Name SOLUTIONS Financial Economerics Jeffrey R. Russell Miderm Winer 009 SOLUTIONS You have 80 minues o complee he exam. Use can use a calculaor and noes. Try o fi all your work in he space provided. If you find you need more space coninue on he back of he page. Some problems are hard and some are easy so budge your ime accordingly. Mos imporanly, don sress, have fun! Sudens in my class are required o adhere o he sandards of conduc in he GSB Honor Code and he GSB Sandards of Scholarship. The GSB Honor Code also require sudens o sign he following GSB Honor pledge, "I pledge my honor ha I have no violaed he Honor Code during his examinaion. Please sign here o acknowledge
. Consider he following plo ACF and PACF for a series wih 400 observaions. a. Based on hese saisics, wha is your bes guess for an ARMA(p,q) model ha would fi he daa. Give a value for p and a value for q. Sae your reasoning o ge full credi. ARMA(,0), since PACF cus off afer lag and he ACF exhibis slowly decaying behavior b. If you were o fi an MA model o his daa, wha can you say abou he number of righ hand side variables ha would be needed? Explain. We would need many righ hand side variables, i.e. los of lags (high q) Recall ha an AR model can be wrien as an MA( ) model by Wold s heorem. c. If you were o fi an AR model o his daa, wha can you say abou he number of AR erms ha would be needed? Explain. We would need wo AR erms, i.e. we would fi an AR() model since he PACF cus off afer lag. d. So which of he wo models (in pars b. and c.) would you prefer and why? We would prefer an AR() since i has fewer parameers and would yield beer forecass.
. When financial asse prices are observed over shor ime horizons he microsrucure of he marke s induces some paerns or predicabiliy in he reurns. Suppose ha a model for rade by rade reurns is given by: r =.3ε + ε where ε is iid N(0,.00). a. Wha is he one sep ahead condiional disribuion of r if ε =.0007? Condiioning on he informaion se a ime, we have ha ( ) and ha var ( ) 0.00 E r ε = ( 0.3)( 0.0007) r ε =. Thus he one sep ahead condiional disribuion is r F ~ N(( 0.3)( 0.0007), 0.00) b. Wha is he uncondiional reurn? The uncondiional reurn is given by μ = 0 c. Wha is he average size of he squared reurn (i.e. he uncondiional variance of r )? This is formally given by: ( r ) = + ( ) var 0.00 0.3 (0.00)
3. Consider he AR() model for quarerly dividend yields given by: d =.008 +.8d + ε where ( ) ε ~ iid N 0,.0 a. Wha is he average dividend yield? 0.008 The average dividend yield is given by μ = = 0.04 0.8 b. If d =.03, find he k sep ahead forecas as a funcion of k. This should be expressed as a funcion of k and some known numbers. This is formally given by y k = (0.8) k (0.03) + ( 0.8 k )(0.04) c. Find he k sep ahead forecas error variance as a funcion of k. This is formally given by ( 0.8 ) k k ( 0.8 ) var( e ) = (0.0 ) d. If d =.03, find a 95% predicive inerval for he dividend yield,, and 3 quarers ahead. You will have 3 differen inervals here. sep ahead forecas is given by ( 0.8 ) (0.8) (0.03) + ( 0.8 )(0.04) ± (0.0 ) ( 0.8 ) sep ahead forecas is given by 4 ( 0.8 ) (0.8) (0.03) + ( 0.8 )(0.04) ± (0.0 ) ( 0.8 ) 3 sep ahead forecas is given by 6 3 3 ( 0.8 ) (0.8) (0.03) + ( 0.8 )(0.04) ± (0.0 ) ( 0.8 ) e. When k becomes large wha does he forecas inerval converge o? Give he inerval ha i converges o and give an inerpreaion of ha inerval. When k becomes large he forecas inerval converges o 0.04 (0.0 ) ± ( 0.8 ) Inerpreaion: 95% of he ime, we can expec forecas o fall in ha inerval
f. Suppose ha you made a misake and ha he rue model doesn follow and AR() bu you assumed in your modeling ha i did. Your residual analysis suggess ha he residual s follow an AR() model hemselves ε =.ε + γ where γ is iid N ( 0,.0 ). Wha is he k sep ahead forecas error variance forecas now? var k ( e ) k ( β ) k k = var β ε+ k j = σ ε j= 0 β cov ε, ε 0. k σ Where ( ) k = γ Hence his inerval will be larger + (all possible covariances) g. When k becomes large, wha value does he variance converge o now? I converges o he uncondiional variance of This variance is larger han he case where ε is iid y
4. The following graph is of he daily dollar euro foreign exchange rae. The able on he righ presens saisics for a uni roo es..7.6.5.4.3...0 0.9 0.8 99 00 0 0 03 04 05 06 07 08 DOLL_EURO a. Based on he graph o he lef, does he exchange rae daa appear o follow a random walk or be saionary? Explain. Exchange rae appears o wanders around, no araced o a mean level OR Difficul o say since i could be highly persisen saionary AR model, i.e. β < b. Consider he following oupu and perform a es for a random walk. Clearly sae your null hypohesis, p value, and he conclusion of your es. H0 : β = Since he p value = 0.79, we fail o rejec he null hypohesis I.E. he daa appears o follow a random walk
5. Consider he following ime series daa. Describe which series you would expec o follow a random walk and which should be saionary. You mus explain your reasoning o ge full credi: a. Unemploymen Raes Saionary, since raes are bounded and canno wander oo far from he mean b. Foreign exchange Raes Non saionary since raes vary freely OR Could be locally saionary if governmen arges or pegs is currency. c. Dow Jones Indusrial average reurns. Saionary, in fac, he efficien marke hypohesis suggess ha hey are iid over ime. d. Absolue value of Dow Jones Indusrial average reurns. Saionary. Volailiy has hisorically been mean revering, albei highly persisen.
6. Consider he following saisics for he reurns on a financial asse. a. Do he reurns show any signs of predicabiliy? Explain. No, reurns do no show any signs of predicabiliy since he values of he auocorrelaions are very small OR A shor lags here is a small negaive correlaion, hus here is some predicabiliy in he reurns a small horizons b. Perform hypohesis es o disinguish if he reurns are correlaed or uncorrelaed hrough ime. Sae your null, he p value, and he conclusion. H0 : ρ = 0 The corresponding p value is 0.05, hus we rejec he null a he 5% level, and hus he daa appears o be predicable OR H0 ρ ρ ρ k : = =... = = 0 (pick your value of k) The corresponding p value is 0.053 (for a choice of k=0), hus we marginally fail o rejec he null a he 5% level, and hus he daa appears o be unpredicable Noe ha he ρ i are he auocorrelaion coefficiens, lised underneah he AC column in he compuer oupu.
The following are he ACF and PACF for he absolue value of he reurns. c. Do he absolue values of he reurns show any signs of predicabiliy? Explain. Yes, he absolue values of he reurns show signs of predicabiliy since here are significan auocorrelaions up hrough he 0 h lag d. Perform hypohesis es o disinguish if he absolue value of he reurns are correlaed or uncorrelaed hrough ime. Sae your null, he p value, and he conclusion. H0 ρ ρ ρ k : = =... = = 0 The corresponding p value is 0.000, hus we rejec he null a he % level, and hus he daa appears o be predicable Noe ha he ρ i are he auocorrelaion coefficiens, lised underneah he AC column in he compuer oupu. e. Do he reurns appear o be iid? If so, sae why, if no, sae why he reurns are no iid. No he reurns do no appear o be iid. The above oupu is indicaive of ime varying volailiy in he reurn series.
7. Equiy reurns are forecasable using pas dividend price raios. Tha is, r =.06 +.003d + ε where ε is iid N ( 0, σ ε ) and d is he dividend price raio. Dividend price raios are very persisen. Suppose ha he dividend price raio follows he AR() model d =.008 +.8d + ξ where ξ is iid N ( 0, σ ξ ). a. If σ ξ 0, wha univariae process do he reurns follow. Wha is he ARMA model for reurns? (HINT: you should sar by subracing.8r from boh sides of he reurn equaion) Firs, we use he hin and find ha: r 0.8r = 0.06 + 0.003d + ε 0.8( 0.06 + 0.003d + ε ), implying ha r 0.8r = 0.06( 0.8) + 0.003( d 0.8 d ) + ε 0.8ε Drawing on he observaion ha d 0.8d = 0.008 + ξ, we simplify he above o r 0.8r = 0.0 + (0.008)(0.003) + 0.003ξ + ε 0.8ε, implying ha r = 0.004 + 0.003ξ 0.8ε + ε + 0.8r Thus if σ 0, we have derived an ARMA(,) model for reurns ξ b. If σ 0, wha univariae process do he reurns follow. Wha is he ARMA model for ξ reurns? (HINT: You need o look a he correlaion srucure of he errors) Firs, wrie r = 0.004 + 0.8r + 0.003ξ 0.8ε + ε η We use he hin o esablish he correlaion srucure of he errors. A lag, we have ( ηη ) = ([ 0.003ξ 0.8ε + ε][ 0.003ξ 0.8ε + ε ] ) = 0.8σ E E ε A lag, we have ha E ( ηη ) E( [ ξ ε ε][ ξ 3 ε 3 ε ] ) = 0.003 0.8 + 0.003 0.8 + = 0 Similarly, lags beyond lag have auocorrelaions equal o 0. Thus we conclude ha he reurns follow an ARMA(,) model since he errors (η ) have non zero s order auocorrelaions, bu zero afer lag.