Econometrics with Fin Apps (Multiple choice) exercises
Econ G28T
MULTIPLE CHOICE TEST
2009
Note that more than one answer may be correct.
If you change your selection please indicate which answer is your final choice.
The correct answers are highlighted in red
1) Consider the process {x t} generated by, x t = x t-1 + u t, with u t serially independent and
E(u t) = 0 and Var(u t) = 1.5, and x0= 0.
i)In the long run, the expected value of x t is:
a) 1-1.5t b) 0c) 1d) 1/t e) None of these
ii)The variance of {x t} is:
1t-) d) 1/(1-t) e) None of these a) 1.5t b) 1 c) 1.5/(2
iii) Suppose we subtract x t-1 from both sides of {x t} we obtain ?x t = u t. Then E(?x t) is: a) t b) E(u t) c)?x t d) 0e) none of these iv) and E[?x t- E(?x t)]2 is:
1t-) e) none of these a) t b) -0.5 c) 1.5d) 1/(2
2)
Consider the general ARMA(p,q) model j t q
j j t p i i t i t u u x x -==-∑∑++=1
1
θβ .
Suppose that you are trying to determine the appropriated order to describe some actual data, with 200 observations available. You have the following figures for the log of the estimated residual variance (i.e log(σ2)) for various candidate models
MODEL ORDER log(σ2) (0,0) 0.932 (1,0) 0.864 (0,1) 0.902 (2,0) 0.801 (0,2) 0.836
i) Calculation of the Akaike Information Criterion will select an ARMA model of order:
a) (0,0) b) (2,0) c) (0,2) d) (0,1) e) (1,0) ii) The Schwartz Bayesian Criterion will select an ARMA model of order
a) (0,0) b) (2,0) c) (0,2) d) (0,1) e) (1,0) iii) Suppose that regardless of the results above you have fitted an ARMA(1,1) to the data obtaining the
following estimates:
t t t t u u x x +++=--1142.069.0036.0
Suppose that you have data for time to (t-1) i.e. you know that y t-1=3.4, and that 3.1?1-=-t u
. The t, t+1, t+2 forecasts are:
a) 715
.0054.1753
.13,12,11,1===---t t t f f f
b) 135
.1542.1436
.13,12,11,1===---t t t f f f
c) 935
.0302.1836
.13,12,11,1===---t t t f f f
d) 735
.0902.0031
.13,12,11,1===---t t t f f f
e) 545
.0312.1666
.13,12,11,1===---t t t f f f
iv) If the actual values for the series turned out to be –0.032, 0.961, 0.203 for t, t+1, t+2, the out of sample mean
squared error is
a) 1.380 b) 1.887 c)2.876
d) 2.003
e) 0.998
3) Consider the Dickey-Fuller test.
i) In the context of a simple AR(1) model such as t t t u x x ++=-1ρμ, the Dickey Fuller test is a t -test of: a) H 0: ρ =0 b)H 0: μ =1 c)H 0: μ+ρ=1 d)H 0: ρ =1 e)None of them ii) Assume that we have run an Augmented Dickey-Fuller test, including a constant a trend and 4 lags of first differences (4321,,,----????t t t t x x x x ), on the log of the variable FTSE100 price index (log(FTSE100)) comprising a sample of 100 observations and we obtained the following value for the ADF statistic:
Variable ADF(4) log(FTSE100) -3.282
You conclude that: a) Log(FTSE100) is definitely stationary b) Log(FTSE100) is definitely non-stationary c) Log(FTSE100) is stationary at the 10% confidence level but non-stationary at the 5% level d) Log(FTSE100) is non-stationary at the 10% confidence level but stationary at the 5% level e) None of the above iii) It is sometimes advisable to use the KPSS test which assumes stationarity under the null because: a) unit root tests such as ADF have low power near the alternative b) unit root tests such as ADF have size distortion in large samples c) the KPSS test perform better when there are structural breaks d) although the ADF test works better when there are structural breaks the KPSS works better in general e) none of the above
4) Consider the following bivariate Vector Auto regression of order 2, VAR (2) in standard form:
t t t t υX ΠX ΠδX +++=--2211,
where ),0(~,,,,212.222.212.122.1121.221.211.121.1112121ΩυυΠΠδX iidN x x t t t t t t ??
????=??????=??????=???
???=??????=υυππππππππδδ
it is common practice to rewrite the model as
)()( ,)(221L L L where L t t ΠΠI ΠυδX Π--=+=is a matrix polynomial in the lag operator. i) This VAR is covariance-stationary if the values of z satisfying 0221=--z z ΠΠI
a) lie on the unit circle (are = 1) b) lie inside the unit circle (are < 1) c) lie outside the unit circle (are > 1) d) some lie on the unit circle but others outside it (some are = 1, some are > 1) e) none of the above ii) Suppose that we have estimated the parameters of this VAR process, and that we found
2.0,2.0,0,0,5.0,1.0,2.0,4.0,2,5.12.222.212.122.111.221.211.121.1121==========ππππππππδδ. As far as Granger-causality is concern, we can say that a) x 1 Granger causes x 2 but x 2 does not Granger cause x 2 b) x 2 Granger causes x 1 but x 1 does not Granger cause x 2 c) x 1 Granger causes x 2 and x 2 Granger causes x 1 d) x 1 does not Granger cause x 2 and x 2 does not Granger cause x 1 e) there is not enough information. ii) If we want to test the hypothesis that one lag of the x 1t and x 2t would be sufficient, so that our H 0 is that our system is a VAR(1) against the alternative H 1 that the system is a VAR(2). The Likelihood Ratio test (corrected for k=1+np 1, the number of parameters estimated per equation) for this hypothesis is:
a) {}
10?log ?log )(Ω-Ω-k T b) {}
10?log ?log Ω-Ωk T c) {}
1?log ?log 2Ω-Ωk
T d) {
}
)?log ?(log 2?log ?log 10012ΩΩ-Ω-Ωk T e) None of the above.
iii) In an attempt to determine the lag order of the VAR using a sample of size T=46 we estimate the model under H 0: the correct model is a VAR(1), and H 1: The correct model is a VAR(2). Assume that we have
obtained the logarithms of the determinants of the covariance matrices for the two cases as: log|Ω0|=1.386, and log|Ω1|=1.147. Also remember that the number of parameters estimated per equation is k=1+np 1, with p 1 being the number of lags of the variables in the unrestricted model. The Likelihood ratio test statistic is then equal to:
a)8.99 b)9.58 c)11.58 d)9.799 e)11.76 v) Recall that the LR statistic is distributed ))(( to equal freedom of degrees with 0122p p n -χunder the null (H 0). Here n is equal to the number of variables in the system, p 0 and p 1 are number of lags under the null and the alternative respectively. Using the results from part iv) and the table for the χ2 distribution we conclude that a) Only one lag is necessary b) two lags are necessary c) three lags are necessary d) two lags are enough but three would be better e) none of the above
5) Consider the following Vector Autoregression in error correction form
t t t U Z Z +∏=?-1,
where
??????+???????????
?--=????????--t t t t t t u u z z z z 2112112145.03.075.05.0
i) The eigenvalues of ∏ are:
a) (0, 0) b) (0, -0.7) c) (0, -0.95) d) (0, -0.78)
e) (-0.88, -0.2) ii) Recall that if z 1t and z 2t are cointegrated, the impact matrix ∏ can be factored into ∏=αβ’. The cointegrating vector β is:
a) [1 -1.5] b) [1 -0.8] c) [1 2] d) [1 -0.6] e)[1 -2] iii) the vector containing the adjustment coefficients α is:
a)??????06.0 b) ??????-6.05.0 c) ??????-5.03.0 d) ??????00 e) ??????-3.05.0 iv) The Johansen tests are tests for
a) The rank of ∏ b) The number of non-zero eigenvalues of ∏
c) The number of cointegrating vectors d) The number of common trends
e) None of the above
6) A researcher is attempting to test for uncovered interest parity (UIP) between The US, UK, and Japan. Let
x 1 = German 3 months treasury bill rate x 2 = UK 3 months treasury bill rate x 3 = French 3 months treasury bill rate.
The researcher has tested for order of integration of the three interest rates and found them to be all I(1) variables, so she decides to test for UIP using the Johansen technique and she obtains the following results.
Eigenvalue 0.262415 0.241907 0.083186
Ho:rank=p -T log(λ
?1-) using T-nm 95% -T ∑log(λ?1-) using T-nm 95% p == 0 27.26 24.35 25.5 55.46 53.45 42.4 p <= 1 22.99 19.16 19.0 30.2 25.50 25.3 p <= 2 7.209 6.948 12.3 7.2 6.94 12.3
i) From inspection of the above table we conclude that the cointegrating relations linking the interest rates of US, UK, and Japan most likely...
a) 1 b) 2 c) zero or 1 d) zero e) 2 or 3
The researcher decides to impose 2 cointegrating relations, and conducts some tests on the cointegration parameters and the adjustment coefficients obtaining the following specifications for α and β. β'
x 1 x 2 x 3 Constant 1.0000 0.00000 -1.0000 -0.089263 0.0000 1.00000 -1.0000 -0.023400 Standard errors of β'
x 1 x 2 x 3 Trend
0.00000 0.00000 0.00000 0.013479
0.00000 0.00000 0.00000 0.010364 α
x 1 0.00000 0.00000 x 2 0.00000 0.0000
x3 0.2276 0.1001
Standard errors of α
x1 0.00000 0.00000
x2 0.00000 0.00000
x3 0.08900 0.03890
ii) From the first column of α she concludes that,
a)There is feedback between the German and the French rates
b)There is feedback between the French and the UK rates
c)The German rate is exogenous and Granger-causes the French rate
d)The UK rate is exogenous and causes the German rate
e) The information given is not sufficient to draw any definite conclusions
iii) Examining the second column of α she concludes that,
a)There is two-way feedback between the UK and the French rates
b)There is two-way feedback between the French and the UK rates
c)The French rate is exogenous and causes the UK rate, but the German rate can also be affected.
d) The French rate receive feedback from the UK rate
e) The information given is just confused.
iv) To test for hypotheses about the parameters in the cointegrating vectors we use:
a) The Akaike Information Criterion
b) A likelihood ratio test.
c) The Lagrange Multiplier (LM) test
d) An F-test
e) None of the above
7) Researchers claim that financial data displays certain specific characteristics which should be treated by means of specific models. i) What stylised features cannot be explained by linear time series models with Gaussian disturbances? a) Homoscedasticity b) Volatility clustering c) Stationarity d) Leverage effects e) None of the above ii) Which of these features are captured by a GARCH(1,1) a) Homoscedasticity b) Volatility clustering c) Non-stationarity d) Non-normality e) Leverage effects
iii) Suppose you encounter the following model
21
5.01),0(~-+=+=t t t t t t u h h u u y μ
Let the unconditional variance of u t be ))]())(([()(t t t t t u E u u E u E u Var --=. In this case, its value will be
a) 10 b) 3
c) 2
d)1
e) 0.5.
iv) If we believe that operators want to be rewarded to hold riskier assets, to model this hypothesis we will need to use a f) MV-GARCH model g) GARCH-GJR model h) E-GARCH model i) GARCH-M model j) Any of the above
v) Suppose we believe that there are asymmetries in the response of volatility to positive or negative returns. To model this phenomenon we parameterise our model as
2+βh t-1
a) y t = μ +other terms + δσt-1+ u t , u t~ N(0,h t) h t = α0 + α1u
t-1
2+βh t-1+γu t-12I t-1
b) y t = μ +other terms + u t , u t~ N(0,h t) h t = α0 + α1u
t-1
2+βh t-1
c)y t = μ +other terms + u t , u t~ N(0,h t) h t = α0 + α1u
t-1
2
d) y t = μ +other terms +δσt+ u t , u t~ N(0,h t) h t = α0 + α1u
t-1
e) Any of the above
压合制程介绍
压合制程目的: 将铜箔(Copper Foil),胶片(Prepreg)与氧化处理(Oxidation)后的内层线路板,压合成多层基板. 内层氧化处理(Black/Brown Oxide Treatment) 氧化反应 A. 增加与树脂接触的表面积,加强二者之间的附着力(Adhesion). B. 增加铜面对流动树脂之润湿性,使树脂能流入各死角而在硬化后有更强的抓地力。 C. 在裸铜表面产生一层致密的钝化层(Passivation)以阻绝高温下液态树脂中胺类(Amine)对铜面的影响。 还原反应 目的在增加氧化层之抗酸性，并剪短绒毛高度，至恰当水准以使树脂易于填充并能减少粉红圈( pink ring ) 的发生。 黑化及棕化标准配方: 表一般配方及其操作条件 上表中之亚氯酸钠为主要氧化剂,其余二者为安定剂,其氧化反应式。 此三式是金属铜与亚氯酸钠所释放出的初生态氧，先---生成中间体氧化亚铜 ----2Cu+[O] →Cu2O,再继续反应成为氧化铜CuO,若反应能彻底到达二价铜的境界,则呈现黑巧克力色之"棕氧化"层,若层膜中尚含有部份一价亚铜时，则呈现无光泽的墨黑色的"黑氧化"层。
制程操作条件( 一般代表),典型氧化流程及条件。 棕化与黑化的比较 黑化层因液中存有高碱度而杂有Cu2O,此物容易形成长针状或羽毛状结晶。此种亚铜之长针在高温下容易折断而大大影响铜与树脂间的附着力,并随流胶而使黑点流散在板中形成电性问题,而且也容易出现水份而形成高热后局部的分层爆板。棕化层则呈碎石状瘤状结晶贴铜面,其结构紧密无疏孔,与胶片间附着力远超过黑化层,不受高温高压的影响,成为聚亚醯胺多层板必须的制程。 B. 黑化层较厚,经PTH后常会发生粉红圈(Pink ring),这是因PTH中的微蚀或活化或速化液攻入黑化层而将之还原露出原铜色之故。棕化层则因厚度很薄.较不会生成粉红圈。内层基板铜箔毛面经锌化处理与底材抓的很牢。但光面的黑化层却容易受酸液之侧攻，而现出
压合制程简介教育内容
壓合製程簡介教育內容 一. 壓合流程介紹 二. 壓合製程原物料介紹 三. 壓合機介紹 四. 影響壓合品質因子 五. 常見壓合之不良項目及改善對策 六. 結論
一. 壓合流程介紹 1-1壓合流程： 黑化(或棕化) 打鉚釘 組合 疊板 上熱壓 上泠壓 拆板 裁半 X-RAY打靶 CNC外框. 1-2 檢驗重點： (1)黑化部份： ?藥液濃度、溫度 . ? Oxide Weight . ?烤箱溫度 . 黑化拉力test. (2)鉚釘對準度. (3)組合之P.P 是否正確. (4)疊板時的對準度. (5)熱壓的溫度.壓力設定. (6)裁半之板面檢查：針孔凹陷、鄒折、氣泡(击出)…… (7)漲縮值. (8)外框之尺寸及粗糙度. (9) 壓合後板厚測試.
二. 壓合製程原物料介紹 2-1 主要原料 2-1-1內層基板(Thin Laminate) (1)檢查重點： ?拉力強度. ?抗酸強度.(抗化學測試) ?表面檢查. 蝕銅後檢查板內. 爆板測試. (2) 一般內層基事板31mil 以下板為不含銅箔. 31 mil (含)以上板原為含銅箔. 2-1-2 膠片(prepreg) (1)檢查重點： ?膠流量(Resin flow) ?膠化時間(gel time) ?膠含量(Resin Content) (2)本廠常用的膠片(殘銅率100%) ? 7628 壓合後厚度7.1±0.7mil ? 2116 壓合後厚度4.2±0.4mil ? 7630 壓合後厚度8.2±0.8mil 1080 壓合後厚度2.5±0.25mil 1506 壓合後厚度6.0±0.6mil 2-1-3 銅箔(copper foil) (1)檢查重點： ?拉力強度.