Generalized Cost Function Based Forecasting for Periodically ...

4

It may be noted that the sets P U and P L may be empty.

In other words, U(˜z t , z t ) and L(˜z t , z t ) may be defined

in such a way that their partial derivatives with respect

to ˜z t is continuous **for** z t in the respective intervals.

Lemma 1: Let U(˜z t , z t ) and L(˜z t , z t ) be defined in

such a way that both Assumptions 1 and 2 hold, then

the first derivative of cost function C t (˜z t ) is a strictly

increasing continuous function of ˜z t in the interval

[0, ∞].

Proof: From (2), using Leibniz rule, we have

dC t (˜z t )

= [U(˜z t , ˜z t ) − L(˜z t , ˜z t )]f t−1 (˜z t )

d˜z t

+

+

∫ ˜zt

0

∫ ∞

∂U(˜z t , z t )

∂ ˜z t

f t−1 (z t ) dz t

˜z t

∂L(˜z t , z t )

∂ ˜z t

f t−1 (z t ) dz t . (4)

Following condition (i) of Assumptions 1 and 2, the first

term in (4) vanishes. Then, (4) is simplified as

dC t (˜z t )

d˜z t

= I U (0, ˜z t ; ˜z t ) + I L (˜z t , ∞; ˜z t ). (5)

Following Remark 3 and condition (v) of Assumption

2, both I U (0, ˜z t ; ˜z t ) and I L (˜z t , ∞; ˜z t ) are Riemann integrable.

There**for**e, it is clearly evident that the first

derivative of C(˜z t ) is a continuous function of ˜z t .

Let c 1 , c 2 ∈ [0, ∞] with c 1 < c 2 and define

D = dC t(˜z t )

d˜z t

∣ ∣∣

˜zt=c 2

− dC t(˜z t )

d˜z t

∣ ∣∣

˜zt=c 1

.

Then, it follows from (5) that

D = I U (0, c 2 ; c 2 ) + I L (c 2 , ∞; c 2 ) − I U (0, c 1 ; c 1 )

−I L (c 1 , ∞; c 1 )

= I U (0, c 1 ; c 2 ) + I U (c 1 , c 2 ; c 2 ) + I L (c 2 , ∞; c 2 )

−I U (0, c 1 ; c 1 ) − I L (c 1 , c 2 ; c 1 ) − I L (c 2 , ∞; c 1 )

= {I U (0, c 1 ; c 2 ) − I U (0, c 1 ; c 1 )}

+ {I U (c 1 , c 2 ; c 2 ) − I L (c 1 , c 2 ; c 1 )}

+ {I L (c 2 , ∞; c 2 ) − I L (c 2 , ∞; c 1 )} . (6)

It may be noted that both I U (a, b; m) and I L (a, b; m)

where ˜z t = m **for** some m > 0 are evaluated as follows:

∫ b

∂U(˜z t , z t )

∣ ∣∣

I U (a, b; m) =

∂ ˜z f t−1(z t ) dz t

t ˜zt=m

and

I L (a, b; m) =

a

∫ b

a

∂L(˜z t , z t )

∂ ˜z t

∣ ∣∣

˜zt=m f t−1(z t ) dz t .

From condition (iii) of Assumption 1, we have

I U (0, c 1 ; c 2 ) − I U (0, c 1 ; c 1 ) ≥ 0 (7)

and from condition (iii) of Assumption 2, we have

I L (c 2 , ∞; c 2 ) − I L (c 2 , ∞; c 1 ) ≥ 0. (8)

Then, we also infer from condition (ii) of Assumption 1

that

I U (c 1 , c 2 ; c 2 ) > 0

and from condition (ii) of Assumption 2 that

implying

I L (c 1 , c 2 ; c 1 ) < 0

I U (c 1 , c 2 ; c 2 ) − I L (c 1 , c 2 ; c 1 ) > 0. (9)

Following (7), (8) and (9), we have D > 0. Since, c 1

and c 2 are chosen arbitrarily, the lemma holds true.

Theorem 1: Under the Assumptions 1 and 2, there

exists a unique z ∗ t such that ˜z t = z ∗ t minimizes the cost

function C(˜z t ) among all ˜z t ∈ [0, ∞].

Proof: Since the first derivative of the cost function

C(˜z t ) is a strictly increasing continuous function in the

interval [0, ∞] (from Lemma 1), it should attain its

minimum at ˜z t = 0 and its maximum at ˜z t = ∞ in

the interval [0, ∞]. Below, we investigate these boundary

values in order to prove the desired result.

Recall from (5) that

dC(˜z t )

d˜z t

= I U (0, ˜z t ; ˜z t ) + I L (˜z t , ∞; ˜z t ).

Then, we have

dC(˜z t )

d˜z t

∣ ∣∣

˜zt=0 = I U(0, 0; 0) + I L (0, ∞; 0). (10)

The first term in (10) vanishes, while the second term

I L (0, ∞; 0) < 0 (11)

following condition (ii) of Assumption 2 implying that

the minimum value of the first derivative of cost function

is negative.

Similarly,

dC(˜z t )

d˜z t

∣ ∣∣

˜zt=∞ = lim

˜z t→∞ I U(0, ˜z t ; ˜z t ) + lim

˜zt→∞ I L(˜z t , ∞; ˜z t ).

(12)

The second term in (12) vanishes, whereas the first term

lim I U(0, ˜z t ; ˜z t ) > 0 (13)

˜z t→∞

as a result of condition (ii) of Assumption 1 implying

thereby that the maximum value of the first derivative of

cost function is positive.

Then, from Weierstrass intermediate value theorem

[13], there exists at least one zt

∗ ∈ (0, ∞) such

that the derivative vanishes. However, from Lemma 1, it

follows that zt ∗ should be unique thereby, guaranteeing

a unique minimizer **for** the cost function C(˜z t ).