JEAN-PAUL PAQUIN, ANNICK LAMBERT AND ALAIN CHARBONNEAU 295
To simplify the notation, let us defineαt=(1+kc)−t. Equation (1) becomes:
P ̃=
∑n
t= 1
αtε ̃t
We must therefore verify, in accordance with the CLT, whether this weighted average
of random error terms converges towards a normal probability distribution. Given our
initial assumptions we deduct:
E(P ̃)=
∑n
t= 1
αtE(ε ̃t)= 0
whereas
V(P ̃)=
∑n
t= 1
α^2 tV(ε ̃t)=
∑n
t= 1
α^2 t.
Therefore, to verify the CLT we must demonstrate that:
nlim→∞
∑n
t= 1
αt ̃εt
√
∑n
t= 1
α^2 t
→N(0, 1)
LetφX ̃(h)=E(eihX ̃)= 1 +
∑∞
t= 1
(ih)t
t!μtbe the characteristic function of any random vari-
ableX ̃. Given that theε ̃t’s are independent in probability, we may write the characteristic
function of their weighted sum in term of their argument as:
φ∑n
t= 1
αtε ̃t
√∑n
t= 1
α^2 t
(h)=E
(
e
ih√∑∑αt ̃εt
α^2 t
)
=
∏n
t= 1
e
√ihα∑tε ̃t
α^2 t=
∏n
t= 1
φ ̃ε
√αth
∑
α^2 t
Let us take the logarithm of the characteristic function in term of its arguments and thus
define thefunction:
√∑αt ̃εt
∑α 2
t
=
∑n
t= 1
logφε ̃t
αth
√∑
α^2 t
=
∑n
t= 1
log
1 +i√αt
∑
α^2 t
hμ 1 −
1
2
√αt
∑
α^2 t
2
h^2 μ 2
−
i
3!
√αt
∑
α^2 t
3
h^3 μ 3 +
1
4!
√αt
∑
α^2 t
4
h^4 μ 4 +...