points
1. The alternating current (AC) generated by a machine has a Uniform(−2, 2) distribution.
90th anniversary.
This current — after having been converted by a half-wave rectifier1 — can be represented
• Requested probability
by the r.v. Y = max{0, X}.
P (S196 > 90)
After obtaining the c.d.f. of Y , use it to find the expectation of this r.v.
• Requested probability — approximate value
• R.v., distribution, range, p.d.f. and c.d.f.
The Lindeberg-L´evy CLT allows us to add that, when we deal with a sufficiently
X = alternating current
large number n of i.i.d. r.v. X1 , . . . , Xn , with common mean µ and common positive
X ∼ Uniform(−2, 2)
and finite variance σ 2 , the c.d.f. of the sum of these r.v. can be approximate as
follows:
1
1
A rectifier is an electrical device that converts alternating current (AC) to direct current (DC).
2
(2.0)
IRX = [−2, 2]
(
1
, −2 ≤ x ≤ 2
4
fX (x) =
0, otherwise
2. Without evaluating the expectation,
∼
show that if X
E[X ln(X)] ≥ 0.


 0,
)x
)x 1
FX (x) = P (X ≤ x) = −∞ fX (t) dt =
dt =
−2 4


1,
x+2
,
4
Uniform (0,2) then (1.0)
• R.v.
x < −2
−2 ≤ x ≤ 2
x>2
X ∼ Uniform (0,2)
• Requested lower limit
f orm. 0+2
2
d [ln(x)+x/x]
= x1 > 0, x
dx
Since E(X) =
• Another r.v., range, c.d.f.
Y = direct current = max{0, X}
= 1 and g(x) = x ln(x) is convex function (because
d2 x ln(x)
dx2
=
∈ (0, 2)), we can apply Jensen’s inequality to obtain:
E[g(X)] = E[X ln(X)]
IRY = [0, 2]
FY (y) = P (Y ≤ y)


0,



 P (Y = 0) = P (X ≤ 0) = F (0) =
X
=
y+2

P
(X
≤
y)
=
F
(y)
=
,

X
4


 1,
≥ g[E(X)]
y<0
0+2
4
= 12 , y = 0
0<y≤2
y>2
[We are dealing with a non negative mixed r.v.!]
• Requested expected value
. +∞
Y ≥0, T heo. 4.65
E(Y )
=
[1 − FY (y)] dy
"
.0 2 !
y+2
=
1−
dy
4
0
. 2
2−y
=
dy
4
0
!
"/
2
y y 2 //
=
−
2
8 /0
1
=
.
2
• [Requested expected value — without using the c.d.f. of Y
Since Y = max{0, X} is a (non negative) Borel measurable function and X is an
absolutely continuous r.v., we can invoke Corollary 4.81 and get:
E(Y ) = E[max{0, X}]
. 2
1
=
max{0, x} × dx
4
−2
. 0
. 2
1
1
=
0 × dx +
x × dx
4
4
−2
0
/
2 /2
x
= 0 + //
8 0
1
.]
=
2
3
= g(1) = 1 × ln(1)
= 0.
3. Let X and Y be two r.v. in L2 such that E(X) = E(Y ) = 0, V (X) = V (Y ) = 1 and (1.5)
corr(X, Y ) = ρ ∈ (0, 1). Now define
Xθ = X cos(θ) − Y sin(θ)
Yθ = X sin(θ) + Y cos(θ),
where θ is a constant in [0, 2 π].
Determine all the values of θ for which Xθ and Yθ are uncorrelated.
• Randomvector
 E(X) = E(Y ) = 0


V (X) = V (Y ) = 1
(X, Y ) :


 corr(X, Y ) = √ cov(X,Y )
V (X)×V (Y )
= cov(X, Y ) = ρ ∈ (0, 1)
• Determining θ
Capitalizing on the properties of this random vector and taking advantage of the fact
that the covariance is a bilinear operator (Proposition 4.159), we get
θ ∈ [0, 2 π] : corr(Xθ , Yθ ) = 0
cov (X cos(θ) − Y sin(θ), X sin(θ) + Y cos(θ)) = 0
sin(θ) cos(θ) V (X) + cos2 (θ) cov(X, Y ) − sin2 (θ) cov(X, Y )
0 2
1
cos (θ) − sin2 (θ) × ρ = 0
tan(θ) = ±1
π 3π 5π 7π
θ= , , , .
4 4 4 4
4
− sin(θ) cos(θ) V (Y ) = 0
(2.5)
4. (a) Let:
If these two chess players played 12 games, what is the probability that Player A would
win 7 games, given that 3 games ended in draw?
• N = (N1 , . . . , Nd ) ∼ Multinomiald−1 (n, p);
• Random vector
• N −j = (N1 , . . . , Nj−1 , Nj+1 , . . . , Nd ).
N = (N1 , N2 , N3 )
Prove that
N1 = number of times player A wins in n games
N −j |Nj = nj ∼ Multinomiald−2 (n − nj , p−j|j ),
N2 = number of times player B wins in n games
N3 = number of draws in n games
!
"
where
p1
pj−1 pj+1
pd−1
pd
,...,
,
,...,
,
.
p−j|j =
1 − pj
1 − pj 1 − pj
1 − pj 1 − pj
• Distribution
N ∼ Multinomiald−1 (n, p)
• Random vector/distribution
• Parameters
N = (N1 , . . . , Nd ) ∼ Multinomiald−1 (n, p)
• Joint p.f.
P (N = n)
d=3
n = 12
p = (p1 , p2 , p3 ) = (0.40, 0.35, 0.25)
=
P (N1 = n1 , . . . , Nd = nd )
d
8
n!
Def. 4.208
=
×
pni i ,
2d
i=1 ni !
i=1
3
for ni ∈ {0, . . . , n}, i = 1, . . . , d, such that di=1 ni = n.
• Joint p.f.
P (N = n)
P (N1 = n1 , N2 = n2 , N3 = n3 )
12!
=
× 0.40n1 × 0.35n2 × 0.25n3 ,
23
n
!
i
i=1
3
for ni ∈ IN0 , i = 1, 2, 3 such that 3i=1 ni = 12.
• Another random vector
N −j = (N1 , . . . , Nj−1 , Nj+1 , . . . , Nd )
• Distribution and p.f. of N3
• Distribution of Nj
N3 ∼ Binomial(12, 0.25), according to Prop. 4.220.
4 5
P (N3 = n3 ) = n123 × 0.25n3 × (1 − 0.25)12−n3 , n3 = 0, 1, . . . , 12
Nj ∼ Binomial(n, pj ), according to Prop. 4.220.
• P.f. of N −j conditional to Nj = nj
P (N −j = n−j , Nj = nj )
P (N −j = n−j |Nj = nj ) =
P (Nj = nj )
P (N = n)
=
P (Nj = nj )
2
n!
Qd
× di=1 pni i
i=1 ni !
= 4 n 5 nj
pj (1 − pj )n−nj
nj
!
"ni
d
8
(n − nj )!
pi
= 2d
×
,
1 − pj
i=1, i%=j ni !
i=1, i%=j
3
for ni ∈ {0, . . . , n}, i = 1, . . . , d, i )= j, such that di=1,i%=j ni = n − nj and nj ∈
• Requested probability
N −j |Nj = nj ∼ Multinomiald−2 (n − nj , p−j|j ), where p−j|j =
7
pd−1
. . . , 1−p
, pd .
j 1−pj
6
pj−1 pj+1
p1
, . . . , 1−p
,
1−pj
j 1−pj
P (N1 = 7, N2 = 2, N3 = 3)
P (N3 = 3)
12!
×
0.407 × 0.352 × 0.253
7! 2! 3!
=
12!
× 0.253 × (1 − 0.25)12−3
3! (12−3)!
0.0248371
&
0.258104
& 0.096229
P (N1 = 7, N2 = 12 − (7 + 3)|N3 = 3) =
[It coincides with
(12 − 3)!
×
7! 2!
{0, . . . , n}.
• Conclusion
=
Def. 4.208
,
!
0.40
1 − 0.25
"7 !
0.35
1 − 0.25
"2
,
because N −3 |N3 = n3 ∼ Multinomiald−2 (n − n3 , p−3|3 ) where p−3|3 =
6
p1
, p2
1−p3 1−p3
7
.2 ]
(b) Suppose that two chess players had played numerous games and it was determined that (2.0)
the probability that Player A would win is 0.40, the probability that Player B would
win is 0.35, and the probability that the game would end in a draw is 0.25.
5
2
d
p1
This distributional result can be read as follows: ((N1 , N2 )|N3 = n3 ) = (N1 |N3 = n3 ) ∼ Binomial(n−n3 , 1−p
).
3
6
Group III — Convergence of sequences of r.v.
8.0 points
• R.v.
i.i.d.
Xi ∼ X, i ∈ IN
P
1. (a) Let {X1 , X2 , . . . } be a strictly decreasing sequence of positive r.v. Prove that if Xn → 0 (1.5)
a.s.
X ∼ Cauchy(0, 1)
then Xn → 0.
• To prove
P
FX (x) = 1 −
a.s.
If {X1 , X2 , . . . } such that 0 > X1 ≥ X2 ≥ . . . and Xn → 0 then Xn → 0.
• Proof
1
π
• Another r.v.
Yn =
P
Since Xn → 0 means that
1
n
× X(n)
• C.d.f. of Yn
∀& > 0, lim P (|Xn − 0| > &) = 0
FYn (y) = P (Y ≤ y)
n→+∞
= P [X(n) ≤ ny]
and 0 > X1 ≥ X2 ≥ . . . implies that
= [FX (ny)]n
! ":n
9
1
1
= 1 − arctan
,y∈R
π
ny
sup |Xk − 0| = |Xn − 0| [= Xn ],
k≥n
we can add that
∀& > 0, lim P (sup |Xk − 0| > &) =
n→+∞
arctan( x1 )
• Note
lim P (|Xn − 0| > &)
n→+∞
k≥n
FYn (y) ought to belong to [0, 1], regardless of the value of n or y.
= 0,
1
a.s.
i.e., Xn → 0, according to Proposition 5.37.
the limit of FYn (y) when y ≤ 0. In fact, limn→+∞ FYn (y) = 0, for y ≤ 0.
Let us remind the reader that
• Checking the convergence in distribution
(0.5)
(b) Comment this convergence result.
• Comment on the convergence result
a.s.
Moreover:
1
limn→+∞ FYn (y) ∈ [0, 1], for any y; e− π y )∈ (0, 1), y ≤ 0. Consequently, e− π y cannot be
We have:
P
→ ⇒ →
P
a.s.
(i) for y > 0,
→ )⇒ →.
but
The result which we have just proved is an implication of restricted validity stating
9
! ":n
1
1
1 − arctan
n→+∞
π
ny
;
9
:<n
1 1
1
1
= lim 1 −
−
+
−
.
.
.
n→+∞
π ny 3(ny)3 5(ny)5
=
>n
− π1y
= lim 1 +
n→+∞
n
lim FYn (y) =
n→+∞
that if we combine the strictly decreasing and positive character of a sequence of
r.v. with the convergence in probability to zero of that same sequence, then we end
up with the almost sure convergence to zero (see Exercise 5.80).
2. The Cauchy distribution, named after Augustin Cauchy, is often used in statistics as the (2.5)
lim
1
canonical example of a pathological distribution — both its mean and its variance are
undefined.
(ii) FY (y) =
(
= e− π y ;
1
e− π y , y > 0
0,
otherwise,
is the c.d.f. of a(n absolutely continuous) r.v., say Y ;3
Let:
• {X1 , X2 , . . . } be a sequence of i.i.d. r.v. with a standard Cauchy distribution, i.e., with
c.d.f. given by FX (x) = 1 −
• Yn =
1
π
(iii) limn→+∞ FYn (y) = FY (y), for all the continuity points of the c.d.f. of Y , R+ .
arctan( x1 ), x ∈ R;
Therefore
1
n
× maxi=1,...,n Xi be the maximum of the n first terms of the sequence.
(
1
e− π y , y > 0
d
Show that Yn → Y , where FY (y) =
0,
otherwise.
Hint: It may be helpful to know that arctan(y) = y −
7
y3
3
+
y5
5
−
y7
7
+ ....
d
Yn → Y.
3
After all, this is mentioned in the text. Furthermore, FY (y) is non-decreasing, right-continuous and has the
following limits: limy→0+ FY (y) = 0, limy→+∞ FY (y) = 1.
8
◦ V (Y¯n ) =
3. Let:
⊥⊥Y¯n
¯ n − Y¯n ) X¯n=
◦ V (X
• {X1 , X2 , . . . } and {Y1 , Y2 , . . . } be two independent sequences of i.i.d. r.v. to X ∼
Poisson(λX ) and Y ∼ Poisson(λY ), respectively;
¯ n = 1 3n Xi be the mean of the first n terms of {X1 , X2 , . . . } (Y¯n is defined
• X
i=1
n
¯ n −Y¯n )−(λX −λY )
(X
q
¯
¯n
X
+ Ynn
n
d
→ Normal(0, 1).
• Sequence of r.v.
(3.0)
i.i.d.
d
→ Normal(0, 1).
i.i.d.
• Convergence of Vn
• Other r.v.
¯ n = 1 3n Xi , n ∈ IN
X
i=1
n
Y¯n =
1
n
3n
i=1
We can invoke the WLLN for i.i.d. r.v. in L2 (Theorem 5.129)4 and state the
following convergences in probability:
P
¯n →
X
λX
Yi , n ∈ IN
P
Y¯n → λY .
• Another sequence of r.v.
{Zn , n ∈ IN }, where Zn =
λY
n
(Theorem 5.186) leads to the conclusion that
¯ n − Y¯n ) − (λX − λY )
(X
%
Un =
λX
+ λnY
n
¯ n − Y¯n ) − E(X
¯ n − Y¯n )
(X
@
=
¯ n − Y¯n )
V (X
{Y1 , Y2 , . . . }, where Yi ∼ Y ∼ Poisson(λY ), i ∈ IN
⊥⊥
+
Combining the auxiliary results and the mere application of Lindeberg-L´evy CLT
{X1 , X2 , . . . }, where Xi ∼ X ∼ Poisson(λX ), i ∈ IN
⊥⊥
λX
n
• Convergence of Un
similarly).
(a) Show that
λY
n
Invoking now the closure of convergence in probability under addition (Theorem
¯ n −Y¯n )−(λX −λY )
(X
q
.
¯n
¯
X
+ Ynn
n
5.83) and under continuous mappings (Theorem 5.81), we get:
P
¯ + Y¯n →
X
λX + λ Y ;
? n
¯ n + Y¯n P
X
Vn =
→ 1.
λX + λ Y
• Asymptotic distribution of Zn
d
To prove that Zn → Normal(0, 1) we need to:
– rewrite Zn as follows,
Un
Zn =
,
Vn
where
¯ n − Y¯n ) − (λX − λY )
(X
%
Un =
λX
+ λnY
n
?
¯n
¯
X
+ Ynn
n
Vn =
λX
+ λnY
? n
¯ n + Y¯n
X
=
;
λX + λ Y
• Conclusion
Finally, we apply Slutsky’s theorem to justify the preservation of the convergence in
distribution under (restricted) division (Remark 5.95) to obtain the desired result:
Zn =
Un d
→ Normal(0, 1).
Vn
(0.5)
(b) Discuss the pertinence of the previous result.
• Pertinence of the convergence result
d
¯ n −Y¯n )−(λX −λY )
(X
q
¯n
¯
X
+ Ynn
n
d
→ Normal(0, 1)
– show that Un → Normal(0, 1);
This result is extremely relevant because it provides a pivotal quantity which
– apply Slutsky’s Theorem.
the unknown expected values of two independent r.v. X ∼ Poisson(λX ) and
P
is essential to derive asymptotic confidence intervals for the difference between
– prove that Vn → 1;
Y ∼ Poisson(λY ).
• Auxiliary results
¯ n ) = λX
◦ E(X
◦ E(Y¯n ) = λY
¯ n − Y¯n ) = λX − λY
◦ E(X
¯ n ) = λX
◦ V (X
4
¯ n ) = 0 and
We could have also invoked Markov’s Theorem (Theorem 5.136) because limn→+∞ V (X
limn→+∞ V (Y¯n ) = 0.
n
9
10

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