**R E S E A R C H**

**Open Access**

## Probability inequality concerning a

## chirp-type signal

### Kanthi Perera

**_{Correspondence:}

kanthip@pdn.ac.lk Department of Engineering Mathematics, University of Peradeniya, Peradeniya, Sri Lanka

**Abstract**

Probability inequalities of random variables play important roles, especially in the theory of limiting theorems of sums of independent random variables. In this note, we give a proof of the following:

Let{

*t*}be a sequence of independent random variables such that

*E*(

*t) = 0,*

*E*(2

*t*) =*v*<∞and*E*(*t*4) =

*σ*

<∞. Then
max
0≤*ω*<*π*
0≤*α*<*π*

*n*

*t=1*

*tβ*

*tei(ωt+αt*

2_{)}

2=*Op(nβ*+78

_{).}

_{).}

It is shown that this result will be useful in estimating the parameters of a chirp-type statistical model and in establishing the consistency of estimators.

**Keywords:** probability inequality; independent random variables; chirp-type signal;
consistency

**1 Introduction**

In [] Whittle () considered the problem of estimating the parameters of a sine wave:

*Xt*=*A*cos*ωt*+*B*sin*ωt*+*t*,

where*Xt*’s are the observations and*t*’s are independent, identically distributed random

variables with mean zero and ﬁnite(unknown) variance. Whittle’s solution to the problem
of estimating the parameters and the proof of consistency of parameters of the above sine
wave used arguments which are not mathematically rigorous. In , a rigorous solution
to Whittle’s problem was given by Walker []. In his proof of consistency of parameters,
he used the following*Op*result:

max
≤*ω*≤*π*

*n*

*t=*

*Yteiωt*

=*Op*

*n*_{,}

where{Y*t*}is a linear process.

In this paper, we extend the above*Op*result to obtain

max
≤*ω*<*π*

≤*α*<*π*

*n*

*t=*

*tβtei(ωt+αt*

_{)}

=*Op*

*nβ*+_{.} _{()}

We came across this problem while attempting to establish the consistency of the pa-rameters of the model

*Xt*=*A*cos

*ωt*+*αt*+*B*sin*ωt*+*αt*+*t*,

where*Xt*’s and*t*’s are as mentioned above. Models of this type are referred to as ‘chirp’

models [, ] and [], and they have drawn the attention of many researchers [, ] and []. Although we tried to ﬁnd our main result given in () in the literature, we were unable to ﬁnd one. We make use of the following basic results to establish our main result.

**2 Basic results**

**Deﬁnition ** Let{*Xn*}be a sequence of random variables. We say that*Xn*=*Op*() if for

given> , there exists*M*such that

*P*|X*n*|>*M*

< for all*n*.

Let (*an*) be a sequence of nonzero real numbers. We say that*Xn*=*Op*(*an*) if*Xn*/*an*=*Op*().

**Lemma ** *Let*{*Xn*} *be a sequence of random variables*,*and let*{*an*}*and*{*bn*}*be two *

*se-quence of nonzero real numbers*.*Then*

*Xn*=*Op*(*bn*) ⇒ *anXn*=*Op*(*anbn*).

**Lemma ** *Let*{X*n*}*be a sequence of random variables*,*let*{a*n*}*be a sequence of real *

*num-bers*,*and let q be a positive real number*.*Then*

*Xn*=*Op*(*an*) ⇒ *Xnq*=*Op*

*aq _{n}*.

**Deﬁnition ** Let{X*n*}be a sequence of random variables. We say that*Xn*=*op*() if*Xn*→

in probability. (So,*Xn*=*op*() if for each≥, we havelim*n*→∞*P*(|*Xn*|>) = .)

**Lemma ** *Let*{*Xn*}*be a sequence of random variables*,*and let*{*an*}*be a sequence of real*

*numbers with an*→*as n*→ ∞.*If Xn*=*Op*(),*then anXn*
*P*

–→.

**Lemma ** *Let*{*Xn*}*be a sequence of random variables*,*and letβbe a positive real number*.

*Then*

*Xn*=*Op*

*nβ* ⇒ *Xn*=*op*().

**Lemma ** *Let*{X*n*}*be a sequence of random variables*,*and letβ* *be any real number*.*If*

*Xn*=*Op*(*nβ*)*and Yn*=*Op*(*nβ*),*then Xn*+*Yn*=*Op*(*nβ*).

**Lemma ** *Letμbe any real number*.*Let*{X*n*}*be a sequence of random variables such that*

*E*(|X*n*|) =*O*(*nμ*).*Then Xn*=*O*(*nμ*).

**Lemma ** *Let at,u*;*t*= , , . . . ,*n*,*u*= , , . . . ,*n be complex numbers*.*Then*
*n*
*t=*
*n*
*u=*

*at,u*=
*n–*
*s=*
*n–s*
*t=*

[*at,t+s*+*at+s,t*] +
*n*

*t=*

*at,t*.

*Proof* The*n _{t=}n_{u=}at,u*is the sum of the terms in the table:

*a*, *a*, *a*, · · · *a*,n– *a*,n

*a*, *a*, *a*, · · · *a*,n– *a*,n

..

. ... ... ... ... ...

*an–,* *an–,* *an–,* · · · *an–,n–* *an–,n*

*an,* *an,* *an,* · · · *an,n–* *an,n*.

We can add by summing along the diagonals. So, the sum is equal to

*n–*
*s=*
*n–s*
*t=*

*at,t+s*+
*n–*
*p=*
*n–p*
*u=*

*au+p,u*+
*n*

*t=*

*at,t*.

Therefore
*n*
*t=*
*n*
*u=*

*at,u*=
*n–*
*s=*
*n–s*
*t=*

[*at,t+s*+*at+s,t*] +
*n*

*t=*

*at,t*. _{}

**Lemma ** *Let bt*;*t*= , , . . . ,*n be complex numbers*.*Then*

*n*
*t=*
*bt*
≤
*n–*
*s=*
*n–s*
*t=*

*btb*¯*t+s*

+
*n*
*t=*

|*bt*|.

*Proof* Since|*n _{t=}bt*|=

*n*
*t=*

*n*

*u=btb*¯*u*, by using*at,u*=*btb*¯*u*in Lemma , then

*n*
*t=*
*bt*
=
*n–*
*s=*
*n–s*
*t=*

[*btb*¯*t+s*+*bt+sb*¯*t*] +
*n*

*t=*

*btb*¯*t*

=
*n–*
*s=*
*n–s*
*t=*

b*tb*¯*t+s*+
*n*

*t=*

|b*t*|

≤
*n–*
*s=*
*n–s*
*t=*

*btb*¯*t+s*

+
*n*
*t=*

|b*t*|. _{}

**3 Main result**

**Theorem ** *Let*{*t*}*be a sequence of independent random variables such that E*(*t*) = ,

*E*(

*t*) =*v*<∞*and E*(*t*) =*σ*<∞.*Then*

max
≤*ω*<*π*

≤*α*<*π*
*n*
*t=*
*tβ _{}*

*tei(ωt+αt*

_{)}

=*Op*

*Proof* Letting*bt*=*tβtei(ωt+αt*

_{)}

in Lemma , we see that

*n*
*t=*

*tβtei(ωt+αt*

_{)}
≤
*n–*
*s=*
*n–s*
*t=*

*tβ*(*t*+*s*)*βtt+sei(ωt+αt*

_{)}

*e*–i(*ω*(t+s)+*α*(t+s))
+
*n*
*t=*
*tβ _{}*

*tei(ωt+αt*

_{)}
=
*n–*
*s=*
*n–s*
*t=*

*tβ*(*t*+*s*)*βtt+se*–i*ωse*–i*αs*

*e*–i*αts*

+
*n*
*t=*

*t**β _{t}*.

Using the triangle inequality, we obtain

*n*
*t=*

*tβtei(ωt+αt*

_{)}
≤
*n–*
*s=*
*n–s*
*t=*

*tβ*(*t*+*s*)*βtt+se*–i*αts*

+
*n*
*t=*

*t**β _{t}*. ()

Fix*s*and consider|*n–s _{t=}tβ*

_{(}

_{t}_{+}

_{s}_{)}

*β*

_{}*tt+se*–i*αts*|. By Lemma with*bt*=*tβ*(*t*+*s*)*βtt+se*–i*αts*,

we obtain
*n–s*
*t=*

*tβ*(*t*+*s*)*β _{}*

*tt+se*–i*αts*

≤
*n–s–*_{}
*p=*
*n–s–p*_{}
*t=*

*tβ*(*t*+*s*)*β*(*t*+*p*)*β*(*t*+*p*+*s*)*βtt+se*–i*αtst+pt+p+se*i*αs(t+p)*

+
*n–s*
*t=*

*tβ*(*t*+*s*)*βtt+se*–i*αts*,

and it follows from the triangle inequality that

*n–s*
*t=*

*tβ*(*t*+*s*)*βtt+se*–i*αts*

≤
*n–s–*_{}
*p=*
*n–s–p*_{}
*t=*

*tβ*_{(}_{t}_{+}_{s}_{)}*β*_{(}_{t}_{+}_{p}_{)}*β*_{(}_{t}_{+}_{p}_{+}_{s}_{)}*β _{tt+st+pt+p+s}*
+

*n–s*

*t=*

*tβ*_{(}_{t}_{+}_{s}_{)}*β _{}*

*tt+s* .

Therefore
*n–s*
*t=*

*tβ*(*t*+*s*)*βtt+se*–i*αts*

≤√
_{n–s–}*p=*
*n–s–p*_{}
*t=*

*tβ*(*t*+*s*)*β*(*t*+*p*)*β*(*t*+*p*+*s*)*βtt+st+pt+p+s*

+
*n–s*
*t=*

*t**β*_{(}_{t}_{+}_{s}_{)}*β _{}*

*tt+s*

/

Substituting this in equation (), we obtain

*n*

*t=*

*tβtei(ωt+αt*

_{)}

≤

*n–*

*s=*

*n–s–*

*p=*

*n–s–p*_{}

*t=*

*tβ*(*t*+*s*)*β*(*t*+*p*)*β*(*t*+*p*+*s*)*βtt+st+pt+p+s*

+

*n–s*

*t=*

*t**β*(*t*+*s*)*β** _{t}_{t+s}*

/

+

*n*

*t=*

*t**β _{t}*.

It follows that

max
≤*ω*<*π*

≤*α*<*π*

*n*

*t=*

*tβ _{}*

*tei(ωt+αt*

_{)}

≤√

*n–*

*s=*

_{n–s–}

*p=*

*n–s–p*_{}

*t=*

*tβ*(*t*+*s*)*β*(*t*+*p*)*β*(*t*+*p*+*s*)*βtt+st+pt+p+s*

+

*n–s*

*t=*

*t**β*_{(}_{t}_{+}_{s}_{)}*β _{}*

*tt+s*

/

+

*n*

*t=*

*t**β _{}*

*t*.

Taking the expectation of both sides of the above inequality, we obtain

*E*

max
≤*ω*<*π*

≤*α*<*π*

*n*

*t=*

*tβtei(ωt+αt*

_{)}

≤√

*n–*

*s=*

*E*
_{n–s–}

*p=*

*n–s–p*_{}

*t=*

*tβ*(*t*+*s*)*β*_{(}
*t*+*p*)*β*_{(}

*t*+*p*+*s*)*β _{}*

*tt+st+pt+p+s*

+

*n–s*

*t=*

*t**β*(*t*+*s*)*β** _{t}_{t+s}*

/

+

*n*

*t=*

*t**βE _{t}*.

We claim that

*E*

max
≤*ω*<*π*

≤*α*<*π*

*n*

*t=*

*tβ _{}*

*tei(ωt+αt*

_{)}

=*On**β*+/_{.}

We note that since

*n*

*t=*

*t**β _{E}_{}*

*t*

=

*n*

*t=*

*t**β _{v}*

_{=}

**

_{O}_{n}*β*+

_{≤}

**

_{O}_{n}*β*+/

_{,}

we can ignore the term*n _{t=}t*

*β*

_{E}_{(}

**

_{}Taking*x*=|z|/_{in Schwarz’s inequality}_{E|x| ≤}_{(}* _{E|x|}*

_{)}/

_{yields}

_{E}_{(}

_{|z|}/

_{)}

_{≤}

_{(}

_{E|z|}_{)}/

_{. Using}

this on the right-hand side of the above inequality, we get

*E*

max
≤*ω*<*π*

≤*α*<*π*

*n*

*t=*

*tβtei(ωt+αt*)

≤

*n–*

*s=*

_{n–s–}

*p=*

*E*

*n–s–p*_{}

*t=*

*tβ*(*t*+*s*)*β*(*t*+*p*)*β*(*t*+*p*+*s*)*βtt+st+pt+p+s*

+

*n–s*

*t=*

*t**β*(*t*+*s*)*βE _{t}*

**

_{t+s}/

.

Using Schwarz’s inequality again on the ﬁrst term on the right, we get

*E*

max
≤*ω*<*π*

≤*α*<*π*

*n*

*t=*

*tβtei(ωt+αt*

_{)}

≤

*n–*

*s=*

_{n–s–}

*p=*

*E*

_{n–s–p}

*t=*

*tβ*_{(}_{t}_{+}_{s}_{)}*β*_{(}_{t}_{+}_{p}_{)}*β*_{(}_{t}_{+}_{p}_{+}_{s}_{)}*β _{}*

*tt+st+pt+p+s*

/

+

*n–s*

*t=*

*t**β*(*t*+*s*)*βE _{t}*

**

_{t+s}/

=

*n–*

*s=*

_{n–s–}

*p=*

*E*

_{n–s–p}

*t=*

*t**β*(*t*+*s*)*β*(*t*+*p*)*β*(*t*+*p*+*s*)*β _{t}*

* *

_{t+s}**

_{t+p}_{t+p+s}

+*E*

*n–s–p*_{}

*u,v=*
*u<v*

*uβ*(*u*+*s*)*β*(*u*+*p*)*β*(*u*+*p*+*s*)*βvβ*(*v*+*s*)*β*(*v*+*p*)*β*(*v*+*p*+*s*)*β*

*uu+su+pu+p+svv+sv+pv+p+s*

/

+

*n–s*

*t=*

*t**β*(*t*+*s*)*βE _{t}*

**

_{t+s}/

.

So,

*E*

max
≤*ω*<*π*

≤*α*<*π*

*n*

*t=*

*tβ _{t}_{e}i(ωt+αt*

_{)}

≤

*n–*

*s=*

_{n–s–}

*p=*

_{n–s–p}

*t=*

*t**β*(*t*+*s*)*β*(*t*+*p*)*β*(*t*+*p*+*s*)*βE _{t}*

**

_{t+s}**

_{t+p}**

_{t+p+s}+

_{n–s–p}

*u,v=*
*u<v*

*uβ*_{(}_{u}_{+}_{s}_{)}*β*_{(}_{u}_{+}_{p}_{)}*β*_{(}_{u}_{+}_{p}_{+}_{s}_{)}*β _{v}β*

_{(}

_{v}_{+}

_{s}_{)}

*β*

_{(}

_{v}_{+}

_{p}_{)}

*β*

_{(}

_{v}_{+}

_{p}_{+}

_{s}_{)}

*β*

*E*(*uu+su+pu+p+svv+sv+pv+p+s*)

/

+

*n–s*

*t=*

*t**β*_{(}_{t}_{+}_{s}_{)}*β _{E}_{}*

*tt+s*

/

Consider*E*(*uu+su+pu+p+svv+sv+pv+p+s*). Since*u*≤*v*,*v*+*p*+*s*is the unique largest

sub-script, the*t*’s are independent and*E*(*t*) = , we have

*E*(*uu+su+pu+p+svv+sv+pv+p+s*) =*E*(*uu+su+pu+p+svv+sv+p*)*E*(*v+p+s*) = .

Therefore

*E*

max
≤*ω*<*π*

≤*α*<*π*

*n*

*t=*

*tβtei(ωt+αt*

_{)}

≤

*n–*

*s=*

_{n–s–}

*p=*

_{n–s–p}

*t=*

*t**β*(*t*+*s*)*β*(*t*+*p*)*β*(*t*+*p*+*s*)*βE _{t}*

**

_{t+s}**

_{t+p}**

_{t+p+s}/

+

*n–s*

*t=*

*t**β*(*t*+*s*)*βE _{t}*

**

_{t+s}/

.

Now consider the term*E*(

*tt+s* ). Since*E*(*t*) =*v*≤ ∞, and the*t*’s are independent,

*n–s*

*t=*

*t**β*(*t*+*s*)*βE _{t}*

* ≤*

_{t+s}*Kn*

*β*+.

Similarly, since*E*(* _{t}*) =

*v*≤ ∞,

*E*(

*) =*

_{t}*σ*≤ ∞, and the

*t*’s are independent,

*n–s–p*_{}

*t=*

*t**β*(*t*+*s*)*β*(*t*+*p*)*β*(*t*+*p*+*s*)*βE** _{t}_{t+s}*

* *

_{t+p}*=*

_{t+p+s}*Kn*

*β*+.

Thus

*E*

max
≤*ω*<*π*

≤*α*<*π*

*n*

*t=*

*tβ _{}*

*tei(ωt+αt*

_{)}

≤

*n–*

*s=*

_{n–s–}

*p=*

*Kn**β*+/

+*Kn**β*+

/

=

*n–*

*s=*

*nKn**β*+/+*Kn**β*+/

=

*n–*

*s=*

*Kn**β*+//

=*nKn**β*+/

=*On**β*+/_{.}

It follows by virtue of Lemma that

max
≤*ω*<*π*

≤*α*<*π*

*n*

*t=*

*tβtei(ωt+αt*

_{)}

=*Op*

Using Lemma , we obtain

max
≤*ω*<*π*

≤*α*<*π*

*n*

*t=*

*tβtei(ωt+αt*

_{)}

=*Op*

*nβ*+/,

which completes the proof of the theorem.

**Competing interests**

The author declares that she has no competing interests.

Received: 21 August 2012 Accepted: 11 March 2013 Published: 27 March 2013
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doi:10.1186/1029-242X-2013-131