ec40 (2018/19) ltpersonal.lse.ac.uk/sabharw5/ec402ltps3and4.pdf · 2019-03-21 · so the...
TRANSCRIPT
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EC40 2 (2018/19) - LT
Ps3 Q7 } TIME SERIESPS4 05,02
RAGVIR 's SPEAKING NOTES !
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Ps3
JQ71 log We = light - it kit - ①
left = plague + ref -20
=wµ,
re,e)
' EL9E ) * structural shocks.
a) Define Xt fight, 1g G)? Rewrite the model given by the system in ① and @in terms of a VAR model for Xt .
We need to specify A,
and
4It
.
X A, Xt . It %
ii.
fight:) = Ailing t k
So let's try to wile bogle as a function of Kyle ,and 1g Uh
Igh .
-- plight . It prw.tt rat - ③
Thisgives
us what we need. . .
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let A, (poo) and Ct ÷
¥ ) 't
µ VAR errors =/ I
m.IT/egiIfIudYjaYdAR%f7byXt--AiXt-i-④is consistent with
me : E-+⇒ ⇒ EE41 = ftp.fpif
,
(b) Describe the SVAR approach .
Comment on advantages and disadvantages .
Step I : VARY ) - Xt-AXt.it Et with4810
,Ig )
and Ee=
ftp.?pJSHpL.hverttheVAR4)-iXt--B4)EtwhereBl4=fI-A,LY
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Step 3 . Recover the structural shocks using the relationship
G. = ME where M := ¥1 .
ie. if = Nice
letIndeed
, if we corpore the above to the Sims Kato) methodologyof picking a rotation matrix P t.tt
(a) PD = Eid and
(b) I is lower triangular,
- Iwe the that M exactly satisfies the above two conditions
.
check : a) HIM -
- fo.f) Kpof iff) = EI'
4) hit = M = is lower In angular .
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So the advantage of the SVAR approach is that we can indeed recover the
structural shocks,
but this hinges on using the appropriate rotation matrix P = M - t.
For anyother choice of P Fri ,
we would potentially mis - identify the true structuralshocks by computing different linear combinations of the elements of q and thisnot recover rt .
4) Prove that elements of vector XE are cointyraled with Coinlegrahng vector 8 : = Ep,1)
'
SHE= - flight + lglt =- plague + fight + rat Iusing equation @ )
= VGE .
Since ffs is an ID pours and has finite mean and finite variance,
it is of course so. stationary .
The stated result follows .
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COMMENT :
I'm writing a comment here because I actually think is a
verynice question for
intuition on SVARS. If I had read it last week
,I would have discussed it when
I wrote you mynotes on
"
intro to SVARS ?
So let's understand what we've just seen lie . in part (a) and (b) ) :
• Normally,
we would start with a VARA like XE-AIXt.it Ee with ETWNIQE e)We would want to identify the shnrthral shocks if so we'd pick a P
and ho# that PEE gives us re .Indeed if A- NT
,we'd indeed recover rt .
However, for any other choice of Rfm
- '
et.
PP = EE'
,we'd get some orthogonal stochastic
disturbances say twhich we'd then interpret as our structural shocks but
we'd bewrong.
As? result,
our IRFI BL4P would be wrong and this would be dangerous for policyanalyses
,
. for example , if we predict that contraction ay monetary shocks lead to expansions in output.
Theonly reason we know exactly how to recover if
inthisquestion
isbecausewe know equations ① and ② but in real . life we would not know the exact relation
between stand I .
• This question is also useful to understand how the Chdesky decomposition may make
sense .Look at the IRF BC I M
.. .
tie if you are struggling for intuition on that.
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/This is the MA rep of Xt in terms of the primitive the trial shocks
.
4gthe) = ('
p 9)(III ) t Bi ME - I t Bam rent . - -
⇐E. t
does net impact light instantaneously
Well,
indeed, if you
stare at equations ① and ③, you see exactly
this : vw,
thits both byWE and logCt on impact but vat won't affect bgwt .
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PS4=
951¥.
: 5¥ .
, * °
, bz I 0
with fat,%)
'
n WN10,
Iz )Is there a eointegraton vector that includes all 3 It ) variables 1kt
, KY43 ,with
non - two weight?
Consider the - bike = GtYit - by Y3T = Elt
i. Yit t the - bzYzt - by Y3T = Get Ezt
=) Yet f-be) Yet - b) 1st = Get hit is It ) .
So 4,I - bz
,
- bz )'
is a coin legrating vector for Heike,be) !
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B4
IQ2 ) in 4TH-4'
Xt -- Et
,G- unto ,
where 4121 ,r > 0
.
Show that Gif is I14 .
let r=o . 4-pl )Xt= 4
EH I - pZ has roots outside that
⇒¥3 is If ) .
let at : 4-pl) 4- L) Xt -- Et
⇒ YP4H- X = Et
⇒ 474 Dxt = Et
Eft ) ÷ 1-fZ has roots outside Htt ⇒ ⑨Xx ) is IG ) so Xt is It ) .
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ht r= 2 : Itpl) f-LYXT = Et
⇒ f-pl) II - alt E) Xt
= 4-f) He- 2X t.it Xt - 2)
= f-f)
IXEXH-kt.it/t-if)--f-p4IDXt-bXtn)
= ifpl ) D4E = Et .
If I = I -
piehas roots outside HE I ⇒ LD4H is It ) so ④Xe) is IN
and this It ) is It ) .
We conclude that for general r, ④-4%3=4%3 is I and {XD is IL4 .
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End the integration order of the AR4) process :
( I - 2542 E- 05 E) ye = 4,
4N WN6,02)
( "
somethin
;)It-4 ye = Et → going by logic in party ,
I'm just
( I - ist test ) It y yt= q
factorising this cubic polynomial .
("
something else"
)EL5Y a= Et → same again .
(I - 0.5L ) ⇐L5J t= Et
Note : OI ⑦ = I -Ote has roots outside HE I so Hye) is It) and this Lye)is It ) .
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EXTRA MATERIAL
let's look at 3 types of stochastic processes :
d) yt HttpXt where ¥3 is a zero - mean SO.
stationary process .
Also, 9ps f- 0
.
) yt-y.ly + Et where 4) is a zero - mean WN process with variance off to.
It. 4¥; I
for some fixed point in time I.
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RAW 1RI Quiz - Discuss AMONGST Your SELVES=
Explain in each case whether EXES is or is not SO. stationary .
.
lit ×t Exp 4) ,d so
.
lii ) Xt=Xt. ItZt,
Zt NL91) with Ztt Xs for sat.
Iii )Xt=tffor all t,
where I - N10, 1) .
K )Xt=Bfor all t,
wherePg~t, .
µ ) ×+
= y,
where He } is S -0. stationary .
I know I'm being mean
⇐ I Xt =/ the;E5Ebut thisis a reality check
Ye for people who don't visit me
if they have questions .
This
for some fixed point = / is all week 1 material ! !
NOTE : I WILL NIT BE PROVIDING SOLUTIONS - EVER !! IF You ARE
UNIUREABOUT ANY OF THE ABOVE
,You HAVEN'T STUDIED 5.0
. STATIONARYENOUGH .
IF You NEED HELP,
ASK EACHOTHER PLEASE,
NOT ME !