Problem Set 6 Solution

Transcription

1 Problem Set 6 Solution May st, 009 by Yang. Causal Expression of AR Let φz : αz βz. Zeros of φ are α and β, both of which are greater than in absolute value by the assumption in the question. By the theorem mentioned in the lecture, we can write y t as causal expression Note that for z <, and thus y t j0 c j ε t j φl ε t φz αz βz + αz + α z + + βz + β z + y t + αl + α L + + βl + β L + ε t + α + βl + α + αβ + β L + ε t j α k β j k L j ε t j0 k0 j α k β j k ε t j j0 k0 Therefore, j c j α k β j k k0. Estimation of AR We have observations y, y. We need to obtain the distribution of y to do the MLE. Let us guess y Nµ y, σy. Since y αy + ε t, y N αµ y, α σy + σ The AR model is stationary, so every observation has the same mean and variance. This implies that the unconditional mean and variance of y is the same with those of y. Therefore, αµ y µ y α σ y + σ σ y

2 which implies that The likelihood function is now y N σ 0, α Lα, σ fy ; α, σ fy y ; α, σ πσ / α exp y σ / α exp y φy πσ σ and thus Lα, σ log fy 0 ; α, σ + log fy y ; α, σ log π log σ α y σ / α + log π log σ + log α y α σ y αy σ Obtain the FOC s as follows. The equations can be simplified as follows. α α + y α σ + y αy y σ 0 σ + y α σ 4 + y αy σ 4 0 ασ + α y y 0 σ + y αy y + y 0 log π log σ y αy From the second equation, obtain an expression of σ in terms of y, y and α as σ y αy y + y and plug in into the first equation, then y α αy y + y or + α y y 0 α y + y + y y 0 σ and thus Plug this back into the equations, then α y y y + y σ y y y + y

3 3. Size of T-test in AR Model Let c µ, then y t c + αy t + u t. The OLS estimator is given by ĉ α T t y t yt T y t t y t and the variance is estimated by T var α σ t y t yt where σ : T T t y t ĉ αy t. The nominal 95% two-sided CI for α is obtained as follows. CI 0.95 α.96 var α, α +.96 var α Then the empirical coverage probability Prα CI 0.95 R R r α CIr 0.95 is given as follows. True alpha Mean Bias Median Bias St. Dev R.M.S.E Cov. Prob In fact, there seems to exist no problem. Note that, however, the given process is nonstationary, since the unconditional mean and variance of y t depends on t. Correct this by letting c µ α so that then y t N µ, σ α y t µ α + αy t + u t for any t as easily verified. Use this to generate data and estimate α, then we get the following result. Now we can see that when α, the empirical coverage probability is strictly less than This means that we obtain too narrow confidence intervals as α. Simply inverting a t-test is thus not a good idea to obtain CI when α is believed to be close to. True alpha Mean Bias Median Bias St. Dev R.M.S.E Cov. Prob

4 4. Inclusion of Irrelevant Parameter Note first that in the hints, θ is a row vector, and thus x t is also a row vector. For completeness, let us begin from the scratch. The conditional likelihood of an MAq process is given by where Lθ T log π T log σ t ε t θ σ ε t θ : y t θ ε t θ θ ε t q θ θ : θ,, θ q The first order conditions with respect to θ read as 0 Lθ σ T t ε t θ ε t σ where x t : ε tθ. So the information matrix is given by L L E T σ 4 E T x tε t x s ε s σ 4 T t t s s E x tx s ε t ε s T σ 4 E x tx t Eε t σ t The third equality holds by the following. Note that x t ε tθ x t ε t t T t Ex tx t ε t ε t q ε t,, ε t q + θ + + θ q ε t,, ε t q θ x t θ q x t q x t is an ARq process and thus can be expressed as a weigthed sum of past errors under some condition. For t > s, x t, x s and ε s are independent of ε t, so Ex tx s ε t ε s Ex tx s ε s Eε t 0. For t < s, x t, x s and ε t are independent of ε s, so Ex tx s ε t ε s Ex tx s ε t Eε s 0. Since the asymptotic variance of θ MLE is the limit of T times the inverse of the information matrix, AV θ MLE lim T T σ Ex tx t lim T σ σ E plim T t T x tx t t x tx t σ Ex tx t t The condition is invertibility of the MAq process of y t, which is equivalent to causality of the ARq process of x t. If x t is noncausal stationary, we may use a noncausal expression to claim Ex tx s ε t ε s 0 for t s. 4

5 where the third equality holds by the dependent version of law of large numbers and the continuous mapping theorem, and the last equality holds by identical distribution of x t. i MA model Consider y t ε t + θ ε t, then x t : ε t the variance of the AR process is so the asymptotic variance of θ is θ x t + ε t is an AR process. As is well known, σ Ex t θ AV θ σ Ex t θ ii MA model with θ 0 Consider y t ε t + θ ε t + θ ε t. We estimate both θ and θ, although θ is actually 0. x t : ε tθ ε t, ε t θ x t θ x t is an AR process with θ 0. This implies that x t is actually an AR process. x t θ x t + ε t, ε t So Ex tx t θex t x t + E ε t ε t ε t ε t ε t ε t θ E x t ε t, ε t εt θ E ε t x t We may obtain Ex tx t from an AR process, and use θ 0 later. Since x tx t ε t, ε t θ x t θ x t ε t, ε t θ x t θ x t, Ex tx t θex t x t + θex t x t + θ θ Ex t x t + θ θ Ex t x t Also since x tx t ε t, ε t θ x t θ x t x t, Ex tx t Eε t, ε t x t θ Ex t x t θ Ex t x t Arranging the above two equations and making use of stationarity of x t, we have θ θex tx t θ θ Ex tx t + θ θ Ex σ θ σ t x t + θ σ σ Ex tx t θ Ex tx t θ Ex 0 0 t x t + σ 0 Ex t x t θ Ex tx t θ Ex 0 σ tx t Solving the system of equations simultaneously, Ex tx t σ θ + θ θ + θ θ θ + θ Now use θ 0, then this simplifies to Ex tx t σ θ θ θ, which is the same as we get using the other method. 5

6 By stationarity of x t, Ex tx t Ex t x t. Note also that E ε t ε t ε t ε t ε t ε t σ 0 0 σ and that εt E ε t x t εt E ε t εt ε t 3 εt ε E t ε t ε t 3 ε t ε t ε t σ 0 θ E εt ε t x t }{{} 0 x t ε t,ε t Therefore or equivalently, θex σ θ tx t σ θ σ σ σ Ex tx t θ θ θ So the asymptotic variance of θ MLE is AV θ MLE σ Ex tx t θ θ The asymptotic variance of θ is. Asymptotic Relative Efficiency From the above result, the asymptotic relative efficiency of θ obtained from MA model with θ 0 to that obtained from MA model is ARE θ θ This is less than. relatively inefficient. Estimating irrelevant parameter θ increases the variance of θ, and thus is 6