An Introduction to Statistical Signal Processing

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Московский государственный юридический университет им. О.Е. Кутафина

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Социология

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Integrals of the form

∞ xke−αx dx

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A.6. PROBLEMS

415

(b)	0	∞ dx	0x dye−3y

25. Evaluate the following integrals:
(a)	x,y:\|x\|+\|y\|≤r dx dy

(b)	0	∞ dx	x∞ dye−y

(c)	x,y:0≤x2≤y≤1 dx dy

416	APPENDIX A. PRELIMINARIES

Appendix B

Sums and Integrals

In this appendix a few useful deﬁnitions and results are gathered for reference.

B.1 Summation

The sum of consecutive integers.

n
	n(n + 1)
k =	2	,	(B.1)
	2

k=1

Proof: The result is easily proved by induction, which requires demonstrating the truth of the formula for n = 1 (which is obvious) and showing

that if the formula is true for any positive integer n, then it must also be

true for n + 1. This follows since if Sn = k=1 k and we assume that Sn = n(n + 1)/2, then necessarily

Sn+1 = Sn + (n + 1)

=(n + 1)(n2 + 1)

=(n + 1)(n + 2), 2

proving the claim.

The sum of consecutive squares of integers.

n
	1			1			1
k2 =		n3	+		n2	+		n,	(B.2)
	3			2			6

k=1

417

418	APPENDIX B. SUMS AND INTEGRALS

The sum can also be expressed as

n k2 = (2n + 1)(n + 1)n. 6

k=1

Proof: This can also be proved by induction, but for practice we note another approach. Just as in solving di erential or di erence equations, one can guess a general form of solution and solve for unknowns. Since summing k up to n had a second order solution in n, one might suspect that solving for a sum up to n of squares of k would have a third order solution in n, that is, a solution of the form f(n) = an3 + bn2 + cn + d for some real numbers a, b, c, d. Assume for the moment that this is the case,

then if f(n) = n k2, clearly n2 = f(n) − f(n − 1) and hence with a

k=1

little algebra

n2 =	an3 + bn2 + cn + d − a(n − 1)3 + b(n − 1)2 + c(n − 1) + d
=	3an2 + (2b − 3a)n + (a − b + c).

This can only be true for all n hover if 3a = 1 so that a = 1/3, if 2b−3a = 0 so that b = 3a/2 = 1/2, and if a − b + c = 0 so that c = b − a = 1/6. This leaves d, but the initial condition that f(1) = 1 implies d = 0.

The geometric progression

Given a complex constant a,	−

n−1 ak =	1 − an	,	(B.3)
	1 a

k=0

and if |a| < 1 this sum is convergent and

∞	−
	−
ak =	1		.	(B.4)
1		a

k=0

Proof: There are, in fact, many ways to prove this result. Perhaps the

simplest is to deﬁne the sum with n terms Sn = n−1 ak and observe that

k=0

	n−1	n−1
(1 − a)Sn =
(1 − a)Sn =	ak − a ak
	k=0	k=0
	n−1	n
=	ak −	ak
	k=0	k=1
= 1 − an,

B.1. SUMMATION

419

proving (B.3). Other methods of proof include induction and solving the di erence equation Sn = Sn−1 + an−1. Proving the ﬁnite n result gives the inﬁnite sum since if |a| < 1,

∞	−
	−
ak = lim Sn =	1		.	(B.5)
n→∞	1	a

k=0

For the reader who might be rusty with limiting arguments, this follows

since

|a|n

n − 1

− a|

|1 − a|

|1 − a| →

as n → ∞ since by assumption |a| < 1.

First moment of the geometric progression

Given q (0, 1),

∞

kqk−1 =

(B.6)

−

q)2

k=0

Proof: Since kqk−1

qk and since we can interchange di erentiation

and summation,

∞

kqk−1 =

qk =

− q)−1 ,

k=0

where we have used the geometric series sum formula.

Second moment of the geometric progression

Given q (0, 1),

∞

(B.7)

k2qk−1

(1 q)3

q)2

k=0

−

Take a second derivative of a geometric progression to ﬁnd

∞

k(k

−

1)qk−2

dq2

k=0

=1q ∞ k(k − 1)qk−1

k=0

=1q ∞ k2qk−1 − 1q ∞ kqk−1

	k=0			k=0
1	∞				1

= q		k2qk−1 −	(1		−	q)2

k=0

420

APPENDIX B. SUMS AND INTEGRALS

and

∞

qk =

−

dq2

k=0

q)3

so that

∞

−

q)3

+ (1

−

q)2

k2qk−1 =

k=0

proving the claim.

B.2 Double Sums

The following lemma provides a useful simpliﬁcation of a double summation that crops up when considering sample averages and laws of large numbers.

Lemma B.1 Given a sequence {an},

N−1 N−1	N−1

ak−l =	(N − \|n\|)an.
k=0 l=0	n=−N+1

Proof: This result can be thought in terms of summing the entries of a matrix A = {Ak,l; k, l ZN } which has the property that all elements along any diagonal are equal, i.e., Ak,l = ak−l for some sequence a. (As mentioned in the text, a matrix of this type is called a Toeplitz matrix. To sum up all of the elements in the matrix note that the main diagonal has N equal values of a0, the next diagonal up has N − 1 values of a1, and so on with the nth diagonal having N − n equal values of an. Note there is only one element aN−1 in the top diagonal.

The next result is a limiting result for sums of the type considered in

the previous lemma.
Lemma B.2 Suppose that {an;			n Z} is an absolutely summable se-
quence, i.e., that		∞
		∞
			\|an\| < ∞.
			\|an\| < ∞.
Then		n=−∞
Then		N−1					∞
		N−1					∞
lim			\|n\|	)a
lim		(1		)a	n	=	a .
N	→∞		− N		n		n
	→∞	n=−N+1					n=−∞
		n=−N+1					n=−∞

Comment: The limit should be believable since the multiplier in the summand tends to 1 for each ﬁxed n as N → ∞.

B.3. INTEGRATION		421
Proof: Absolute summability implies that the inﬁnite sum exists and
∞			N−1
	an = lim
			an
n=−∞	N→∞
		n=−N+1
so the result will follow if we show that
lim	N−1	\|n\|an = 0.

N→∞		N
	n=−N+1

Since the sequence is absolutely summable, given an arbitrarily small * > 0 we can choose an N0 large enough to ensure that for any N ≥ N0 we have

|an| < *.

n: n|≥N

For any N ≥ N0 we can then write

N−1

|n|a

N−1

|n| a

≤

n=−N+1

|n|

n:N0≤|n|≤N−1

| n|

n: n|≤N0−1

|n|

≤

| n|

n: n|≤N0−1

n:|n|≥N0

≤

|n|

+ *.

n|≤N0−1

Letting N → ∞ the remaining term can be made arbitrarily small, proving the result.

B.3 Integration

A basic integral in calculus and engineering is the simple integral of an exponential, which corresponds to the sum of a “discrete time exponential,” a geometric progression. This integral is most easily stated as

∞

e−r dr = 1.

(B.8)

422 APPENDIX B. SUMS AND INTEGRALS

If α > 0, then making a linear change of variables as r = αx or x = r/α implies that dr = αdx and hence

can be evaluated by parts, or by using the same trick that worked for the geometric progression. Take the kth derivative of both sides of B.9 with respect to α:

	dk	0	∞		dk
			e−αx dx =			α−1
	dαk				dαk
0∞(−x)ke−αx dx = (−1)kk!α−k−1
0		∞ xke−αx dx =		k!α−k−1.			(B.10)
Computations using a Gaussian pdf follow from the basic integral
			∞	2
			I = −∞ e−x		dx.

This integral is a bit trickier than the others considered. It can of course be found in a book of tables, but again a proof is provided to make it seem a bit less mysterious. The proof is not di cult, but the initial step may appear devious. Simplify things by considering the integral

I = ∞ e−x2 dx

2 0

and note that this one dimensional integral can also be written as a two dimensional integral:


I					∞
		=	8( 0		e−x2 dx)2
	2
			8
		=		( ∞ e−x2 dx)( ∞ e−y2 dy)
			0		0
			8
				∞ ∞

=e−(x2+y2 dx dy.

B.4. THE LEBESGUE INTEGRAL

423

This subterfuge may appear to actually complicate matters, but it allows

us to change to polar coordinates using r =

+ y2

x = r cos(θ), y =

r sin(θ), and dx dy = rdr dθ to obtain

π/2

∞

(

)2 =

re−r

dr dθ

∞

re−r

dr.

Again this might appear to have complicated matters by introducing the

extra factor of r, but now a change of variables of u = r2 or r =

√

implies

that dr = du/2√

so that

∞ 1

(

e−u du =

using (B.8). Thus

∞

−∞ e−x

dx = √π.

(B.11)

√

This is commonly expressed by changing variables to r/ 2 = x so that dx = dr/sqrt2 and the result becomes

∞	r2
∞	r2
−∞ e−	2	dr =	√2π,	(B.12)

from which it follows that a 0 mean unit variance Gaussin pdf has unit integral. The general case is handled by a change of variables. In the following integral change variables by deﬁning r = (x − m)/σ so that dx =

σdr

∞	1			(x−m)2		1			∞
−∞	√		e−		dx =	√			−∞ e−r2 σdr	(B.13)
				2σ2
		2σ2						2σ2
						√
				=			2π			(B.14)
						√
							2π
				=		1.				(B.15)

B.4 The Lebesgue Integral

This section provides a brief introduction to the Lebesgue integral, the calculus that underlies rigorous probability theory. In the authors view the Lebesgue integral is not nearly as mysterious as it is sometimes suggested in

424	APPENDIX B. SUMS AND INTEGRALS

the engineering literature and that, in fact, it has a very intuitive engineering interpretation and avoids the rather clumsy limits required to study the Riemann integral. We here present a few basic deﬁnitions and properties without proof. Details can be found in most any book on measure theory or integration and in many books on advanced probability, including the ﬁrst author’s Probability, Random Processes, and Ergodic Properties[22].

Suppose that (Ω, F, P ) is a probability space as deﬁned in chapter 2. For simplicity we focus on real-valued random variables, the extensions to complex random variables and more general random vectors are straightforward. The integral or expectation of a random variable f deﬁned on this probability space is deﬁned in a sequence of steps treating random variables of increasing generality.

First suppose that f takes on only a ﬁnite number of values, for example

N
i	(x); x ,	(B.16)
f(x) = ai1Fi	(x); x ,	(B.16)
=1

where it is assumed that Fi F for all i. A discrete random variable of this form is sometimes called a simple function. The (Lebesgue) integral of

f or expectation of f is then deﬁned by
	N
f dP =
f dP =	aiP (Fi).	(B.17)

i=1

The integral is also written as f(x) dP (x) and is also denoted by E(f). It is easy to see that this deﬁnition reduces to the Riemann integral.

The deﬁnition is next generalized to all nonnegative random variables by means of a sequences of quantizers which map the random variable into an ever better approximation with only a ﬁnite possible number of outputs. Deﬁne for each real r and each positive integer n the quantizer

(k − 1)2−n qn(r) = −(k − 1)2−n

−n

r ≥ n

(k − 1)2−n ≤ r < k2−n, k = 1, 2, . . . , n2n −(k − 1)2−n > r ≥ k2−n, k = 1, 2, . . . , n2n r < −n

(B.18)

The sequence of quantizers is asymptotically accurate in the sense that

f(x) = lim qn(f(x))	(B.19)
n→∞

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