Math 2b, Winter 2009: Solutions to Homework 5
1. The Raleigh distribution is given by
f
(
r

θ
) =
r
θ
2
exp

r
2
2
θ
2
.
Find the mle for a sample of size
n
from this distribution.
Solution.
Let
X
1
, . . . , X
n
be independent random variables from the Raleigh distribution for
some parameter
θ
>
0. Since these variables are independent, their joint density function is the
product of the individual density functions. Therefore we have
f
(
x
1
, . . . , x
n
,
θ
) =
n
i
=1
x
i
θ
2
exp

x
2
i
2
θ
2
.
A maximum likelihood estimate of
θ
is a value of
θ
which maximizes
f
(
x
1
, . . . , x
n
,
θ
).
First
note that if any
x
i
is 0, this expression will be 0 and hence any
θ
>
0 will maximize.
We
now assume that
x
i
>
0 for all
i
.
To show that
f
(
x
1
, . . . , x
n
,
θ
) has a maximum, we must
verify that the limits as
θ
goes to 0 and
∞
exist and are not maxima. We first have that that
lim
θ
→∞
f
(
x
1
, . . . , x
n
,
θ
) = 0 since the first part of the product goes to 0 and the exponential
part goes to 1. Also,
lim
θ
→
0
x
i
θ

2
exp(
x
2
i
/
2
θ
2
)
= lim
θ
→
0

2
x
i
θ

3
(

x
2
i
/
θ
3
) exp(
x
2
i
/
2
θ
2
)
= lim
θ
→
0
2
x
i
exp

x
2
i
2
θ
2
= 0
for all
i
by l’Hˆopital’s Rule, so therefore
lim
θ
→
0
f
(
x
1
, . . . , x
n
,
θ
) =
n
i
=1
lim
θ
→
0
x
i
θ

2
exp(
x
2
i
/
2
θ
2
)
=
n
i
=1
0 = 0.
Since
f
(
x
1
, . . . , x
n
,
θ
) takes positive values in general, 0 is not a maximum.
Therefore there
must be a maximum at some
θ
∈
(0
,
∞
), and we will now find it. Since log is a monotonically
increasing function,
f
(
x
1
, . . . , x
n
,
θ
) will have maxima as the same points as
log
f
(
x
1
, . . . , x
n
,
θ
) =
n
i
=1
log(
x
i
)

log(
θ
2
)

x
2
i
2
θ
2
=
n
i
=1
log(
x
i
)

n
log(
θ
2
)

1
2
θ
2
n
i
=1
x
2
i
.
Di
ff
erentiating log
f
(
x
1
, . . . , x
n
,
θ
) with respect to
θ
, we obtain

2
n
θ
+
1
θ
3
n
i
=1
x
2
i
.
Since this is defined for all
θ
>
0, maxima must occur at points where this equals 0. Multiplying
through by
θ
3
, this means

2
n
θ
2
+
n
i
=1
x
2
i
= 0.
1
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Thus
θ
=
±
1
2
n
∑
n
i
=1
x
2
i
, and since
θ
>
0, we must have
θ
=
1
2
n
∑
n
i
=1
x
2
i
. Since we’ve shown
that a maximum must exist, it must occur at this point, so therefore the maximum likelihood
estimate of
θ
is
ˆ
θ
=
1
2
n
n
i
=1
x
2
i
.
2. The results of 600 independent trials of an experiment that yields an outcome 1, 2, 3, or 4 are
as follows
Outcome
1
2
3
4
# of trials
280
199
93
28
Assume that for some
q
between 0 and 1, the probabilities of outcomes 1 to 4 on each trial are
(1

q
)
3
, 3
q
(1

q
)
2
, 3
q
2
(1

q
), and
q
3
. Calculate a maximum likelihood estimator of
q
. (Hint:
This is similar to Example A in Rice’s section 8.5.1.)
Solution.
Let
n
denote the total number of trials, and let
X
1
,
X
2
,
X
3
, and
X
4
be random
variables which count the number of outcomes 1, 2, 3, and 4, respectively.
The joint density
function of
X
1
, . . . , X
n
is
f
(
x
1
, . . . , x
n
, q
) =
n
!
4
i
=1
x
i
!
(
(1

q
)
3
)
x
1
(
3
q
(1

q
)
2
)
x
2
(
3
q
2
(1

q
)
)
x
3
(
q
3
)
x
4
.
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 Winter '08
 Makarov,N
 Math, Differential Equations, Statistics, Equations, Probability, probability density function, Maximum likelihood, Estimation theory, Likelihood function, mean square error, Bias of an estimator

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