HANDOUT 2 Flashcards
Serial correlation =
the presence of some form of linear dependence over time for some series Zt.
Auto-Correlation function =
a pictorial representation of this linear dependency over time. It measures the correlation between Zt and Zt-k for different values of k.
Corr(Zt, Zr-k) formula
= COV(Zt, Zt-k)/sqrt[V(Zt) V(Zt-k)]
V(Zt) = V(Zt-k) = gamma 0 by stationarity
COV(Zt, Zt-k) = gamma k since only depends on distance apart by stationarity.
So Corr = gamma k / gamma 0 = pk
If p=1 this means
Shock today, on a new path & never return to old equilibrium as never forget about the shock.
If p=0 this means
Shock today, next period no memory of it = immediate adjustment back to equilibrium.
A change in X1t on Yt in a model with no lags means…
change in X1t only affects Y today
At t+1, immediately adjust back to equilibrium.
White noise process formula
Zt = €t
E(Zt) for white noise
E(Zt) = 0
V(Zt) for white noise
V(Zt) = V(€t) = sigma^2
COV(Zt, Zt-k) for white noise
= 0 since E(€t.€t-k)=0
P1 to Pk for white noise
P0 = 1; all Pk = 0
ACF for white noise
Shock at period 0, immediately return back to equilibrium by period 1.
What does an ACF show?
The proportion of a shock remaining k periods later. The correlation compared to period 0 when the shock hits.
AR1 model formula
Zt = phi Zt-1 + €t
one lag of dependent variable
What condition must we impose on phi for AR1 and why?
I phi I < 1
For stationarity - otherwise shock will never dissipate out of system.
E(Zt) for AR1
E(Zt) = phi E(Zt-1) + E(€t)
E(Zt) = E(Zt-1) by stationarity
(1-phi)E(Zt) = 0
Since phi<1, must be E(Zt) = 0
V(Zt) for AR1
V(Zt) = phi^2 V(Zt-1) + V(€t) + 2COV(Zt-1, €t)
V(Zt) = V(Zt-1) by stationarity
(1 - phi^2)V(Zt) = sigma^2
V(Zt) = sigma^2 / (1 - phi^2)
COV(Zt, Zt-1) for AR1
E(Zt.Zt-1) as zero mean E[Zt-1(phiZt-1 + €t)] = phi V(Zt) gamma 1 = phi gamma 0 = phi sigma^2 / (1 - phi^2)
Corr(Zt, Zt-1) for AR1
Corr = gamma 1 / gamma 0 = Phi
P2 for AR1
P2 = Phi^2
Pk for AR1
Pk = Phi^k
2 shapes for AR1 look like
In both cases mod Phi< 1
If Phi>0, smooth decay to 0
If Phi<0, oscillations by still end up at 0.
Roots for AR1 using lag operator
(1 - phiL)Zt = €t
Solve 1 - phi L = 0
V = L^-1
V = phi - AR1 has 1 root
Why when solving for roots, do we solve for L^-1 and not L?
Because if we solved for L, the root would > 1 so wouldn’t be stationary.
AR2 model formula
Zt = phi1 Zt-1 + Phi2 Zt-2 + €t
Roots for AR2
1 - phi1 L - phi2 L^2 = 0
V = L^-1; V^2 = L^-2
V^2 - phi1 V - phi2 = 0
solve by quadratic formula - 2 roots for AR2
E(Zt) for AR2
E(Zt) = 0
V(Zt) for AR1
gamma 0 = phi1 gamma 1 + phi2 gamma 2 + sigma^2
Yule walker equations x 2 for AR
- gamma 1 = phi1 gamma 0 + phi2 gamma1
2. gamma 2 = phi1 gamma 1 + phi2 gamma0
How many possible ACFs for AR2?
4
P1 = Corr(Zt, Zt-1) for AR2
phi1 / (1 - phi2)
MA1 model formula
Zt = theta €t-1 + €t
Average of 2 shocks
1 period memory only
What conditions do we need on theta for MA1? Why?
NO conditions - an MA model is stationary by definition.
E(Zt) for MA1
0