Download Solutions to Final Exam - Applied Probability | STAT 331 and more Exams Probability and Statistics in PDF only on Docsity!
Stat 331/Elec 331, Solutions to Final Exam
1a. Since f is non-negative and
∫ (^) ∞
0
f (x)dx =
∫ (^) ∞
1
x^2
dx = 1
f is a possible pdf.
b.
E[X] =
∫ (^) ∞
0
xf (x)dx =
∫ (^1)
0
xdx +
∫ (^) ∞
1
x
dx = ∞.
For 0 ≤ x ≤ 1, the cdf is
F (x) =
∫ (^) x
0
f (t)dt =
x 2 and for x ≥ 1,
F (x) =
∫ (^) x
0
f (t)dt =
∫ (^) x
1
t^2
dt =
x
2 x
d. The range of Y is (0, ∞). For 0 ≤ y ≤ 1, we get
FY (y) = P (Y ≤ y) = P (
X
≤ y) = P (X ≥
y
) = 1−FX (
y
2(1/y)
y 2
since 1/y is greater than 1. For y ≥ 1, we get
FY (y) = 1 − FX (
y
2 y
since 1/y is now bewteen 0 and 1. Hence, Y has the same distribution as X.
2a. Since C = 2X + 2Y
E[C] = 2E[X] + 2E[Y ] = 30
E[C|X = x] = 2x + 2E[Y |X = x]
where
E[Y |X = x] = 0. 8
(x − 5) = 1. 6 x + 2
and hence
E[C|X = x] = 5. 2 x + 4.
b. Since A = XY we get
E[A] = E[XY ] = Cov[X, Y ] + E[X]E[Y ] = 0. 8
E[A|X = x] = xE[Y |X = x] = x(1. 6 x + 2) = 1. 6 x^2 + 2x.
c. The probaility that a plate is useful is
P (29 ≤ C ≤ 31) = Φ
( 31 − 30 √
- 328
) − Φ
( 29 − 30 √
- 328
) = Φ(1.75) − (1 − Φ(1.75))
d. Let X=the number of useful plates. Then X has a binomial distribution with n = 10 and p = P (useful) = 0.92. Hence
P (X ≥ 9) = P (X = 9) + P (X = 10) =
( 10 9
)
- 9290 .08 + 0. 9210 = 0. 81.
3a. The likelihood function is
L(a) =
∏^ n
k=
fa(Xk) = (a + 1)n
∏n
k=
Xka
which gives
l(a) = log L(a) = n log(a + 1) + a
∑^ n
k=
log Xk
and hence
c. The first moment is μ 1 = E[N] = 52e−λ, and hence
λ = − log
( (^) μ 1 52
)
and the moment estimator is
̂ λ = − log
X
) .
In our case, X¯ = N, and we finally get the moment estimator
λ̂ = − log
( N
) .
5a. The state space is { 0 , 1 , 2 , 3 } and the rates are
0 → 1 : αp 0 → 2 : α(1 − p) 1 → 2 : αp 1 → 3 : α(1 − p) 1 → 0 : β 2 → 3 : αp 2 → 1 : β 3 → 2 : β
b. With p = 1/2, and β = α, the balance equations are
State 0: απ 0 = απ 1 State 1: (2α)π 1 = (α/2)π 0 + απ 2 State 2: (3α/2)π 2 = (α/2)π 0 + (α/2)π 1 + απ 2
which together with the condition that π 0 + ... + π 3 = 1 gives stationary distribution
π 0 = π 1 =
, π 2 =
, π 3 =
- The plots in c,d,f,g, and h are from bivariate normal distributions. The correlation coefficient is 0 in f and g (in g, Y has a larger variance than X), -0.99 in c, 0.5 in d, and 0.9 in h. The plot in a has X normal and Y binomial
(notice the horisontal lines), the plot in b is uniform on the unit disk, and the plot in e has X normal and Y also normal but centered around the curve y = x^3 (notice the shape of the plot).
- The elements yttrium, erbium, terbium and ytterbium are all named for the village Ytterby outside Stockholm, Sweden.
