which gives , and the result follows.

Let U be a random variable, uniformly distributed on {0, 1}. We define the process (Y _t ) _{t = 0, 1, …} by

for any ω ∈ Ω and any t ≥ 0. The process (Y _t ) is stationary. We have in particular EY _t  = 0 and Cov(Y _t , Y _t + h ) = (−1) ^h . With probability 1/2, the realisation of the stationary process (Y _t ) will be the sequence {(−1) ^t } (and with probability 1/2, it will be {(−1) ^t + 1}).

This example leads us to think that it is virtually impossible to determine whether a process is stationary or not, from the observation of only one trajectory, even of infinite length. However, practitioners do not consider {(−1) ^t } as a potential realisation of the stationary process (Y _t ). It is more natural, and simpler, to suppose that {(−1) ^t } is generated by the non‐stationary process (X _t ).

Let Ω^* = {ω ∣ X _2t  = 1, X _2t + 1 = 0, ∀ t}. If (X _t ) is ergodic and stationary, the empirical means and both converge to the same limit P[X _t  = 1] with probability 1, by the ergodic theorem. For all ω ∈ Ω^* these means are, respectively, equal to 1 and 0. Thus P(Ω^*) = 0. The probability of such a trajectory is thus equal to zero for any ergodic and stationary process.

This value can be arbitrarily larger than 1, which is the value of the asymptotic variance of the empirical autocorrelations of a strong white noise.

where ∣b∣ < 1 and (u _t ) is a white noise of variance σ ² . The coefficients b and σ ² are determined by

which gives and σ ² = 2/b .

Since , for k ≠ h the asymptotic variance can be arbitrarily smaller than 1, which corresponds to the asymptotic variance of the empirical autocorrelations of a strong white noise.

1. We have
   
   when n > m and m → ∞. The sequence {u _t (n)} _n defined by is a Cauchy sequence in L ² , and thus converges in quadratic mean. A priori,
   

   exists in ℝ ∪  + {∞}. Using Beppo Levi's theorem,

   

   which shows that the limit is finite almost surely. Thus, as n → ∞, u _t (n) converges, both almost surely and in quadratic mean, to . Since

   

   we obtain, taking the limit as n → ∞ of both sides of the equality, u _t  = au _t − 1 + η _t . This shows that (X _t ) = (u _t ) is a stationary solution of the AR(1) equation.

   Finally, assume the existence of two stationary solutions to the equation X _t  = aX _t − 1 + η _t and u _t  = au _t − 1 + η _t . If , then

   

   which entails

   

   This is in contradiction to the assumption that the two sequences are stationary, which shows the uniqueness of the stationary solution.

2. We have X _t  = η _t  + aη _t − 1 + ⋯ + a ^k η _t − k  + a ^k + 1 X _{t − k − 1} . Since ∣a ∣  = 1,
   

   as k → ∞. If (X _t ) were stationary,

   

   and we would have

   

   This is impossible, because by the Cauchy–Schwarz inequality,

   
3. The argument used in Part 1 shows that
   
   almost surely and in quadratic mean. Since
   
   for all n , ( ) is a stationary solution (which is called anticipative, because it is a function of the future values of the noise) of the AR(1) equation. The uniqueness of the stationary solution is shown as in Part 1.
4. The autocovariance function of the stationary solution is
   

   We thus have Eε _t  = 0 and, for all h > 0,

   

   which confirms that ε _t is a white noise.

With the change of index h = i − ℓ, we obtain

which gives (B.14), using the parity of the autocovariance functions.

for (i, j) ≠ (0, 0) and

Thus

In formula (B.15), we have _ij  = 0 when i ≠ j and _ii  = 1. We also have when i ≠ j and for all i ≠ 0. Since , we obtain

For significance intervals C _h of asymptotic level 1 − α , such that , we have

By definition of C _h ,

Moreover,

We have used the convergence in law of to a vector of independent variables. When the observed process is not a noise, this asymptotic independence does not hold in general.

For m = 20 and α = 5%, this limit is equal to 0.36. The probability of not rejecting the right model is thus low.

Then, step (B.9) yields

Finally, step (B.8) yields

> # reading the SP500 data set
> sp500data <- read.table("sp500.csv",header=TRUE,sep=",")
> sp500<-rev(sp500data$Close) # closing price
> n<-length(sp500)
> rend<-log(sp500[2:n]/sp500[1:(n-1)]); rend2<-rend∧2
> op <- par(mfrow = c(2, 2)) # 2 × 2 figures per page
> plot(ts(sp500),main="SP 500 from 1/3/50 to 7/24/09",
+                ylab="SP500 Prices",xlab="")
> plot(ts(rend),main="SP500 Returns",ylab="SP500 Returns",
+                xlab="")
> acf(rend, main="Autocorrelations of the returns",xlab="",
+                ylim=c(-0.05,0.2))
> acf(rend2, main="ACF of the squared returns",xlab="",
+                ylim=c(-0.05,0.2))
> par(op)

For the multiplicative norm ‖A‖ =  ∑  ∣ a _ij ∣, we have The result follows immediately.

When A is any square matrix, the Jordan representation can be used. Let n _i be the multiplicity of the eigenvalue λ _i . We have the Jordan canonical form A = P ⁻¹ JP , where P is invertible, and J is the block‐diagonal matrix with a diagonal of m matrices J _i (λ _i ), of size n _i  × n _i , with λ _i on the diagonal, 1 on the superdiagonal, and 0 elsewhere. It follows that A ^t  = P ⁻¹ J ^t P , where J ^t is the block‐diagonal matrix whose blocks are the matrices . We have , where N _i is such that . It can be assumed that ∣λ ₁ ∣  >  ∣ λ ₂ ∣  > ⋯ >  ∣ λ _m  ∣ . It follows that

as t → ∞, and the proof easily follows.

and thus

Using Eq. (2.21) and the ergodic theorem, we obtain

Consequently, γ < 0 if and only if ρ(A) < exp(−E log  ∣ z _t ∣).

which completes the proof of 1.

We have shown that, for any , the stationary sequences and have the same top Lyapunov exponent, i.e.

The convergence follows by showing that .

To show that the norm N ₁ is not multiplicative, consider the matrix A whose elements are all equal to 1: we then have N ₁(A) = 1 but N ₁(A ²) > 1.

and

(see Theorem 2.5 and Remark 2.6(1)). The strictly stationary solution satisfies

in ℝ ∪ {+∞}. Moreover,

which gives

Using this relation in the previous expression for , we obtain

If , then the term in brackets on the left‐hand side of the equality must be strictly positive, which gives the condition for the existence of the fourth‐order moment. Note that the condition is not symmetric in α ₁ and α ₂ . In Figure E.2, the points (α ₁, α ₂) under the curve correspond to ARCH(2) models with a fourth‐order moment. For these models,

where is a (weak) white noise. The author correlation of thus satisfies

E.1

Using the MA(∞) representation

we obtain

and

It follows that the lag 1 autocorrelation is

The other autocorrelations are obtained from (E.1) and . To determine the autocovariances, all that remains is to compute

which is given by

We have

The eigenvalues of A ⁽²⁾ are 0, 0, 0 and 3α ² + 2αβ + β ² , thus I ₄ − A ⁽²⁾ is invertible (0 is an eigenvalue of I ₄ − A ⁽²⁾ if and only if 1 is an eigenvalue of A ⁽²⁾ ), and the system (2.63) admits a unique solution. We have

The solution to Eq. (2.63) is

As first component of this vector, we recognise , and the other three components are equal to . Equation (2.64) yields

which gives , but with tedious computations, compared to the direct method utilised in Exercise 2.8.

and the result follows.

1. Subtracting the ( q + 1)th line of (λI _p + q  − A) from the first, then expanding the determinant along the first row, and using Eq. (2.32), we obtain
   
   and the result follows.
2. When the previous determinant is equal to zero at λ = 1. Thus ρ(A) ≥ 1. Now, let λ be a complex number of modulus strictly greater than 1. Using the inequality ∣a − b ∣  ≥  ∣ a ∣  −  ∣ b∣, we then obtain
   
   It follows that ρ(A) ≤ 1 and thus ρ(A) = 1.

The convergence follows from the Borel–Cantelli lemma.

Now, let (X _n ) be an iid sequence of random variables with density f(x) = x ⁻² _x ≥ 1 . For all K > 0, we have

The events {n ⁻¹ X _n  > K} being independent, we can use the counterpart of the Borel–Cantelli lemma: the event {n ⁻¹ X _n  > K for an infinite number of n} has probability 1. Thus, with probability 1, the sequence (n ⁻¹ X _n ) does not tend to 0.

the first line of EB _t + 1 B _t …B ₁ is thus the product of the first line of EB _t + 1 and of EB _t …B ₁ . The conclusion follows.

1. For any fixed t , the sequence converges almost surely (to ) as K → ∞. Thus
   
   and the first convergence follows. Now note that we have
   

   The first inequality uses (a + b) ^s  ≤ a ^s  + b ^s for a, b ≥ 0 and s ∈ (0, 1]. The second inequality is a consequence of . The second convergence then follows from the dominated convergence theorem.

2. We have . The convergence follows from the previous question, and from the strict stationarity, for any fixed integer K , of the sequence .
3. We have
   

   for any i ^′ = 1, …, ℓ, j ^′ = 1, …, m . In view of the independence between X _n and Y , it follows that almost surely as n → ∞. Since is a strictly positive number, we obtain almost surely, for all i ^′, j ^′ . Using (a + b) ^s  ≤ a ^s  + b ^s once again, it follows that

   
4. Note that the previous question does not allow us to affirm that the convergence to 0 of entails that of E(‖A _k A _k − 1…A ₁‖ ^s ), because has zero components. For k large enough, however, we have
   

   where is independent of A _k A _k − 1…A _N + 1. The general term a _{i, j} of A _N …A ₁ is the (i, j)th term of the matrix A ^N multiplied by a product of variables. The assumption A ^N  > 0 entails a _{i, j}  > 0 almost surely for all i and j . It follows that the i th component of Y satisfies Y _i  > 0 almost surely for all i . Thus . Now the previous question allows to affirm that E(‖A _k A _k − 1…A _N + 1‖ ^s ) → 0 and, by strict stationarity, that E(‖A _k − N A _{k − N − 1}…A ₁‖ ^s ) → 0 as k → ∞. It follows that there exists k ₀ such that

5. If α ₁ or β ₁ is strictly positive, the elements of the first two lines of the vector are also strictly positive, together with those of the ( q + 1)th and ( q + 2)th lines. By induction, it can be shown that under this assumption.
6. The condition can be satisfied when α ₁ = β ₁ = 0. It suffices to consider an ARCH(3) process with α ₁ = 0, α ₂ > 0, α ₃ > 0, and to check that .

The last inequalities imply β ₁ ≥ 0. Finally, the positivity constraints are

If q = 2, these constraints reduce to

Thus, we can have α ₂ < 0.

If the last equality is true, it remains true when h is replaced by h + 1 because . Since , it follows that for all h ≥ 0. Moreover,

Since , if then we have, for all h ≥ 1, We have thus shown that the sequence is decreasing when . If , it can be seen that for h large enough, say h ≥ h ₀ , we have , again because of . Thus, the sequence is decreasing.

Since in probability, there exist K ₀ ∈ ℝ and n ₀ ∈ ℕ such that P(X _n  < K ₀/2) ≤ ς < 1 for all n ≥ n ₀ . Consequently,

as n → ∞, for all K ≤ K ₀ , which entails the result.

as n → ∞. If γ < 0, the Cauchy rule entails that

converges almost surely, and the process (ε _t ), defined by , is a strictly stationary solution of model (2.7). As in the proof of Theorem 2.1, it can be shown that this solution is unique, non‐anticipative and ergodic. The converse is proved by contradiction, assuming that there exists a strictly stationary solution . For all n > 0, we have

It follows that a(η ₋₁)…a(η _−n )ω(η _{−n − 1}) converges to zero, almost surely, as n →  ∞ , or, equivalently, that

E.2

We first assume that E log {a(η _t )} > 0. Then the strong law of large numbers entails almost surely. For (E.2) to hold true, it is then necessary that log ω(η _{−n − 1}) →  − ∞ almost surely, which is precluded since (η _t ) is iid and ω(η ₀) > 0 almost surely. Assume now that E log {a(η _t )} = 0. By the Chung–Fuchs theorem, we have with probability 1 and, using Exercise 2.17, the convergence (E.2) entails log ω(η _{−n − 1}) →  − ∞ in probability, which, as in the previous case, entails a contradiction.

Regardless of the value of , fixed or even random, we have almost surely

using the law of large numbers and Jensen's inequality. It follows that almost surely as t → ∞.

1. Since the φ _i are positive and A ₁ = 1, we have φ _i  ≤ 1, which shows the first inequality. The second inequality follows by convexity of x ↦ x log x for x > 0.
2. Since A ₁ = 1 and A _p  < ∞, the function f is well defined for q ∈ [p, 1]. We have
   
   The function q ↦ log E|η ₀|^2q is convex on [p,1] if, for all λ ∈ [0, 1] and all q, q ^* ∈ [p, 1],
   

   which is equivalent to showing that

   

   with X = |η ₀|^2q , . This inequality holds true by Hölder's inequality. The same argument is used to show the convexity of . It follows that f is convex, as a sum of convex functions. We have f(1) = 0 and f(p) < 0, thus the left derivative of f at 1 is negative, which gives the result.

3. Conversely, we assume that there exists p ^* ∈ (0, 1] such that and that condition (2.52) is satisfied. The convexity of f on [p ^*, 1] and (2.52) implies that f(q) < 0 for q sufficiently close to 1. Thus condition (2.41) is satisfied. By convexity of f and since f(1) = 0, we have f(q) < 0 for all q ∈ [p, 1]. It follows that, by Theorem 2.6, E|ε _t | ^q  < ∞ for all q ∈ [0, 2].

since the condition for the existence of is . Note that when the GARCH effect is weak (that is, α ₁ is small), the part of the variance that is explained by this regression is small, which is not surprising. In all cases, the ratio of the variances is bounded by 1/κ _η , which is largely less than 1 for most distributions (1/3 for the Gaussian distribution). Thus, it is not surprising to observe disappointing R ² values when estimating such a regression on real series.

1. (a) Since P _ε admits a positive density on ℝ, the probability measure P(x, .) is, for all x ∈ E , absolutely continuous with respect to λ and its density is positive on ℝ. Thus any measure ϕ which is absolutely continuous with respect to λ is a measure of irreducibility: ∀x ∈ E ,
   

   Moreover, λ is a maximal measure of irreducibility.

2. (b) Assume, for example, that ε _t is uniformly distributed on [−1, 1]. If θ > 1 and X ₀ = x ₀ > 1/(θ − 1), we have x ₀ < X ₁ < X ₂ < …, regardless of the ε _t . Thus there exists no irreducibility measure: such a measure should satisfy ϕ([ −  ∞ , x]) = 0, for all x ∈ ℝ, which would imply ϕ = 0.

Thus μ is an invariant probability measure.

Conversely, suppose that μ is invariant. Using the Chapman–Kolmogorov relation, by which ∀t ∈ ℕ,     ∀ s,    0 ≤ s ≤ t,    ∀ x ∈ E,     ∀ B ∈ ℰ,

we obtain

Thus, by induction, for all t , ℙ[X _t  ∈ B] = μ(B) (∀B ∈ ℬ ). Using the Markov property, this is equivalent to the strict stationarity of the chain: the distribution of the process (X _t , X _t + 1, …, X _t + k ) is independent of t , for any integer k .

Thus π is invariant. The third equality is an immediate consequence of the Fubini and Lebesgue theorems.

Now let B ∈ ℰ . Then for all x ∈ C ,

The measure ν is non‐trivial since ν(E) = δλ(C) = 2δc > 0.

Thus if K ₁ < 1, we have, for K ₁ < K < 1 and for g(x) > (K ₂ + 1 − K ₁)/(K − K ₁),

If we put A = {x; g(x) = 1 +  ∣ x ∣  ≤ (K ₂ + 1 − K ₁)/(K − K ₁)}, the set A is compact and the conditions of Theorem 3.1 are satisfied, with 1 − δ = K .

It follows that

because V ≥ 1. Thus, there exists κ > 0 such that

Note that the positivity of δ is crucial for the conclusion.

The inequality is justified by (i) and the fact that P f is a continuous positive function. It follows that for f =  _C , where C is a compact set, we obtain

which shows that,

(that is, π is subvarient) using (ii). If there existed B such that the previous inequality were strict, we should have

and since π(E) < ∞ we arrive at a contradiction. Thus

which signifies that π is invariant.

where

Inequality (A.8) shows that d ₇ is bounded by

By an argument used to deal with d ₆ , we obtain

and the conclusion follows.

is continuous at x when g is continuous.

To show that the irreducibility condition (ii) is not satisfied, consider the set of numbers in [0,1] such that the sequence of decimals is periodic after a certain lag:

For all h ≥ 0, if and only if . We thus have,

and,

This shows that there is no non‐trivial irreducibility measure.

The drift condition (iii) is satisfied with, for instance, a measure φ such that φ([−1, 1]) > 0, the energy V(x) = 1 +  ∣ x∣ and the compact set A = [−1, 1]. Indeed,

provided

Using the independence between and the other variables of , we have, for all ,

when the distribution is symmetric.

and, by continuity of the exponential,

is finite if and only if the series of general term converges. Using the inequalities , we obtain

Since the tend to 0 at an exponential rate and , the series of general term converges absolutely, and we finally obtain

which is finite under condition (4.12).

with probability 1. The integral of a positive measurable function being always defined in , using Beppo Levi's theorem and then the independence of the , we obtain

which is of course finite under condition (4.12). Applying the dominated convergence theorem, and bounding the variables by the integrable variable , we then obtain the desired expression for .

and . With the notation , it follows that

It then suffices to use the fact that is equivalent to , and that is thus equivalent to , in a neighborhood of 0.

where is a white noise with variance . Using and

the coefficients and are such that

and

When, for instance, , , and , we obtain

If the volatility is a positive function of that possesses a moment of order 2, then

under conditions (4.34). Thus, condition (4.38) is necessarily satisfied. Conversely, under (4.38) the strict stationarity condition is satisfied because

and, as in the proof of Theorem 2.2, it is shown that the strictly stationary solution possesses a moment of order 2.

and

Using , we obtain

We then obtain the autocovariances

and the autocorrelations . Note that for all , which shows that is a weak ARMA process. In the standard GARCH case, the calculation of these autocorrelations would be much more complicated because is not a linear function of .

and this solution possesses a moment of order 2 when

which is the case, in particular, for . In the Gaussian case, we have

and

using the calculations of Exercise 4.5 and

Since is an increasing function, provided , we observe the leverage effect .

and the conclusion follows from the Borel–Cantelli lemma.

and

as h → 0. The conclusion follows.

Using the indication, one can check that and

We thus conclude by noting that

The sequence (ε _t ε _t + h , ℱ _t + h ) _t is thus a stationary sequence of square integrable martingale increments. We thus have

where . To conclude, ¹ it suffices to note that

in probability (and even in L ² ).

Its fourth‐order moment is

Thus,

Moreover,

Using Exercise 5.1, we thus obtain

By the ergodic theorem, the denominator converges in probability (and even a.s.) to γ _ε(0) = ω/(1 − α) ≠ 0. In view of Exercise 5.2, the numerator converges in law to . Cramér's theorem ² then entails

The asymptotic variance is equal to 1 when α = 0 (that is, when ε _t is a strong white noise). Figure E.3 shows that the asymptotic distribution of the empirical autocorrelations of a GARCH can be very different from those of a strong white noise.

In view of Exercises 5.1 and 5.3,

for any h ≠ 0.

Similarly, Eε _t ε _t + 1ε _s ε _s + 2 = 0 when t + 1 > s + 2. When t + 1 = s + 2, we have

because ε _t − 1 σ _t  ∈ ℱ _t − 1 , , E(η _t  ∣ ℱ _t − 1) = Eη _t  = 0 and . Using (7.24), the result can be extended to show that Eε _t ε _t + h ε _s ε _s + k  = 0 when k ≠ h and (ε _t ) follows a GARCH(p, q), with a symmetric distribution for η _t .

with ω = 1, α = 0.3 and β = 0.55. Thus γ _ε(0) = 6.667, , . Thus

for i = 1, …, 5. Finally, using Theorem 5.1,

Since , γ _ε(0) = ω/(1 − α) and

(see, for instance, Exercise 2.8), we have

Note that as i → ∞.

where ε _t (θ) = Y _t  − F _θ (W _t ). We thus have

and, when σ ² does not depend on θ ,

up to a constant. The constrained estimator is , with . The constrained score and the Lagrange multiplier are related by

On the other hand, the exact laws of the estimators under H ₀ are given by

and

with

For the case , we can estimate I ²² by

The test statistic is then equal to

E.3

with

and where R ² is the coefficient of determination (centred if X ₁ admits a constant column) in the regression of on the columns of X ₂ . For the first equality of (E.3), we use the fact that in a regression model of the form Y = Xβ + U , with obvious notation, Pythagoras's theorem yields

In the general case, we have

Since the residuals of the regression of Y on the columns of X ₁ and X ₂ are also the residuals of the regression of on the columns of X ₁ and X ₂ , we obtain LM _n by:

1. computing the residuals of the regression of Y on the columns of X ₁ ;
2. regressing on the columns of X ₂ and X ₁ , and setting LM _n  = nR ² , where R ² is the coefficient of determination of this second regression.

Since T is invertible, we have , where Col(Z) denotes the vectorial subspace generated by the columns of the matrix Z , and

If e ∈ Col(X) then and

Noting that , we conclude that

for h = 0, …, q . We then put

and then, for k = 2, …, q (when q > 1),

With standard notation, the OLS estimators are then

and

with equality if and only if , and we are done.

for all t , and consequently , which is not possible.

Using the data, we obtain , and thus . Therefore, the constrained estimate must coincide with one of the following three constrained estimates: that constrained by α ₂ = 0, that constrained by α ₁ = 0, or that constrained by α ₁ = α ₂ = 0. The estimate constrained by α ₂ = 0 is , and thus does not suit. The estimate constrained by α ₁ = 0 yields the desired estimate .

and this estimator satisfies

Under the assumptions of the exercise, the ergodic theorem entails the almost sure convergence

and thus the almost sure convergence of to φ ₀ . For the consistency, the assumption suffices.

If , the sequence (ε _t X _t − 1, ℱ _t ) is a stationary and ergodic square integrable martingale difference, with variance

We can see that this expectation exists by expanding the product

The CLT of Corollary A.1 then implies that

and thus

When , the condition suffices for asymptotic normality.

1. Let Then solves the model
   
   The parameter ω ₀ vanishing in this equation, the moments of do not depend on it. It follows that
2. and 3. Write to indicate that a matrix M is proportional to . Partition the vector into Z _t − 1 = (1, W _t − 1)^′ and, accordingly, the matrices A and B of Theorem 6.2. Using the previous question and the notation of Exercise 6.7, we obtain
   

   We then have

   

   Similarly,

   

   It follows that C = A ⁻¹ BA ⁻¹ is of the form

   

1. Let Let us show the existence of x ^* . Let (x _n ) be a sequence of elements of C such that, for all n > 0, ‖x − x _n ‖² < α ² + 1/n . Using the parallelogram identity ‖a + b‖² + ‖a − b‖² = 2‖a‖² + 2‖b‖² , we have
   
   the last inequality being justified by the fact that (x _m  + x _n )/2 ∈ C , the convexity of C and the definition of α . It follows that (x _n ) is a Cauchy sequence and, E being a Hilbert space and therefore a complete metric space, x _n converges to some point x ^* . Since C is closed, x ^* ∈ C and ‖x − x ^*‖ ≥ α . We have also ‖x − x ^*‖ ≤ α , taking the limit on both sides of the inequality which defines the sequence (x _n ). It follows that ‖x − x ^*‖ = α , which shows the existence.

   Assume that there exist two solutions of the minimisation problem in C , and . Using the convexity of C , it is then easy to see that satisfies

   

   This is possible only if (once again using the parallelogram identity).

2. Let λ ∈ (0, 1) and y ∈ C . Since C is convex, (1 − λ)x ^* + λy ∈ C . Thus
   

   and, dividing by λ ,

   

   Taking the limit as λ tends to 0, we obtain inequality (6.17).

   Let z such that, for all y ∈ C , 〈z − x, z − y〉 ≤ 0. We have

   

   the last inequality being simply the Cauchy–Schwarz inequality. It follows that ‖x − z‖ ≤ ‖x − y‖, ∀ y ∈ C . This property characterising x ^* in view of part 1, it follows that z = x ^* .

1. It suffices to show that when C = K , (6.17) is equivalent to (6.18). Since 0 ∈ K , taking y = 0 in (6.17) we obtain 〈x − x ^*, x ^*〉 ≤ 0. Since x ^* ∈ K and K is a cone, 2x ^* ∈ K . For y = 2x ^* in (6.17) we obtain 〈x − x ^*, x ^*〉 ≥ 0, and it follows that 〈x − x ^*, x ^*〉 = 0. The second equation of (6.18) then follows directly from (6.17). The converse, (6.18) ⇒ (6.17), is trivial.
2. Since x ^* ∈ K , then z = λx ^* ∈ K for λ ≥ 0. By (6.18), we have
   
   It follows that (λx)^* = z and (a) is shown. The properties (b) are obvious, expanding ‖x ^* + (x − x ^*)‖² and using the first equation of (6.18).

Since and , it follows that , and thus that X _n converges to the zero vector of ℝ ^k .

1. When j < 0, all the variables involved in the expectation, except ε _t − j , belong to the σ ‐field generated by {ε _{t − j − 1}, ε _{t − j − 2}, …}. We conclude by taking the expectation conditionally on the previous σ ‐field and using the martingale increment property.
2. For j ≥ 0, we note that is a measurable function of and of Thus is an even function of the conditioning variables, denoted by .
3. It follows that the expectation involved in the property can be written as
   
   The latter equality follows from of the nullity of the integral, because the distribution of η _t is symmetric.

using Markov's inequality, strict stationarity and the existence of a moment of order s > 0 for .

When κ → ∞, the variable increases to X ₁ . Thus by Beppo Levi's theorem converges to E(X ₁) =  + ∞. It follows that tends almost surely to infinity.

1. The assumptions made on f and Θ guarantee that is a measurable function of η _t , η _t − 1, …. By Theorem A.1, it follows that (Y _t ) is stationary and ergodic.
2. If we remove condition (7.94), the property may not be satisfied. For example, let Θ = {θ ₁, θ ₂} and assume that the sequence (X _t (θ ₁), X _t (θ ₂)) is iid, with zero mean, each component being of variance 1 and the covariance between the two components being different when t is even and when t is odd. Each of the two processes (X _t (θ ₁)) and (X _t (θ ₂)) is stationary and ergodic (as iid processes). However, is not stationary in general because its distribution depends on the parity of t .

1. In view of (7.30) and of the second part of assumption A1, we have
   E.7
   almost surely. Indeed, on a set of probability 1, we have for all ι > 0,
   E.8

   Note that and (7.29) entail that . The limit superior (E.5) being less than any positive number, it is null.

2. Note that is the strong innovation of . We thus have orthogonality between ν _t and any integrable variable which is measurable with respect to the σ ‐field generated by θ₀ :
   
   with equality if and only if ‐almost surely, that is, θ = θ ₀ (by assumptions A3 and A4; see the proof of Theorem 7.1).
3. We conclude that is strongly consistent, as in (d) in the proof of Theorem 7.1, using a compactness argument and applying the ergodic theorem to show that, at any point θ ₁ , there exists a neighbourhood V(θ ₁) of θ ₁ such that
   
4. Since all we have done remains valid when Θ is replaced by any smaller compact set containing θ ₀ , for instance Θ ^c , the estimator is strongly consistent.

For all c > 0, there exists such that for all t ≥ 0. Note that if and only if c ≠ 1. For instance, for a GARCH(1, 1) model, if we have . Let The minimum of f is obtained at the unique point

If , we have . It follows that c ₀ = 1 with probability 1, which proves the result.

In view of (7.41), (7.42), (7.79) and (7.24), we have

and

It follows that

E.6

and ℐ is block‐diagonal. It is easy to see that 풥 has the form given in the theorem. The expressions for J ₁ and J ₂ follow directly from (7.39) and (7.75). The block‐diagonal form follows from (7.76) and (E.6).

1. We have The parameters to be estimated are a and α , ω being known. We have
   
   It follows that
   

   Letting ℐ = (ℐ _ij ), and , we then obtain

   
2. In the case where the distribution of η _t is symmetric we have μ ₃ = 0 and, using (7.24), . It follows that
   

   The asymptotic variance of the ARCH parameter estimator is thus equal to : it does not depend on a ₀ and is the same as that of the QMLE of a pure ARCH(1) (using computations similar to those used to obtain (7.1.2)).

3. When α ₀ = 0, we have , and thus . It follows that
   
   

   We note that the estimation of too complicated a model (since the true process is AR(1) without ARCH effect) does not entail any asymptotic loss of accuracy for the estimation of the parameter a ₀ : the asymptotic variance of the estimator is the same, , as if the AR(1) model were directly estimated. This calculation also allows us to verify the ‘ α ₀ = 0’ column in Table 7.3: for the law we have μ ₃ = 0 and κ _η  = 3; for the normalized χ ²(1) distribution we find and κ _η  = 15.

It follows that

and, since ε can be chosen arbitrarily small, we have the desired result.

In order to give an example where (7.95) is not satisfied, let us consider the autoregressive model X _t  = θ ₀ X _t − 1 + η _t where θ ₀ = 1 and (η _t ) is an iid sequence with mean 0 and variance 1. Let J _t (θ) = X _t  − θX _t − 1 . Then J _t (θ ₀) = η _t and the first convergence of the exercise holds true, with J = 0. Moreover, for all neighbourhoods of θ ₀ ,

almost surely because the sum in brackets converges to +∞, X _t being a random walk and the supremum being strictly positive. Thus (7.95) is not satisfied. Nevertheless, we have

Indeed, converges in law to a non‐degenerate random variable (see, for instance, Hamilton 1994, p. 406) whereas in probability since has a non‐degenerate limit distribution.

Therefore is positive semi‐definite. Thus

Setting x = Jy , we then have

which proves the result.

1. In the ARCH case, we have . It follows that
   
   or equivalently , that is, We also have , and thus
   
2. Introducing the polynomial , the derivatives of satisfy
   
   It follows that
   
   In view of assumption A2 and Corollary 2.2, the roots of ℬ _θ (L) are outside the unit disk, and the relation follows.
3. It suffices to replace θ ₀ by in 1.

1. It suffices to apply (7.38), and then to apply Corollary 2.1.
2. The result follows from the Lindeberg central limit theorem of Theorem A.3.
3. Using Eq. (7.39) and the convergence of to +∞,
   
4. In view of Eq. (7.50) and the fact that , we have
   
5. The derivative of the criterion is equal to zero at . A Taylor expansion of this derivative around α ₀ then yields
   
   where α ^* is between and α ₀ . The result easily follows from the previous questions.
6. When ω ₀ ≠ 1, we have
   
   with
   

   Since d _t  → 0 almost surely as t → ∞, the convergence in law of part 2 always holds true. Moreover,

   

   with

   

   which implies that the result obtained in Part 3 does not change. The same is true for Part 4 because

   

   Finally, it is easy to see that the asymptotic behaviour of is the same as that of , regardless of the value that is fixed for ω .

7. In practice ω ₀ is not known and must be estimated. However, it is impossible to estimate the whole parameter (ω ₀, α ₀) without the strict stationarity assumption. Moreover, under condition (7.14), the ARCH(1) model generates explosive trajectories which do not look like typical trajectories of financial returns.

1. Consider a constant . We begin by showing that for n large enough. Note that
   
   We have
   

   In view of the inequality x ≥ 1 + log x for all x > 0, it follows that

   

   For all M > 0, there exists an integer t _M such that for all t > t _M . This entails that

   

   Since M is arbitrarily large,

   E.7

   provided that . If is chosen so that the constraint is satisfied, the inequalities

   

   and (E.7) show that

   E.8

   We will define a criterion O _n asymptotically equivalent to the criterion Q _n . Since a. s. as t → ∞, we have for α ≠ 0,

   

   where

   

   On the other hand, we have

   

   when α ₀/α ≠ 1. We will now show that Q _n (α) − O _n (α) converges to zero uniformly in . We have

   

   Thus for all M > 0 and any ε > 0, almost surely

   

   provided n is large enough. In addition to the previous constraints, assume that . We have for any , and

   

   for any α ≥ α ₀ . We then have

   

   Since M can be chosen arbitrarily large and ε arbitrarily small, we have almost surely

   E.9

   For the last step of the proof, let and be two constants such that . It can always be assumed that . With the notation , the solution of

   

   is This solution belongs to the interval when n is large enough. In this case

   

   is one of the two extremities of the interval , and thus

   

   This result, (E.9), the fact that min _α Q _n (α) ≤ Q _n (α ₀) = 0 and (E.8) show that

   

   Since is an arbitrarily small interval that contains α ₀ and , the conclusion follows.

2. It can be seen that the constant 1 does not play any particular role and can be replaced by any other positive number ω . However, we cannot conclude that almost surely because , but is not a constant. In contrast, it can be shown that under the strict stationarity condition the constrained estimator does not converge to α ₀ when ω ≠ ω ₀ .

Since at the optimum

the solution is such that Since we obtain , and then the solution is

E.10

Instead of the Lagrange multiplier method, a direct substitution method can also be used.

The constraints can be written as

where H is n × (n − p), of full column rank, and x ^* is (n − p) × 1 (the vector of the non‐zero components of x ). For instance: (i) if n = 3, x ₂ = x ₃ = 0 then x ^* = x ₁ and ; (ii) if n = 3, x ₃ = 0 then x ^* = (x ₁, x ₂)^′ and .

If we denote by Col(H) the space generated by the columns of H , we thus have to find

where ‖.‖ _J is the norm .

This norm defines the scalar product 〈z, y〉 _J  = z ^′ Jy . The solution is thus the orthogonal (with respect to this scalar product) projection of x ₀ on Col(H). The matrix of such a projection is

Indeed, we have P ² = P , PHz = Hz , thus Col(H) is P ‐invariant, and 〈Hy, (I − P)z〉 _J  = y ^′ H ^′ J(I − P)z = y ^′ H ^′ Jz − y ^′ H ^′ JH(H ^′ JH)⁻¹ H ^′ Jz = 0, thus z − Pz is orthogonal to Col(H).

It follows that the solution is

E.11

This last expression seems preferable to (E.10) because it only requires the inversion of the matrix H ^′ JH of size n − p , whereas in (E.11) the inverse of J , which is of size n , is required.

and then

and, using (E.11),

which gives a constrained minimum at

In case (b), we have

E.12

and, using (E.11), a calculation, which is simpler than the previous one (we do not have to invert any matrix since H ^′ JH is scalar), shows that the constrained minimum is at

The same results can be obtained with formula (E.10), but the computations are longer, in particular because we have to compute

E.13

It follows that the solution will be found among (a) λ = Z ,

The value of Q(λ) is 0 in case (a), in case (b), in case (c) and in case (d).

To find the solution of the constrained minimisation problem, it thus suffices to take the value λ which minimizes Q(λ) among the subset of the four vectors defined in (a)–(d) which satisfy the positivity constraints of the two last components.

We thus find the minimum at λ ^Λ = Z = (−2, 1, 2)^′ in case (i), at

and

It follows that

The coefficient of the regression of Z ₁ on Z ₂ is −ω ₀ . The components of the vector (Z ₁ + ω ₀ Z ₂, Z ₂) are thus uncorrelated and, this vector being Gaussian, they are independent. In particular , which gives . We thus have

Finally,

It can be seen that

is a positive semi‐definite matrix.

and the information matrix (written for simplicity in the ARCH(3) case) is equal to

This matrix is invertible (which is not the case for a general GARCH(p, q)). We finally obtain

and thus

In view of Theorem 8.1 and (8.15), the asymptotic distribution of is that of the vector λ ^Λ defined by

We have , thus

Since the components of the Gaussian vector (Z ₁ + ω ₀ Z ₂, Z ₂) are uncorrelated, they are independent, and it follows that

We then obtain

Let f(z ₁, z ₂) be the density of Z , that is, the density of a centred normal with variance (κ _η  − 1)J ⁻¹ . It is easy to show that the distribution of admits the density and to check that this density is asymmetric.

A simple calculation yields . From , we then obtain . And from we obtain . Finally, we obtain

The p ‐value of C ^* is . Since log 2{1 − Φ(x)}∼ − x ²/2 in the neighbourhood of +∞, the asymptotic slope of C ^* is also c ^*(θ) = θ ² for θ > 0. The C and C ^* tests having the same asymptotic slope, they cannot be distinguished by the Bahadur approach.

We know that C is uniformly more powerful than C ^* . The local power of C is thus also greater than that of C ^* for all τ > 0. It is also true asymptotically as n → ∞, even if the sample is not Gaussian. Indeed, under the local alternatives , and for a regular statistical model, the statistic is asymptotically 풩(τ, 1) distributed. The local asymptotic power of C is thus γ(τ) = 1 − Φ(c − τ) with c = Φ⁻¹(1 − α). The local asymptotic power of C ^* is γ ^*(τ) = 1 − Φ(c ^* − τ) + Φ(−c ^* − τ), with c ^* = Φ⁻¹(1 − α/2). The difference between the two asymptotic powers is

and, denoting the 풩(0, 1) density by φ(x), we have

where

Since 0 < c < c ^* , we have

Thus, g(τ) is decreasing on [0, ∞). Note that g(0) > 0 and . The sign of g(τ), which is also the sign of D ^′(τ), is positive when τ ∈ [0, a] and negative when τ ∈ [a, ∞), for some a > 0. The function D thus increases on [0, a] and decreases on [a, ∞). Since D(0) = 0 and , we have D(τ) > 0 for all τ > 0. This shows that, in Pitman's sense, the test C is, as expected, locally more powerful than C ^* in the Gaussian case, and locally asymptotically more powerful than C ^* in a much more general framework.

To justify the score test, we remark that the log‐likelihood constrained by H ₀ is

which gives as constrained estimator of σ ² . The derivative of the log‐likelihood satisfies

at . The first component of this score vector is asymptotically 풩(0, 1) distributed under H ₀ . The third test is of course the likelihood ratio test, because the unconstrained log‐likelihood at the optimum is equal to whereas the maximal value of the constrained log‐likelihood is . Note also that under H ₀ .

The asymptotic level of the three tests is of course α , but using the inequality for x > 0, we have

with almost surely strict inequalities in finite samples, and also asymptotically under H ₁ . This leads us to think that the Wald test will reject more often under H ₁ .

Since is invariant by translation of the X _i , tends almost surely to σ ² both under H ₀ and under H ₁ , as well as under the local alternatives . The behaviour of under H _n (τ) is the same as that of under H ₀ , and because

under H ₀ , we have both under H ₀ and under H _n (τ). Similarly, it can be shown that under H ₀ and under H _n (τ). Using these two results and x/(1 + x)∼ log(1 + x) in the neighbourhood of 0, it can be seen that the statistics L _n , R _n and W _n are equivalent under H _n (τ). Therefore, the Pitman approach cannot distinguish the three tests.

Using for x in the neighbourhood of +∞, the asymptotic Bahadur slopes of the tests C ₁ , C ₂ and C ₃ are, respectively

Clearly

Thus the ranking of the tests, in increasing order of relative efficiency in the Bahadur sense, is

All the foregoing remains valid for a regular non‐Gaussian model.

Note that Var(Z _d )c corresponds to the last column of VarZ = (κ _η  − 1)J ⁻¹ . Thus c is the last column of J ⁻¹ divided by the (d, d)th element of this matrix. In view of Exercise 6.7, this element is . It follows that and . By (8.24), we thus have

This shows that the statistic 2/(κ _η  − 1)L _n has the same asymptotic distribution as the Wald statistic W _n , that is, the distribution in the case d ₂ = 1.

The result then follows from (8.30).

Since and belong to the σ ‐field ℱ _t − 1 generated by {ε _u  : u < t}, and since the distribution of ε _t given ℱ _t − 1 has the density , we have

and the result follows. We can also appeal to the general result that a score vector is centred.

Thus

when X∼풩(θ ₀, σ ²), and

when X∼풩(θ, σ ²). Note that

as in Le Cam's third lemma.

and

Using the ergodic theorem, the fact that (1 − |η _t | ^λ ) is centred and independent of the past, as well as elementary calculations of derivatives and integrals, we obtain

and

almost surely.

where the inequality is strict if σf(ησ)/f(η) is non‐constant. If this ratio of densities were almost surely constant, it would be almost surely equal to 1, and we would have

which is possible only when σ = 1.

Thus and . We then show that κ _η  ≔  ∫ x ⁴ f _p (x)dx = (3 + p)(2 + p)/p(p + 1). It follows that .

To compare the ML and Laplace QML, it is necessary to normalise in such a way that E ∣ η _t  ∣  = 1, that is, to take the double Γ(p, p) as density f . We then obtain 1 + xf ^′(x)/f(x) = p − p ∣ x∣. We always have , and we have . It follows that , which was already known from Exercise 9.6. This allows us to construct a table similar to Table 9.5.

Denoting by c any constant whose value can be ignored, we have

and thus

Now consider the second density,

We have

which gives

Consider the last instrumental density,

We have

and thus

In each case, does not depend on the parameter λ of h . We conclude that the estimators exhibit the same asymptotic behaviour, regardless of the parameter λ . It can even be easily shown that the estimators themselves do not depend on λ .

1. The Laplace QML estimator applied to a GARCH in standard form (as defined in Example 9.4) is an example of such an estimator.
2. We have
   
3. Since
   
   we have
   

   It follows that, using obvious notation,

   

where ϱ =  ∫  ∣ x ∣ f(x)dx . Thus, using Exercise 9.9, we obtain

with .

that of the vectorial model is

that of the CCC model is

that of the BEKK model is

For and we obtain Table E.1.

and

Conversely, it is easy to check that (10.101) implies (10.100).

1. Since and are independent, we have
   
   which shows that is constant.
2. We have
   
   which is nonzero only if , thus and take only one value.
3. Assume that there exist two events and such that and . The independence then entails that
   
   and we obtain a contradiction.

where has the same norm as . Assuming, for instance, that we have

Moreover, this maximum is reached at .

An alternative proof is obtained by noting that solves the maximization problem of the function under the constraint . Introduce the Lagrangian

The first‐order conditions yield the constraint and

This shows that the constrained optimum is located at a normalized eigenvector associated with an eigenvalue of , . Since , we of course have .

The first inequality of (10.69) is a simple application of the Cauchy–Schwarz inequality. The second inequality of (10.69) is obtained by twice applying the second inequality of (10.68).

We now give a second‐order stationarity condition. If exists, then this matrix is symmetric positive semi‐definite and satisfies

that is,

If is positive definite, it is then necessary to have

(E.14)

For the reverse we use Theorem 10.5. Since the matrices are of the form with , the condition is equivalent to (E.14). This condition is thus sufficient to obtain the stationarity, under technical condition (ii) of Theorem 10.5 (which can perhaps be relaxed). Let us also mention that, by analogy with the univariate case, it is certainly possible to obtain the strict stationarity under a condition weaker than (E.14).

when . To show the almost sure convergence, let us begin by noting that, using Hölder's inequality,

with and . Let , for , and which is defined in , a priori. Since

it follows that is almost surely defined in and is almost surely defined in . Since , we have almost surely.

and it suffices to take

The conditional covariance between the factors and , for , is

which is a nonzero constant in general.

Denoting by the th vector of the canonical basis of , we have

and we obtain the BEKK representation with ,

which shows that is an eigenvector associated with an eigenvalue of . Left‐multiplying the previous equation by , we obtain

which shows that must be the largest eigenvalue of . The vector is unique, up to its sign, provided that the largest eigenvalue has multiplicity order 1.

An alternative way to obtain the result is based on the spectral decomposition of the symmetric definite positive matrices

Let , that is, . Maximizing is equivalent to maximizing . The constraint is equivalent to the constraint . Denoting by the components of , the function is maximized at under the constraint, which shows that is the first column of , up to the sign. We also see that other solutions exist when . It is now clear that the vector contains the principal components of the variance matrix .

element by element. This shows that is diagonal, and the conclusion easily follows.

The proof is completed by induction on .

On the right‐hand side of the equality, the term in parentheses is nonnegative and the last term is positive, unless . But in this case and the term in parentheses becomes .

By It 's formula, the SDE satisfied by is

Using the hint, we then have

It follows that

The positivity follows.

this inequality being uniform on any ball of radius . The assumptions of Theorem 11.1 are thus satisfied.

It is then easy to check that the limits in (11.23) and (11.25) are null. The limiting diffusion is thus

The solution of the equation is, using Exercise 11.1 with ,

where is the initial value. It is assumed that and , in order to guarantee the positivity. We have

The result is obtained by multiplying this equality by and taking the expectation with respect to the probability .

with, in particular, In view of Exercise 11.5, we thus have

The maximum likelihood estimator of is then

It is easy to verify that . It follows that

The option buyer wishes to be covered against the risk: he thus agrees to pay more if the asset is more risky.

The property immediately follows.

We are in the framework of model (11.44), with . Thus the risk‐neutral model is given by (11.47), with and

It can easily be seen that if we have, for and for all ,

Writing

we thus obtain

and writing

we have

It follows that

Thus

There are an infinite number of possible choices for and . For instance, if , one can take and with . Then follows. The risk‐neutral probability is obtained by calculating

Under the risk‐neutral probability, we thus have the model

(E.15)

Note that the volatilities of the two models (under historical and risk‐neutral probability) do not coincide unless for all .

Thus, introducing the notation ,

The conditional law of is thus the distribution, and (11.58) follows.

At horizon 2, the conditional distribution of is not Gaussian if , because its kurtosis coefficient is equal to

There is no explicit formula for when .

Using (11.63), the desired equality follows.

Note that

because the two bracketed terms have the same sign. It follows that

The property is thus shown.

It follows that

We have , the inequality being strict because the distribution of is nondegenerate. In view of Theorem 2.1, this implies that a.s., and thus that a.s., when tends to infinity.

is not normal. Indeed, and , but . Similarly, the variable

is centered with variance 2, but is not normally distributed because

Note that the distribution is much more leptokurtic when is close to 0.

We have, using the independence assumptions on the sequences (η _t ) and ( ),

provided that the expectation of the term between accolades exists and is finite. To show this, write

We have

Thus EZ _{t, ∞} < ∞. Similarly, the same arguments show that

Moreover, for all k > 0,

using again the independence between (η _t ) and ( ).

where the last equality holds because η _t and (h _t ) are independent, and because the law of η _t is symmetric. The same arguments show that for x ≥ 0. Thus, there is a one‐to‐one relation between the law of ε _t and that of . In addition the law of ε _t is symmetric. Similarly, for any n ≥ 1, one can show that there is a one‐to‐one relation between the law of (ε _t , …, ε _t + n ) and that of . When the distribution of η _t is not symmetric, the fourth equality of the previous computation fails.

From the independence between (Y _t ) and (Z _t ), (X _t ) is a second‐order process whose mean and autocovariance function are obtained as follows:

Since γ _X (k) = βγ _X (k − 1),  ∀ k > 1, the process (X _t ) admits an ARMA(1,1) representation of the form (12.7). The constant α is deduced from the first two autocovariances of (X _t ). By Eq. (12.7) we have, denoting by , the variance of the noise in this representation

Hence, if , the coefficient α is a solution of

(E.16)

and the solution of modulus less than 1 is given by

Moreover, the variance of the noise in model (12.7) is if β ≠ 0 (and if β = 0). Finally, if the relation γ _X (k) = βγ _X (k − 1) also holds for k = 1 and (X _t ) is an AR(1) (i.e. α = 0 in model (12.7)).

Now, when β ≠ 0 and σ ≠ 0, we get , using (E.16). It follows that either 0 < α < β < 1/α or 0 > α > β > 1/α . In particular ∣α ∣  <  ∣ β∣, which shows that the orders of the ARMA(1,1) representation for X _t are exact.

We also have,

Thus

for ρ ≠ 0 and ∣β ∣  < 1.

Denote by θ ⁽¹⁾ = (0.098, 0.087, 0.84)^′ and θ ⁽²⁾ = (0.012, 0.075, 0.919)^′ the parameters of the two models. The estimated values of ω and β seem quite different. Denote by and the estimated standard deviations of the estimators of ω and β of Model Mi . It turns out that the confidence intervals

and

have empty intersection. The same holds true for the confidence intervals

and

The third graph of Figure E.4 displays the boxplot of the distribution of on 100 independent simulations of Model M1. The difference θ ⁽²⁾ − θ ⁽¹⁾ between the parameters of M1 and M2 is marked by a diamond shape. The difference θ ⁽²⁾ − θ ⁽¹⁾ is an outlier for the distribution of , meaning that estimated GARCH on the two periods are significantly distinct.

(E.17)

The number of balls in urn changes successively from odd to even, and conversely, along the steps. For instance . Thus the chain is irreducible but periodic.

Using the formula , it can be seen that is an invariant law. It follows that for all .

When the initial distribution is the Dirac mass at 0 we have when is odd, and when is even. Thus does not exist.

# one iteration of the EM algorithm
EM <- function(omega,pi0,p,y){
d<-length(omega)
n <- length(y) # y contient les n observations
vrais<-0
pit.t<-matrix(0,nrow=d,ncol=n)
pit.tm1<-matrix(0,nrow=d,ncol=n+1)
vecphi<-rep(0,d)
pit.tm1[,1]<-pi0
for (t in 1:n) {
 for (j in 1:d) vecphi[j]<-{dnorm(y[t],
      mean=0,sd=sqrt(abs(omega[j])))}
 den<-sum(pit.tm1[,t]*vecphi)
 if(den<=0)return(Inf)
 pit.t[,t]<-(pit.tm1[,t]*vecphi)/den
 pit.tm1[,t+1]<-t(p)%*%pit.t[,t]
 vrais<-vrais+log(den)
        }
pit.n<-matrix(0,nrow=d,ncol=n)
pit.n[,n]=pit.t[,n]
for (t in n:2) {
 for (i in 1:d) {
 pit.n[i,t-1]<- {pit.t[i,t-1]*sum(p[i,1:d]*
         pit.n[1:d,t]/pit.tm1[1:d,t])}
         } }
pitm1et.n<-array(0,dim=c(d,d,n))
for (t in 2:n) {
 for (i in 1:d) {
 for (j in 1:d) {
 pitm1et.n[i,j,t]<-p[i,j]*pit.t[i,t-1]*pit.n[j,t]/pit.tm1[j,t]
         } } }
omega.final<-omega
pi0.final<-pi0
p.final<-p
for (i in 1:d)  {
omega.final[i]<-sum((y[1:n]∧2)*pit.n[i,1:n])/sum(pit.n[i,1:n])
pi0.final[i]<-pit.n[i,1]
 for (j in 1:d) {
 p.final[i,j]<-sum(pitm1et.n[i,j,2:n])/sum(pit.n[i,1:(n-1)])
        } }
liss<-{list(probaliss=pit.n,probatransliss=pitm1et.n,
vrais=vrais,omega.final=omega.final,pi0.final=pi0.final,
        p.final=p.final)}
               }

The proof of Theorem 2.4 in Chapter 2 applies directly with this sequence (A _t ), showing that there exists a strictly stationary solution if and only if the top Lyapunov exponent of (A _t ) is strictly negative. The solution is then unique, non‐anticipative, and ergodic, and takes the form (2.18).

In the ARCH(1) case with d regimes, we obtain the necessary and sufficient condition

For the existence of a positive solution to this equation, it is necessary to have

Conversely, under this condition, the process

is a strictly stationary and non‐anticipative solution which satisfies

and the result follows.

The EM algorithm cannot be generalised trivially because the maximisation of Eq. (12.32) is replaced by that of

which does not admit an explicit form like (12.35) but requires the use of an optimisation algorithm.

we have

under conditions entailing the existence of the series. For the alternative model, ε _t  = σ _t (Δ _t )η _t with

we have

Let ℱ _t be the sigma‐field generated by the past observations {ε _u , u < t}, and by the past and present value of the chain {Δ _u , u ≤ t}. We have

but, given the past observations, only depends on Δ _t , whereas h _t depends also on {Δ _u , u < t}.

This entails differences between the two models, in terms of probabilistic properties (the stationarity conditions are easier to obtain for the standard MS‐GARCH model, but they have also been obtained by Liu (2006) for the alternative model), of statistical inference (the fact that only depends on Δ _t renders the alternative model much easier to estimate), and also on the dynamics behaviour and on the interpretation of the parameters.

For instance, for the MS‐GARCH, β(i) can be interpreted as a parameter of inertia of the volatility in regime i : if the volatility h _t − 1 is high and β(i) is close to 1, the next volatility h _t will remain high in regime i . This interpretation is no more valid for the alternative model, since may not be equal to .