Exercise 1 Exercise 2 Exercise 3 Exercise 4
Question 3p.d.f. integrates to 1⇒
∫∞−∞
1√2πe−
x2
2 dx = 1⇒∫ ∞−∞
e−x2
2 dx =√
2π.
E[X] =
∫ ∞−∞
x · 1√2πe−
x2
2 dx
=1√2π
∫ ∞−∞
x · e−x2
2 dx
= 0 (integration by substitution, or by symmetry).
E[X2] =1√2π
∫ ∞−∞
x2 · e−x2
2 dx
= 1 (by parts, and using L’Hopital’s rule).
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Question 4The m.g.f. of X is
φX(t) = E[etX ] =
∞∑k=0
etk · λk
k!e−λ = [
∞∑k=0
(etλ)k
k!] · e−λ = e(et−1)λ.
The m.g.f. of X + Y is
φX+Y (t) = E[et(X+Y )]
= E[etX ] · E[etY ] (by independence)
= φX(t) · φY (t)
= e(et−1)(λ+µ).
The m.g.f. of X + Y is the same as that of Poisson withparameter λ+ µ, and the m.g.f. can uniquely determine thedistribution, so X + Y has Poisson distribution with parameterλ+ µ.
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Question 5The event {Y ≤ y} can be expressed as
{Y ≤ y} = {g(X) ≤ y} = {X ≤ g−1(y)},so the c.d.f. of Y is
FY (y) = P(Y ≤ y) = P(X ≤ g−1(y)) = FX(g−1(y)),
As X has a continuous distribution, its c.d.f. is continuous.Thus the c.d.f. of Y is continuous, which implies Y hascontinuous distribution. Since g(·) is an increasing bijection,g−1(·) is also an increasing bijection. Since g(·) is differentiable,and g(x) 6= 0 for all x ∈ R, g−1(·) is differentiable. We haveFY (y) = FX(g−1(y)), so
fY (y) =d
dyFX(g−1(y)) = fX(g−1(y))· d
dy[g−1(y)] =
fX(g−1(y))
g′[g−1(y)].
X ∼ N(0, 1)⇒ Y = aX + b ∼ N(b, a2), fY (y) = 1√2πa
e−(y−b)2
2a2 .
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Question 6Ti ∼ Exp(1), i = 1, 2, . . . , n. The m.g.f. of T is
φT (θ) = E(eθT ) =
∫ ∞0
e(θ−1)tdt
This integral converges when θ < 1, so the m.g.f. only exists in(−∞, 1), and φT (θ) = 1
1−θ , θ < 1. Thus the m.g.f. of Sn is
φSn(θ) = E(eθSn)
= E(e∑ni=1 θTi)
=
n∏i=1
E(eθTi) (by independence)
=1
(1− θ)n(θ < 1),
which is the same as the m.g.f. of the given distribution, so Snfollows the given distribution.
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Question 6The probability can be calculated as follows.
P(Sn > t) =
∫ ∞t
xn−1 e−x
(n− 1)!dx
=1
(n− 1)![tn−1e−t + (n− 1)
∫ ∞t
e−x · xn−2dx] (by parts)
= · · ·
=1
(n− 1)!· [tn−1 · e−t + (n− 1)tn−2 · e−t + . . .
+ (n− 1)! · te−t + (n− 1)! · e−t]
=tn−1
(n− 1)!· e−t +
tn−2
(n− 2)!· e−t + . . .+ te−t + e−t
=
n−1∑i=0
ti
i!· e−t
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Question 6
The probability can be calculated as follows.
P(N(t) = k) = P(N(t) < k + 1)− P(N(t) < k)
= P(Sk+1 > t)− P(Sk > t)
=
k∑i=0
ti
i!· e−t −
k−1∑i=0
ti
i!· e−t
=tk
k!· e−t
which is the same as shown in the question.
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Question 3Y = X2, X ∼ Exp(1).
P(0 ≤ Y ≤ y) = P(0 ≤ X2 ≤ y)
= P(0 ≤ X ≤ √y)
= 1− e−√y.
So the density of Y is
fY (y) =
{0 if y ≤ 0,
12√ye−√y if y > 0.
The first way to compute E(Y ):
E(Y ) = E(X2) =
∫ ∞0
x2e−xdx = 2
∫ ∞0
xe−xdx (by parts) = 2.
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Question 3
The second way to compute E(Y ):
E(Y ) =
∫ ∞0
y
2√ye−√ydy =
∫ ∞0
x2e−xdx = 2.
Two ways to compute E(Y ) are the same because substitutiongives the same integral to be dealt with.
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Question 4The joint p.d.f. fXY (x, y) is
fXY (x, y) =
{1π if (x, y) ∈ D,0 if otherwise.
P(−1 ≤ a ≤ X ≤ b ≤ 1) =
∫ b
a
∫ ∞−∞
fXY (x, y)dydx
=
∫ b
a
2
π
√1− x2dx.
So X has a distribution with density
fX(x) =
{2π
√1− x2 if − 1 ≤ x ≤ 1,
0 if otherwise.
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Question 4Similarly, Y has a distribution with density
fY (y) =
{2π
√1− y2 if − 1 ≤ y ≤ 1,
0 if otherwise.
X and Y are not independent since fXY (x, y) 6= fX(x) · fY (y).
cov(X,Y ) =1
π
∫ ∫Dxydxdy
=1
π
∫ ∫D1
xydxdy +1
π
∫ ∫D2
xydxdy
+1
π
∫ ∫D3
xydxdy +1
π
∫ ∫D4
xydxdy,
whereD1 = D ∩ {x ≥ 0} ∩ {y ≥ 0},
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Question 4
D2 = D ∩ {x < 0} ∩ {y ≥ 0},
D3 = D ∩ {x < 0} ∩ {y < 0},
D4 = D ∩ {x ≥ 0} ∩ {y < 0}.
These four integrals in the R.H.S have the same absolute value,two being positive and two being negative, so
cov(X,Y ) = 0.
Zero covariance can lead to uncorrelatedness, however theconverse is not necessarily true. Uncorrelatedness doesn’t implyindependence. Independence is a stronger condition thanuncorrelatedness.
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Question 5The event {Mn ≤ c} can be expressed as
{Mn ≤ c} = {Uk ≤ c for all k = 1, 2, . . . , n},
and then for 0 ≤ c ≤ 1,
P(Mn ≤ c) = P(Uk ≤ c for all k = 1, 2, . . . , n)
=
n∏k=1
P(Uk ≤ c)
= cn.
So the c.d.f. of Mn is
FMn(x) =
0 if x ≤ 0,
xn if 0 ≤ x ≤ 1,
1 if x ≥ 1.
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Question 5
Now for x ≥ 0
P(n(1−Mn) ≤ x) = P(Mn ≥ 1− x
n)
= 1− (1− x
n)n.
For large enough n,
1− (1− x
n)n → 1− e−x,
son(1−Mn)
D−→ Z,
where Z has standard exponential distribution.
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Question 6
X,Y ∼ N(0, 1) ⇒ fX(x) = 1√2πe−
x2
2 and fY (y) = 1√2πe−
y2
2 ,
and X ⊥⊥ Y , thus
fX+Y (u) =
∫ ∞−∞
fX(u− v)fY (v)dv
=
∫ ∞−∞
1√2πe−
(u−v)22 × 1√
2πe−
v2
2 dv
=
∫ ∞−∞
1
2πe−v
2+uv−u2
2 dv
=
∫ ∞−∞
1
2πe−(v−u
2)2−u
2
4 dv
=e−
u2
4
2√π
∫ ∞−∞
1√πe−(v−u
2)2dv
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Question 6
1√πe−(v−u
2)2 is the pdf of N(
u
2,1
2) ⇒∫ ∞
−∞
1√πe−(v−u
2)2dv = 1.
So
fX+Y (u) =e−
u2
4
2√π
∫ ∞−∞
1√πe−(v−u
2)2dv
=e−
u2
4
2√π.
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Question 7
We assume the needle has unit length. Fix the coordinatesystem where the origin is one end of the needle. The other endcan be considered as a random point (X,Y, Z) uniformlydistributed on the unit sphere. Projection onto the x-ray isequivalent to projecting (X,Y, Z) onto any plane (e.g. x-yplane). Then the length of projection
R ≤ r ⇔ x2 + y2 ≤ r2.
We need to calculate surface area of that part of the sphere:
Sp = {(x, y, z) ∈ R3 : x2 + y2 + z2 = 1, x2 + y2 ≤ r2}.
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Question 7
The surface area of the sphere is 4π and the area of Sp is
4π(1−√
1− r2),
then
P(R ≤ r) = P(X2 +Y 2 ≤ r2) =4π(1−
√1− r2)
4π= 1−
√1− r2.
So the density of R is
fR(r) =
{r√
1−r2 , if 0 ≤ r ≤ 1
0, if otherwise
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Part Bi) The m.g.f. of Y = [Z2
1 + Z22 , Z
21 − Z2
2 ] is
φY (θ) = E[exp(θ1Y1 + θ2Y2)]
= E[exp(θ1(Z21 + Z2
2 ) + θ2(Z21 − Z2
2 ))]
= E[exp((θ1 + θ2)Z21 + (θ1 − θ2)Z2
2 )]
= E[exp((θ1 + θ2)Z21 )] · E[exp((θ1 − θ2)Z2
2 )] (by independence)
= φZ21(θ1 + θ2) · φZ2
2(θ1 − θ2)
Z1, Z2 ∼ N(0, 1)⇒ Z21 , Z
22 ∼ χ2
1 ⇒ φZ21(θ) = φZ2
2(θ) =
1√1− 2θ
.
So we have
φY (θ) =1√
1− 2(θ1 + θ2)· 1√
1− 2(θ1 − θ2)=
1√(1− 2θ1)2 − 4θ2
2
.
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Part B
The marginal m.g.f. of Y1 is
φY1(θ1) = φY (θ1, 0) =1
1− 2θ1(θ1 <
1
2),
which is the m.g.f. of χ22. Similarly,
φY2(θ2) = φY (0, θ2) =1√
1− 4θ22
(θ2 ∈ (−1
2,1
2)).
ii)
cov(Y1, Y2) = E(Y1Y2)− E(Y1)E(Y2)
=∂2
∂θ1∂θ2φY (0, 0)− ∂
∂θ1φY (0, 0) · ∂
∂θ2φY (0, 0)
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Part B
∂∂θ1
φY (θ) = 2(1− 2θ1)[(1− 2θ1)2 − 4θ22]−
32 ⇒ ∂
∂θ1φY (0, 0) = 2.
∂∂θ2
φY (θ) = 4θ2[(1− 2θ1)2 − 4θ22]−
32 ⇒ ∂
∂θ2φY (0, 0) = 0.
∂2
∂θ1∂θ2φY (θ) = 24(1− 2θ1)θ2[(1− 2θ1)2 − 4θ2
2]⇒∂2
∂θ1∂θ2φY (0, 0) = 0.
Thus, cov(Y1, Y2) = 0⇒ Y1 and Y2 are uncorrelated.
φY (θ) =1√
(1− 2θ1)2 − 4θ22
, φY1(θ1)φY2(θ2) =1
(1− 2θ1)√
1− 4θ22
.
φY (θ) 6= φY1(θ1)φY2(θ2)⇒ Y1 and Y2 are not independent.
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Part C
Objective: to compute the m.g.f. of the r.v. M = X21 +X2
2 .
M = X21 +X2
2
= X · X ′
= (Z · Σ12 ) · (Z · Σ
12 )′
= Z · Σ · Z ′ (by symmetry)
= (Z1, Z2)
(σ1 ρρ σ2
)(Z1
Z2
)= σ1Z
21 + 2ρZ1Z2 + σ2Z
22
= σ1(Z1 +ρ
σ1Z2)2 + (σ2 −
ρ2
σ1)Z2
2
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Part CThe m.g.f. of M is
φM (θ) = E[exp(θM)]
= E[exp(θσ1(Z1 +ρ
σ1Z2)2 + θ(σ2 −
ρ2
σ1)Z2
2 )]
= E[E[exp(θσ1(Z1 +ρ
σ1Z2)2 + θ(σ2 −
ρ2
σ1)Z2
2 )|Z2]]
= E[exp(θ(σ2 −ρ2
σ1)Z2
2 ) · E[exp(θσ1(Z1 +ρ
σ1Z2)2)|Z2]]
= E[exp(θ(σ2 −ρ2
σ1)Z2
2 ) · φ(Z1+ ρσ1Z2)2(θσ1)]
= E[exp(θ(σ2 −ρ2
σ1)Z2
2 ) ·exp(
ρ2Z22
σ21θσ1
1−2θσ1)
(1− 2θσ1)12
]
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Part C
Simplifying the expression of the expectation gives
φM (θ) = E[(1− 2θσ1)−12 exp(θ(
σ1σ2 − ρ2
σ1+
ρ2
σ1(1− 2θσ1))Z2
2 )]
= (1− 2θσ1)−12 · E[exp(θ(
σ1σ2 − ρ2
σ1+
ρ2
σ1(1− 2θσ1))Z2
2 )]
= (1− 2θσ1)−12 · φZ2
2(θ(
σ1σ2 − ρ2
σ1+
ρ2
σ1(1− 2θσ1)))
= (1− 2θσ1)−12 · (1− 2θ(
σ1σ2 − ρ2
σ1+
ρ2
σ1(1− 2θσ1)))−
12
= [(1− 2θσ1)(1− 2θσ2)(1− 4θ2ρ2
(1− 2θσ1)(1− 2θσ2))]−
12
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Question 3
i.i.d. r.v.s X1, . . . , Xn ∼ U(0, 1).
E[logX1] =
∫ 1
0log xdx = x log x|10 −
∫ 1
0x · 1
xdx = −1,
E[(logX1)2] =
∫ 1
0(log x)2dx = x(log x)2|10−
∫ 1
0x·2 log x·1
xdx = 2,
var[logX1] = E[(logX1)2]− [E(logX1)]2 = 2− 1 = 1.
Denote Y = (X1 · · ·Xn)1√n e√n.
log Y =1√n
n∑i=1
logXi +√n =√n · ( 1
n
n∑i=1
logXi + 1)
Since E(logXi) = −1 and var(logXi) = 1,
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Question 3√n · ( 1
n
∑ni=1 logXi + 1)
D−→ Z, where Z ∼ N(0, 1) by usingcentral limit theorem (CLT). Thus,
log YD−→ Z,
and thenYD−→ ln N(0, 1),
where ln N(0, 1) is a log-normal distribution.
Central Limit Theorem:A sequence of i.i.d. r.v.’s Xi with E(Xi) = µ, var(Xi) = σ2
(σ2 <∞),√n · ( 1
n
n∑i=1
Xi − µ)D−→ N(0, σ2).
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Question 4X1, . . . , Xn ∼ Pois(λ), and X ′is are mutually independent, so∑n
i=1Xi has Poisson distribution with parameter nλ, which canbe obtained by using m.g.f.
The support of Xi is {0, 1, . . .} ⇒ the support of∑n
i=1Xi is{0, 1, . . .} ⇒ the support of X is {0, 1
n , . . .}, and the p.m.f. of Xis
fX(k
n) = P(
n∑i=1
Xi = k) = e−nλ · (nλ)k
k!.
E[X − λ] = E[X]− λ
=1
n
n∑i=1
E[Xi]− λ
= 0.
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Question 4
E[(X − λ)2] = var(X − λ)
= var(X)
=1
n2
n∑i=1
var(Xi) (by independence)
=λ
n.
Denote Y =√n(X − λ) =
√n · ( 1
n
∑ni=1Xi − λ).
Xi ∼ Pois(λ)⇒ E(Xi) = λ and var(Xi) = λ.
By CLT,
YD−→ N(0, λ).
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Question 5X,Y ∼ N(0, 1). For r > 0,
P(X2 + Y 2 ≤ r2) =
∫ ∫x2+y2≤r2
1√2πe−
x2
2 · 1√2πe−
y2
2 dxdy
=
∫ ∫x2+y2≤r2
1
2πe−
x2+y2
2 dxdy
=
∫ r
0
∫ 2π
0
1
2πe−
u2
2 ududθ
=
∫ r
0ue−
u2
2 du
= 1− e−r2
2
So x2 + y2 has exponential distribution with mean 2, or χ22, or
Gamma distribution with shape parameter 1 and scaleparameter 2.
Exercise 1 Exercise 2 Exercise 3 Exercise 4
Question 6Y1, . . . , Yn : Yi = α+ βxi + εi,
∑ni=0 xi = 0, εi ∼ N(0, σ2).
α =1
n
n∑i=1
Yi = α+ β · 1
n
n∑i=1
xi +1
n
n∑i=1
εi = α+1
n
n∑i=1
εi
So α ∼ N(α, σ2
n ).
β =
∑ni=1 xiYi∑ni=1 x
2i
=
∑ni=1 xi(α+ βxi + εi)∑n
i=1 x2i
= β +
∑ni=1 xiεi∑ni=1 x
2i
So β ∼ N(β, σ2∑ni=1 x
2i).
Here [α, β] and σ2 are independent, and
nσ2
σ2=
∑ni=1(Yi − α− βxi)2
σ2∼ χ2
n−2.