Download Statistics 351 – Intermediate Probability: Fall 2008 Final Exam Solutions and more Exams Probability and Statistics in PDF only on Docsity!
Statistics 351–Intermediate Probability Fall 2008 (200830) Final Exam Solutions
Instructor: Michael Kozdron
- (a) We see that fX,Y (x, y) ≥ 0 for all 0 < x, y < 1, and that ∫ (^) ∞
−∞
−∞
fX,Y (x, y) dx dy =
0
∫ (^) y
0
10 xy^2 dx dy =
0
5 y^4 dy = y^5
1
0
Thus, fX,Y is a legitimate density.
- (b) We compute
fX (x) =
−∞
fX,Y (x, y) dy =
x
10 xy^2 dy =
10 x(1 − x^3 ) 3 ,^0 < x <^1.
- (c) We compute
E(X) =
−∞
xfX (x) dx =
0
x · 10 x(1^ −^ x
dx =^10 9
=^5
- (d) We compute
fY |X=x(y) = fX,Y^ (x, y) fX (x)
= 10 xy
2 10 x(1−x^3 ) 3
= 3 y
2 1 − x^3
, x < y < 1.
- (e) We compute
E(Y |X = x) =
−∞
yfY |X=x(y) dy =
x
y · 3 y
2 1 − x^3
dy = 3(1^ −^ x
4(1 − x^3 )
- (f ) Using properties of conditional expectation (Theorem II.2.1), we compute
E(Y ) = E( E(Y |X) ) = E
3(1 − X^4 )
4(1 − X^3 )
0
3(1 − x^4 ) 4(1 − x^3 )
· 10 x(1^ −^ x
dx =^5 2
0
x − x^5 dx =^5 6
- (g) If U = XY and V = Y , then solving for X and Y gives X = U/V and Y = V , so that the Jacobian of this transformation is
J =
∂x ∂u
∂x ∂v ∂y ∂u
∂y ∂v
∣∣^1 /v^ −u/v
2 0 1
∣∣ =^1
v
By Theorem I.2.1, the joint density of (U, V )′^ is therefore given by
fU,V (u, v) = fX,Y (u/v, v) · |J| = 10uv−^1 v^2 · v−^1 = 10u
provided that 0 < u < v^2 < 1 and v > 0. It then follows that the density function of U is given by
fU (u) =
−∞
fU,V (u, v) dv =
√u^10 u^ dv^ = 10u(1^ −
u)
provided that 0 < u < 1.
- (a) The density function of X is given by
fX (x) =
−∞
fX,Y (x, y) dx dy =
−∞
fX|Y =y(x) · fY (y) dy.
Substituting in gives
fX (x) =
0
θyxθ−^1 exp{−yxθ} · a
p Γ(p)
yp−^1 e−ay^ dy = a
p Γ(p)
θxθ−^1
0
yp^ exp
−y(a + xθ)
dy.
Properties of the gamma function imply that ∫ (^) ∞
0
yp^ exp
−y(a + xθ)
dy = Γ(p + 1) · (a + xθ)−(p+1)
so that
fX (x) = a
p Γ(p)
θxθ−^1 · Γ(p + 1) · (a + xθ)−(p+1)^ = p a
p (^) θ xθ− 1 (a + xθ)p+
provided that x > 0 (and using the fact that Γ(p + 1) = pΓ(p).)
- (b) If Z = Xθ, then the density function of Z is
fZ (z) = fX (z^1 /θ) · d dz
z^1 /θ^ = p a
p (^) θ z 1 − 1 /θ (a + z)p+^
θ
z^1 /θ−^1 = p a
p (a + z)p+
provided that z > 0.
- There are two ways to compute the resulting integral, depending on the order of integration. One way is much easier than the other.
Solution 1: ( dy dx) We begin by noticing that
P {X < Y } =
{x<y}
fX,Y (x, y) dy dx =
0
x
x 27
exp
− x^ +^ y 3
dy dx
0
x 27 e
−x/ 3
x
e−y/^3 dy dx.
We now find (^) ∫ (^) ∞
x
e−y/^3 dy = 3e−x/^3
which implies that
P {X < Y } =
0
x 9
e−^2 x/^3 dx.
We recognize this as a Gamma function; that is, ∫ (^) ∞
0
x 9
e−^2 x/^3 dx =^1 4
0
4 x 9
e−^2 x/^3 dx =^1 4
0
ue−u^ du =^1 4
Γ(2) =^1
- (b) Using Corollary II.2.3.1 on page 39, we know
Var(Y ) = E(Var(Y |X)) + Var(E(Y |X)).
Since Y |X = x is U (x, 1) we know that E(Y |X = x) = (1 + x)/2 and Var(Y |X = x) = (1 − x)^2 / 12 so that
Var(Y ) = E(Var(Y |X)) + Var(E(Y |X)) = E
(1 − X)^2
1 + X
12 −^
E(X)
6 +^
E(X^2 )
Var(X)
Since X is U (0, 1), we know that E(X) = 1/2 and Var(X) = 1/12 so that E(X^2 ) = 1/3. Thus, we conclude that Var(Y ) = 1 12
+^1
+^1
- (a) Since X 1 and X 2 are independent and identically distributed Γ(2, 1) random variables, they have common density function f (x) = xe−x, x > 0 ,
and common distribution function
F (x) =
∫ (^) x
0
ue−u^ du = 1 − xe−x^ − e−x
for x > 0 (and F (x) = 0 for x ≤ 0) using the integration-by-parts formula given on the first page. Therefore, it follows from Theorem IV.1.2 that the density function of X(1) is
fX(1) (y) = 2(1 − F (y))f (y) = 2ye−y(ye−y^ + e−y) = 2y(y + 1)e−^2 y
provided that y > 0.
- (b) It follows from Theorem IV.1.2 that the density function of X(2) is
fX(2) (y) = 2F (y)f (y) = 2ye−y(1 − ye−y^ − e−y)
provided that y > 0.
- (c) It follows from Theorem IV.2.1 that the density function of (X(1), X(2))′^ is
fX(1),X(2) (y 1 , y 2 ) = 2f (y 1 )f (y 2 ) = 2y 1 y 2 e−y^1 −y^2
provided that 0 < y 1 < y 2 < ∞.
- (d) If U = X(1)/X(2) and V = X(2), then solving for X(1) and X(2) gives X(1) = U V and X(2) = V so that the Jacobian of this transformation is
J =
∂y 1 ∂u
∂y 1 ∂v ∂y 2 ∂u
∂y 2 ∂v
∣∣v^ u 0 1
∣∣ = v.
By Theorem I.2.1, the joint density of (U, V )′^ is therefore given by
fU,V (u, v) = fX(1),X(2) (uv, v) · |J| = 2uv^3 e−uv−v
provided that 0 < u < 1 and 0 < v < ∞.
It now follows that the density function of U is
fU (u) =
−∞
fU,V (u, v) dv =
0
2 uv^3 e−uv−v^ dv = 2u
0
v^3 e−v(u+1)^ dv = 2u · Γ(4)(u + 1)−^4
= 12 u (1 + u)^4
provided that 0 < u < 1 and using properties of the gamma function.
- (a) In order to find the eigenvalues of ΛΛΛ, we must find those values of λ such that det[ΛΛΛ−λI] = 0. Therefore,
det[ΛΛΛ − λI] = det
3 2 −^ λ^ −
√ 3 2 −
√ 3 2
5 2 −^ λ
2 −^ λ
2 −^ λ
− 34 = λ^2 − 4 λ + 3 = (λ − 1)(λ − 3)
so that the eigenvalues of ΛΛΛ are λ 1 = 1 and λ 2 = 3.
- (b) Since λ 1 = 1,
[ΛΛΛ − λ 1 I | 00 0] =
[ 1
√ 3 2 0 −
√ 3 2
3 2 0
]
[
1 2 −
√ 3 2 0 0 0 0
]
[
]
and since λ 2 = 3,
[ΛΛΛ − λ 2 I | 00 0] =
[
√ 3 2 0 −
√ 3 2 −^
1 2 0
]
[
√ 3 2 0 0 0 0
]
[
]
we conclude that eigenvectors for λ 1 and λ 2 are
v 1 =
[√
]
and v 2 =
[
]
respectively. Therefore, the diagonal matrix is
D = diag(λ 1 , λ 2 ) =
[
λ 1 0 0 λ 2
]
[
]
and the orthogonal matrix is
C =
[
v 1 ||v 1 ||
v 2 ||v 2 ||
]
√ 3 2 −^
1 2 1 2
√ 3 2
since ||v 1 || = ||v 2 || = 2.
- (c) Since det[ΛΛΛ] = 3, we see that
ΛΛΛ−^1 =
5 6 −
√ 3 6 −
√ 3 6
1 2
- (a) Let
B =
[
1 /σ 1 −ρ/σ 2 0 1 /σ 2
]
so that Y = BX. By Theorem V.3.1, Y is MVN with mean
Bμμμ =
[
1 /σ 1 −ρ/σ 2 0 1 /σ 2
] [
]
[
]
and covariance matrix
BΛΛΛB′^ =
[
1 /σ 1 −ρ/σ 2 0 1 /σ 2
] [
σ 12 ρσ 1 σ 2 ρσ 1 σ 2 σ^22
] [
1 /σ 1 0 −ρ/σ 2 1 /σ 2
]
[
1 − ρ^2 0 1
]
- (b) Since Y is multivariate normal we know from Definition I that Y 1 and Y 2 are each one- dimensional normals. We also know from Theorem V.7.1 that the components of Y are independent if and only if they are uncorrelated. From (a) we know that Cov(Y 1 , Y 2 ) = 0 so that Y 1 and Y 2 are, in fact, independent.
- We know from Theorem V.9.1 that X′ΛΛΛ−^1 X ∈ χ^2 (3). Thus, the required matrix A is
A = ΛΛΛ−^1 =
- If Mn = S n^3 − 3 nSn, then
Mn+1 = S n^3 +1 − 3(n + 1)Sn+1 = (Sn + Yn+1)^3 − 3(n + 1)(Sn + Yn+1) = S n^3 + 3S n^2 Yn+1 + 3SnY (^) n^2 +1 + Y (^) n^3 +1 − 3(n + 1)Sn − 3(n + 1)Yn+ = Mn + 3Sn(Y (^) n^2 +1 − 1) + 3S n^2 Yn+1 − 3(n + 1)Yn+1 + Y (^) n^3 +1.
Thus, we see that we will be able to conclude that {Mn, n = 0, 1 ,.. .} is a martingale if we can show that E
3 Sn(Y (^) n^2 +1 − 1) + 3S^2 nYn+1 − 3(n + 1)Yn+1 + Y (^) n^3 +1|Sn
Now 3 E(Sn(Y (^) n^2 +1 − 1)|Sn) = 3SnE(Y (^) n^2 +1 − 1) and 3 E(S n^2 Yn+1|Sn) = 3S^2 nE(Yn+1)
by “taking out what is known,” and using the fact that Yn+1 and Sn are independent. Furthermore,
3(n + 1)E(Yn+1|Sn) = 3(n + 1)E(Yn+1) and E(Y (^) n^3 +1|Sn) = E(Y (^) n^3 +1)
using the fact that Yn+1 and Sn are independent. Since E(Yn+1) = 0, E(Y (^) n^2 +1) = 1, and E(Y (^) n^3 +1) = 0, we see that
E(Mn+1|Mn) = Mn + 3SnE(Y (^) n^2 +1 − 1) + 3S n^2 E(Yn+1) − 3(n + 1)E(Yn+1) + E(Y (^) n^3 +1) = Mn + 3Sn · (1 − 1) + 3S n^2 · 0 − 3(n + 1) · 0 + 0 = Mn
which proves that {Mn, n = 0, 1 , 2 ,.. .} is, in fact, a martingale.
- (a) The optional sampling theorem applied to the martingale {Sn, n = 0, 1 , 2 ,.. .} gives E(ST ) = E(S 0 ) = 0. Since
E(ST ) = −aP (ST = −a) + bP (ST = b) = −a[1 − P (ST = b)] + bP (ST = b)
we see that −a[1 − P (ST = b)] + bP (ST = b) = 0
which solving for P (ST = b) gives P (ST = b) =
a a + b as required.
- (b) The optional sampling theorem applied to the martingale {Xn, n = 0, 1 , 2 ,.. .} gives E(XT ) = E(X 0 ) = 0. But E(XT ) = E(S T^2 −T ) and E(X 0 ) = 0 which implies that E(S T^2 )−E(T ) =
- Since
E(S^2 T ) = (−a)^2 P (ST = −a) + b^2 P (ST = b) = a^2
1 − a a + b
a a + b
= ab
using our result from (a), we conclude that
E(T ) = ab
as required.
- (a) Let {Xt, t ≥ 0 } denote the Poisson process with intensity 1 according to which subway trains arrive (where t is measured in units of four minutes). The random number of subway trains that arrive in one hour (i.e., 15 time units) is X 15. Since Xt ∈ Po(t) for all t by the Poisson process assumption, we conclude that E(X 15 ) = 15.
- (b) If we use the fact that a Poisson process resets at fixed times, then the probability that Christian will wait at least 8 minutes (i.e., 2 time units) for the next train to arrive is
P (T 1 ≥ 2) = P (X 2 ≤ 1) = P (X 2 = 0) + P (X 2 = 1) = e
= 3e−^2
using the fact that X 2 ∈ Po(2).
- (b) We again use the fact that a Poisson process resets at fixed times. The probability that at least three trains pass Christian in the 16 minutes (i.e., 4 time units) while he is waiting for Veronica is
P (X 4 ≥ 3) = 1 − P (X 4 < 3) = 1 − P (X 4 = 0) − P (X 4 = 1) − P (X 4 = 2)
= 1 − e
− e
− e
= 1 − 13 e−^4.
