Probability and Statistics Final Solutions
1. Let X1 , X2 , X3 be iid with pdf f (x) = 3x2 for 0 ≤ x ≤ 1. Find the
probability that exactly two of these three variables exceed 1/2.
Solution
R
For any one of them the probability of exceeding 1/2 is 1 − 01/2 3x2 dx =
1 − ((1/2)3 − 03 ) = 7/8. The probability of exactly two exceeding is 3 ×
(1/8)1 × (7/8)2 .
2. If the moment generating function of a random variable X is mX (t) =
t
e2(e −1) , find P (X ≥ 3).
Solution
From the mgf we know that X is a poisson variable with mean m = 2. Using
the table, we find that P (X ≥ 3) = 1 − P (X ≤ 2) = 1 − .677 = .333.
Alternately, we can simply compute by hand:
1 − P (X ≤ 2) = 1 − 20 e−2 − 2e−2 − (22 e−2 /2) = 1 − 5e−2 .
3. Let X ∼ N (µ, σ 2 ) be such that P (X < 40) = .90 and P (x < 46) = .95.
Find µ and σ.
Solution
As Z = X−µ
is the standard normal distribution Z = N (0, 1), and from
σ
the table we have that that P (Z < 1.281) is about .90 and P (Z < 1.645)
is about .95; we get that 40−µ
= 1.281 and 46−µ
.1.645. So 6 = 46 − 40 =
σ
σ
σ(1.645 − 1.281) = σ(.364) giving that σ = 6/.364 (about 16.5). And so
µ = 40 − (1.281) · 6/.364 (about 19).
4. Show that the sample variance S 2 =
P
(Xi −X)2
n−1
2
(for a sample of size n) is
2
an unbiased estimator of the variance σ . (You may use that E(X ) =
µ2 + σ 2 /n).)
Solution
We have (where all sums are for i = 1, . . . , n unless otherwise denoted) that
P
X
X 2
X
(Xi − X)2
1
Xi +
X )
E(S 2 ) = E(
)=
E(
Xi2 − 2X
n−1
n−1
1
n
2
2
2
=
E(n(Xi2 ) − 2nX + nX ) =
E(X 2 − X )
n−1
n−1
n
n
2
=
(E(X 2 ) − E(X )) =
((µ2 + σ 2 ) − (µ2 + σ 2 /n))
n−1
n−1
n (n − 1)σ 2
=
= σ2 .
n−1
n
This shows that S 2 is an unbiased estimator of σ 2 .
5. Let X have a Poisson distribution with parameter µ. A sample of 200
observations of X has a sample mean of 4. Construct an approximate .99
confidence interval for µ.
Solution
Using the Central Limit Theorem we assert that the distribution of Z =
X−µ
√
S/ n
is approximately N (0, 1). Looking up z.005 ≈ 2.575 in the table for N (0, 1)
√
√
we get that (X − 2.58(S/ n), X + 2.58(S/ n)) is an approximate confidence
interval for µ. Our sample mean X is 4 and as distribution with mean µ
has variance σ 2 = µ, we can use our sample mean 4 as our sample variance.
√
So S = 2 and the approximate confidence interval is (4 − 2.58( 2/10), 4 +
√
2.58( 2/10)) ≈ (3.64, 4.36).
6. Let Y1 < Y2 < · · · < Y5 be the order statistics of a random sample of an
RV X with pdf f (x) = e−x for x ≥ 0. Find the joint pdf of Y2 and Y4 .
Solution
The joint pdf g of Y2 and Y4 is
g(y2 , y4 )
=
=
=
=
5!e−y2 e−y4
5!e−y2 e−y4
5!e−y2 e−y4
∞
Z
Z
y4
y4
Z ∞
y2
Z y4
y4
Z ∞
y2
e−y1 e−y3 e−y5 dy1 dy3 dy5
0
e−y3 e−y5 dy3 dy5 (1 − e−y2 )
e−y5 dy5 (e−y2 − e−y4 )(1 − e−y2 )
y4
−y4
5!e−y2 e−y4 (e
y2
Z
)(e−y2 − e−y4 )(1 − e−y2 )
7. Let Y1 < Y2 < Y3 < Y4 be the order statistics of a sample of X having pdf
f (x; θ) = 1/θ for 0 < x < θ. The hypothesis H0 : θ = 1 is rejected and
H1 : θ > 1 is accepted if Y4 > c. Find the constant so that the test has
significance .05. What is the power function for the test?
Solution
The probability of getting Y4 > c if θ = 1 is 1 less the probability that all four
Xi are less than c. As X is uniform on [0, 1] this probability is 1 − (c/1)4 .
Setting this equal to .05 we get that c = .951/4 .
The power γ(θ) is 1 minus the probability of getting Y4 < c. This is 1 −
(c/θ)4 = 1 − .95/θ4 .
8. Let a distribution X have pdf f (x) = 4x3 for 0 < x < 1. Show how you
would generate random observations from X using random values from
the uniform distribution U on [0, 1]. (What value of X would you get
from the value .125 of U ?)
Solution
The cdf FX (x) of X is
Z
FX (x) =
x
4
4
4z 3 dz = z 4 |x
0 =x −0=x
0
−1
So X = FX
(U ) = U 1/4 . Thus a value .125 for U yields a value .1251/4 of
X.