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TA: Yury Petrachenko, CAB 484, yuryp@ualberta.ca, http://www.ualberta.ca/∼yuryp/
Review Questions, Chapters 8, 9
8.15 Suppose that Y 1
, Y
2
,... , Y
n
denote a random sample of size n from a population with an
exponential distribution whose density is given by
f (y) =
(1/θ)e
−y/θ , y > 0
0 , elsewhere.
If Y (1)
= min(Y 1
, Y
2
,... , Y
n
) denotes the smallest-order statistic, show that
θ = nY (1)
is an
unbiased estimator for θ and find MSE(
θ).
Solution. Let’s find the distribution function of Y :
F (y) =
1 − e
−y/θ
, y > 0
0 , elsewhere.
Now we can use the formula F Y (1)
(y) = 1 −
[
1 − F (y)
]
n
or f Y (1)
= n
1 − F (y)
n− 1
f (y) to find
the the density function for Y (1)
: for y > 0,
f Y (1)
= n
e
−y/θ
n− 1 1
θ
e
−y/θ
=
n
θ
e
−yn
θ .
We can recognize this density function to be the density of the exponential distribution with
parameter θ
n, Y (1)
∼ Exp
θ
n
Knowing the distribution of Y (1)
allows us to compute the expectation of
θ = nY (1)
E[
θ] = nE[Y (1)
] =
nθ
n
= θ.
So, E[
θ] = θ, and
θ is an unbiased estimator of θ.
To find MSE(
θ), use the formula MSE(
θ) = V [
θ] +
B(
θ)
2
. Since the estimator is unbiased,
its bias B(
θ) equals zero. For the variance, remember that Y (1)
is exponential. We have
MSE(
θ) = V [
θ] + 0 = n
2
V
[
Y
(1)
]
= n
2
θ
2
n
2
= θ
2
. §
9.7 Suppose that Y 1
, Y
2
,... , Y
n
denote a random sample of size n from an exponential distribution
with density function given by
f (y) =
(1/θ)e
−y/θ
, y > 0
0 , elsewhere.
In Exercise 8.15 we determined that
θ 1
= nY (1)
is an unbiased estimator of θ with MSE(
θ)= θ
2 .
Consider the estimator
θ 2
Y , and find the efficiency of
θ 1
relative to
θ 2
Solution. First compute the variance of
θ 2
V [
θ 2 ] = V [
Y ] = V
[
Y 1 + · · · + Yn
n
]
n
2
V [Y 1 + · · · + Yn] =
n
2
V [Y 1 ] + · · · + V [Yn]
n
2
θ
2
2
n times
nθ
2
n
2
θ
2
n
To find the relative efficiency, we need to find the ratio of two variances:
eff(
θ 1 ,
θ 2 ) =
V (
θ 2 )
V (
θ 1
θ
2
n
θ
2
n
We conclude that
θ 2
is preferable to
θ 1
9.61 Let Y 1
, Y
2
,... , Y
n
denote a random sample from the probability density function
f (y) =
(θ + 1)y
θ
, 0 < y < 1; θ > − 1
0 , elsewhere.
Find an estimator for θ by the method of moments.
Solution. Let’s find the first moment of this distribution:
μ =
1
0
(θ + 1) y
θ+
dy =
(θ + 1) y
θ+
θ + 2
1
0
θ + 1
θ + 2
The method of moments implies
Y =
θ + 1
θ + 2
θ =
Y − 1
Y
Solution. This is a somewhat different problem from the previous one because the support
of the density function depends on θ. Recall the indicator function I(A). It is equal to one
when A is true, and zero if A is false.
We can write the likelihood function in the following way:
L =
n ∏
i=
f (y i
n ∏
i=
2 θ + 1
I(0 ≤ y i
≤ 2 θ + 1) =
(2θ + 1)
n
n ∏
i=
I(0 ≤ y i
≤ 2 θ + 1).
We can simplify this even further if we note that the product of indicator is non-zero only
when all of the underlying conditions fulfill. That is, all y i
are less that 2θ + 1 and positive.
Notice that this statement is equivalent to the following: 0 ≤ y(1) and y(n) ≤ 2 θ + 1. (We use
order statistics y (1)
= min(y 1
,... , y n
) and y (n)
= max(y 1
,... , y n
).) We have
L =
(2θ + 1)
n
I(0 ≤ y(1)) · I(y(n) ≤ 2 θ + 1).
Now look at the first part of the likelihood function L, (2θ + 1)
−n
. Notice that this is a
decreasing (and continuous) function of θ. If we want to maximize L, we should choose the
value of θ as small as possible. Notice that if 2θ + 1 is smaller than y (n)
, then the value of L(θ)
is zero. So, the minimum of 2θ + 1 is y (n)
. This gives the minimum value for θ and maximizes
the likelihood L(θ). We conclude (provided at least one observation in the sample is positive)
Y
(n)
θ + 1 ∴
θ =
Y
(n)
9.80 Let Y 1
, Y
2
,... , Y
n
denote a random sample from the probability density function
f (y) =
(θ + 1)y
θ
, 0 < y < 1; θ > − 1
0 , elsewhere.
Find the maximum-likelihood estimator for θ. Compare your answer to the method of mo-
ments estimator found in Exercise 9.61.
Solution. Define the likelihood function:
L =
n ∏
i=
(θ + 1)y
θ
i
= (θ + 1)
n
( n ∏
i=
yi
θ
Take the logarithms:
ln L = n ln(θ + 1) + θ
n ∑
i=
ln y i
Find critical points:
d
dθ
ln L =
n
θ + 1
n ∑
i=
ln y i
so
θ = −
n
n
i=
ln y i
and finally
θ = −
n
n
i=
ln Y i
This is quite different from the method of moments estimator found in Exercise 9.61. §
