An Introduction to Signal Detection and Estimation - Second Edition Chapter IV: Selected Solutions

8/9/2019 An Introduction to Signal Detection and Estimation - Second Edition Chapter IV: Selected Solutions

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An Introduction to Signal Detection and

Estimation - Second Edition

Chapter IV: Selected Solutions

H. V. PoorPrinceton University

April 26, 2005

Exercise 1:

a.

MAP(y) = arg

max1e

log |y| log

= 1.

b.

MMSE(y) =

e1 e

|y|de1 e

|y|d =

1

|y|+

e|y| e1e|y|

e|y| ee|y|

Exercise 3:

w(|y) = yee0

yeed =

ye(+1)(1 + )y+1

y! .

So:

MMSE(y) = 1

y!

0

y+1e(+1)d(1 + )y+1 = y+ 1

+ 1;

MAP(y) = arg

max>0

y log (+ 1)

= y

+ 1;

andABS(y) solves ABS(y)0

w(|y)d=1

2

which reduces toy

k=0

ABS(y)

kk!

=1

2eABS(y).

Note that the series on the left-hand side is the truncated power series expansion ofexp{ABS(y)} so that the ABS(y) is the value at which this truncated series equals halfof its untruncated value when the truncation point is y.

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Exercise 7:

We have

p(y) = ey+ ify 0 ify <

;

and

w() =

1 if (0, 1)0 if (0, 1)

.

Thus,

w(|y) = ey+min{1,y}

0 ey+d

= e

emin{1,y} 1,

for 0 min{1, y}, and w(|y) = 0 otherwise. This implies:

MMSE(y) = min{1,y}0 e

d

emin{1,y} 1 =

[min{1, y} 1]emin{1,y} + 1

emin{1,y} 1 ;

MAP(y) = arg

max

0min{1,y}e

= min{1, y};

and ABS(y)0

ed=1

2

emin{1,y} 1

which has the solution

ABS(y) = log

emin{1,y} + 1

2

.

Exercise 8:

a. We have

w(|y) =ey+e

eyy0 d

=1

y, 0 y,

and w(|y) = 0 otherwise. That is, given y, is uniformly distributed on the interval[0, y]. From this we have immediately that MMSE(y) =ABS(y) =

y2

.b. We have

MMSE=E{V ar(|Y)} .

Sincew(|y) is uniform on [0, y], V a r(|Y) = Y2

12 . We have

p(y) =ey y0

d= yey, y >0,

from which

MMSE=E{Y2}

12 =

1

12

0

y3ey|dy= 3!

12=

1

2.

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c. In this case,

p(y) =n

k=1

eyk+, if0<


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Exercise 15:

We have

p(y) =ey

y! , y 0, 1, . . . .

Thus

logp(y) =

(+y log ) = 1 +

y

,

and2

2logp(y) =

y

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So the CRLB is2n

k=1s4k

(1+s2k)2

.

c. With s2k

= 1, the likelihood equation yields the solution

ML(y) =

1

n

nk=1

y2k

1,

which is seen to yield a maximum of the likelihood function.d. We have

E

ML(Y)

=

1

n

nk=1

E

Y2k

1 =.

Similarly, since the Y ksare independent,

V ar

ML(Y

= 1n2

nk=1

V ar

Y2k

= 1n2

nk=1

2(! + )2 =2(1 + )

2

n .

Thus, the bias of the MLE is 0 and the variance of the MLE equals the CRLB. (Hence,the MLE is an MVUE in this case.)

Exercise 22:

a. Note that

p(y) =exp

n

k=1

log F(yk)/

,

which impies that the statisticn

k=1log F(yk) is a complete sufficient statistic for viathe Completeness Theorem for Exponential Families. We have

E

n

k=1

log F(Yk)

= nE{log F(Y1)} =

n

log F(Y1) [F(y1)](1)/ f(y1)dy1.

Noting thatd log F(y1) = f(y1)F(y1)

dy1,and that [F(y1)]1/ =exp{log F(y1)/}, we can make

the substitution x = log F(y1) to yield

E

n

k=1 log F(Yk)

=

n

0 xe

x/

dx= n.

Thus, we haveE

MV(Y)

= ,

which implies thatMV is an MVUE since it is an unbiased function of a complete sufficientstatistic.

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and

ys= 1

n

nk=1

yksin

k

2

1

n

n/2k=1

(1)k+1y2k1.

b. Appending the prior to the above problem yields MAP estimates:

MAP =ML,

and

AM AP =AML+

rn

2+ 2(1+)

2

n

1 + ,

where 22

n2 .

c. Note that, when (and the prior diffuses), the MAP estimate ofA doesnot approach the MLE ofA.However, asn , the MAP estimate does approach theMLE.

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An Introduction to Signal Detection and Estimation - Second Edition Chapter IV: Selected Solutions

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