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Econometrics

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Question no. 1
Y1= ∝0+∝1Y2+∝2X1+∈1
Y2= β0+β1Y1+β2X1+β3X3+ϵ2 i. Identification Status:
Equation 1: P1=1, P2=1 so that P1=P2 so, equation is Exactly identified.
Equation 2: P1=0, P2=1 so that P1<P2 so, equation is Unidentified.

ii. Reduced form equations:
Putting Y1 in Y2:
Y2= β0+β1(∝0+∝1Y2+∝2X1+∈1) +β2X1+β3X3+ϵ2
Y2= β0+β1α0+β1α1Y2+β1α2X1+β1ϵ1+β2X1+β3X3+ϵ2
Y21-β1α1= β0+β1α0+X1β1α2+β2+β3X3+β1ϵ1+ϵ2
Y2= β0+β1α01-β1α1+β1α2+β21-β1α1X1+β31-β1α1X3+β1ϵ1+ϵ21-β1α1
Y1=π20+π21X1+π22X3+ν2

Now putting this reduced form equation of Y2 in Y1 equation:
Y1= ∝0+∝1(π20+π21X1+π22X3+V2)∝2X1+∈1
Y1= ∝0+∝1π20+X1α1π21+α2+α1π22X3+α1V2+ϵ1
Y1= π10+π11X1+π12X3+V1

π10= ∝0+∝1π20 α0= π10-π12π22(π20) π11= α1π21+α2 α2= π11- π12π22(π21) π12= α1π22 α1= π12π22
Using STATA the reduced form equation (DATA set 1)

Y2= -1.57953 -.37781X1+1.744 X3
Y1= 14.245+ .67809 X1+ .99181 X3

Estimations of structural parameters
For equation 1:
Run the regression on reduced form equations in STATA and we calculated the following values of structural parameters: 1. α1= π12π22
= .99181 1.744 α1= 0.5688 2. α2= π11- π12π22π21
= .67809 - 0.5688 (-.37781) α2= 0.8930 3. α0= π10-π12π22(π20)
= 14.245-0.5688(-1.57953) α0= 15.1437

Question no. 2
Y1=αo+α1Y2+α2X1+α3X2+e1
Y2=βo+β1Y1+β2X1+β3X3+e2
Status Identification of equations:
For equation 1:
P1=1
P2=1
P1=P2This identifies that the equation 1 is “Exactly-identified”.
For equation 2: P1=1 P2=1
P1=P2 This identifies that the equation 2 is “Exactly-identified”
Transformation into reduced form equation:
Putting Y2 equation in Y1 equation
Y1=αo+α1[βo+β1Y1+β2X1+β3X3+e2]+α2X1+α3X2+e1
Y1 1-α1β1=αo+α1βo+α1β2X1+α1β3X3+α1e2+α2X1+α3X2+e1
Y1 =αo+α1βo1-α1β1+α1β2+α2X11-α1β1+α3X21-α1β1+α1β3X31-α1β1+α1e2+e11-α1β1
Y1=π10+π11X1+π12X2+π13X3+V1

Now putting this reduced form equation of Y1 in Y2 equation:
Y1=βo+β1[π10+π11X1+π12X2+π13X3+V1]+β2X1+β3X3+e2
Y1=βo+β1π10+ β1π11X1+β1π12X2+β1π13X3+β1V1+β2X1+β3X3+e2
Y1=(βo+β1π10)+(β1π11+ β2 )X1+β1π12X2+(β1π13+β3)X3+β1V1+e2

Y2=π20+π21X1+π22X2+π23X3+V2

π20=βo+β1π10 βo=π20- π22π12.π10 π21 = β1π11+β2 β2=π21- π22π12.π11 π22=β1π12 β1= π22π12 π23=β1π13+β3 β3=π23- π22π12.π13 Note: Equation Y2is exactly-identified because its structural parameters can be found out by one way so we can solve this equation through ILS.
Now putting Y1 in Y2 equation:
Y2=βo+β1[αo+α1Y2+α2X1+α3X2+e1]+β2X1+β3X3+e2
Y21-β1α1=βo+β1αo+β1α2X1+β1α3X2+β1e1+β2X1+β3X3+e2
Y2=βo+β1αo1-β1α1+β1α2+β21-β1α1X1+β1α31-β1α1X2+β31-β1α1X3+
β1e1+e2(1-β1α1)
Y2=π30+π31X1+π32X2+π33X3+V3

Putting reduced equation of Y2 in Y1
Y1=αo+α1[π30+π31X1+π32X2+π33X3+V3] +α2X1+α3X2+e1
Y1=π40+π41X1+π42X2+π43X3+V4

Y1=αo+α1π30+α1π31X1+α1π32X2+α1π33X3+α1V3+α2X1+α3X2+e1

π43=α1π33 α1=π43π33 π40=αo+α1π30 αo=π40-π43π33. π30 π41=α1π31+α2 α2=π41-π43π33. π31 π42=α1π32+α3 α3=π42-π43π33. π32
Note: As the equation 1 is exactly identified so we can find out the structural parameters through ILS
Using STATA the reduced form equation (DATA set 1)
Y1=15.519+1.103X1+.384X2+.719X3
Y2=.892+.447X1+ .746X2+1.216X3

Estimations of structural parameters
For equation 1:
Run the regression on reduced form equations in STATA and we get the following values of structural parameters: 4. α1=0.59161 5. α2=0.83886 6. α3=-0.05674 7. α0=14.99196

For equation 2:
Now we compute the values of β from α values 1. β1=1.93972 2. β2=-1.69311 3. β3=1.93972 4. βo=-29.2117

Question # 3: Model
Y1=α0+α1Y2+α2X1+α3X2+ϵ1
Y2=β0+β1Y1+β2X1+β3X3+β4X5+ϵ2 iii. Identification Status:
Equation 1: P1=1, P2=2 so that P1<P2 so, equation is Over identified.
Equation 2: P1=1, P2=1 so that P1=P2 so, equation is Over Exactly identified. iv. Reduced form equations
Putting Y2 in equation Y1
Y1=α0+α1β0+β1Y1+β2X1+β3X3+β4X5+ϵ2+α2X1+α3X2+ϵ1
Y1=α0+α1β0+α1β1Y1+α1β2X1+α1β3X3+α1β4X5+α1ϵ2+α2X1+α3X2+ϵ1
Y1(1-α1β1)=α0+α1β0+(α1β2+α2)X1+α3X2+α1β3X3+α1β4X5+α1ϵ2+ϵ1
Y1=α0+α1β0(1-α1β1)+(α1β2+α2)(1-α1β1)X1+α3(1-α1β1)X2+α1β3(1-α1β1)X3+α1β4(1-α1β1)X5+α1ϵ2+ϵ1(1-α1β1)
Y1=π10+π11X1+π12X2+π13X3+π14X5+υ1
Now putting reduce equation of Y1 in equation Y2
Y2=β0+β1[π10+π11X1+π12X2+π13X3+π14X5+υ1]+β2X1+β3X3+β4X5+ϵ2
Y2=β0+β1π10+(β1π11+β2)X1+β1π12X2+(β1π13+β3)X3+(β1π14+β4)X5+β1υ1+ϵ2
Y2=π20+π21X1+π22X2+π23X3+π24X5+υ2
π20=β0+β1π10 | β0=π20-π10[π22π12] | π21=β1π11+β2 | β1=π22π12 | π22=β1π12 | β2=π21-π11[π22π12] | π23=β1π13+β3 | β3=π23-π13[π22π12] | π24=β1π14+β4 | β4=π24-π14[π22π12] |

v. Estimated reduced equations:
Y1=π10+π11X1+π12X2+π13X3+π14X5+υ1
Y1=-6.860086 +1.301827 (X1)+.7571526 (X2)+.4143187(X3)+.1310392(X5)
Y2=π20+π21X1+π22X2+π23X3+π24X5+υ2
Y2= 6.063845+.4011194 (X1)+.659783 (X2)+ 1.286288(X3)-.0302849 (X5) vi. Structural Parameters of Equation 2 ; by using ILS
Y2=β0+β1Y1+β2X1+β3X3+β4X5+ϵ2
β0=π20-π10[π22π12] β0=6.063845 -6.860086[.659783.7571526] β0=0.085964002 β1=π22π12 β1=.659783.7571526 β1=.8714003 β2=π21-π11π22π12 β2=.4011194-1.301827[.659783.7571526] β2=-.73329304 β3=π23-π13π22π12 β3=1.286288-.4143187 [.659783.7571526] β3=.92525056 β4=π24-π14π22π12 β4=-.0302849-.1310392[.659783.7571526] β4=-.1444725
Question no. 4
Y1=α0+α1Y2+e1
Y2=β0+β1Y1+β2X1+β3X5+e2
Status Identification of equations:
For equation 1:
P1=2
P2=1 P1>P2This identifies that the equation 1 is “over-identified”.
For equation 2: P1=0 P2=1 P1<P2 This identifies that the equation 2 is “un-identified”

Transformation into reduced form equation:
Putting Y1 equation in Y2 equation:
Y2=β0+β1[α0+α1Y2+e1]+β2X1+β3X5+e2
Y2=β0+β1α0+β1α1Y2+β1e1+β2X1+β3X5+e2
Y2[1-β1α1]=β0+β1α0+β2X1+β3X5+β1e1+e2
Y2=β0+β1α01-β1α1+β21-β1α1X1+β31-β1α1X5+β1e1+e2 1-β1α1
Y2=π20+π21X1+π22X5+v2

Now putting this reduced form equation of Y2 in Y1 equation:
Y1=α0+α1[π20+π21X1+π22X5+v2]+e1
Y1=α0+α1π20+α1π21X1+α1π22X5+α1v2+e1 Y1=π10+π11X1+π12X5+v1

π10= α0+α1π20 α0=π10-π11π21*π20 π11=α1π21 α1=π11π21 π12=α1π22 α1=π12π22 Note: Equation Y1is over-identified because its structural parameters can be found out by more than one ways so we can solve this equation through 2 stage least square (2SLS). Estimations of structural parameters
For equation 1:
Run the regression on reduced form equation of Y2 in stata in first step and we get the values of π :
Y2=37.125+2.745X1+.043X5
Then in second step run the regression Y1 on Y2hat then get the following values of structural parameters 1. α0=0.3354 2. α1=0.89298

Question no. 5
Y1=α0+α1Y2+α2X1+e1
Y2=β0+β1Y1+β3X5+e2
Status Identification of equations:
For equation 1:
P1=1
P2=1 P1=P2This identifies that the equation 1 is “exactly-identified”.
For equation 2: P1=1 P2=1 P1=P2 This identifies that the equation 2 is “exactly-identified”
Putting Y1 equation in Y2 equation:
Y2=β0+β1[α0+α1Y2+α2X1+e1]+β3X5+e2
Y2=β0+β1α0+β1α1Y2+β1α2X1+β1e1+β3X5+e2
Y2[1-β1α1]=β0+β1α0+β1α2X1+β3X5+β1e1+e2
Y2=β0+β1α01-β1α1+β1α21-β1α1X1+β31-β1α1X5+β1e1+e2 1-β1α1
Y2=π20+π21X1+π22X5+v2

Now putting this reduced form equation of Y2 in Y1 equation:
Y1=α0+α1π20+π21X1+π22X5+v2+α2X1+e1
Y1=π10+π11X1+π12X5+v1

Y1=α0+α1π20+(α1π21+α2)X1+α1π22X5+α1v2+e1 π10= α0+α1π20 α0=π10-π11π21*π20 π11=α1π21+α2 α2= π11-π12π22 *π21 π12=α1π22 α1=π12π22 Estimations of structural parameters
For equation 1:
ILS
By using STATA:
Y2=37.125+2.745X1+.043X5
Y1=18.263+ 2.243X1+ 0.120X5 1. α0= -85.467 2. α1= 2.749 3. α2= -5.427
2SLS:
Run the regression on reduced form equation of Y2 in STATA in first step and we get the values of π :
Y2=37.125+2.745X1+.043X5
Then in second step run the regression Y1 on Y2hat then get the following values of structural parameters 4. α0= -85.467 5. α1= 2.749 6. α2= -5.427
Thus, prove that ILS and 2SLS provide identical results for exactly identified equations.

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