Econometria

3) Sea Y=Xγ+ ε, donde γ=αγβ1β2
Calcular : ; para el caso en que

Y= Y11Y21Y12Y22 , y X=1101101011100000

Solución:
Y=Xγ, donde ordenando se tiene en términos matriciales:

γ=XιX-1Xι Y …………………………………………..(1)

i. La transpuesta será:

Xι=1111101001100000

luego
XιX= 11111010011000001101101011100000

XιX= 4211220111011001

ii. XιY=1111101001100000Y11Y21Y12Y22

XιY= Y11+Y21+Y12+Y22Y11+ Y12Y21Y11

iii. XιX-1= 4211220111011001-1

Hallando por el método GAUSS –JORDAN

1/4F1

4 | 2 | 1 | 1 | 1 | 0 | 0 | 0 |
2 | 2 | 0 | 1 | 0 | 1 | 0 | 0 |1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 |
1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 |
-2F1+ F2
1 | ½ | ¼ | ¼ | ¼ | 0 | 0 | 0 |
2 | 2 | 0 | 1 | 0 | 1 | 0 | -F1+F3
0 |
1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 |
1 | 1 | 0| 1 | 0 | 0 | 0 | -F1+F4
1 |
1 | 1/2 | 1/4 | 1/4 | 1/4 | 0 | 0 | -1/2F2+F1
0 |
0 | 1 | -1/2 | 1/2 | -1/2 | 1 | 0 | 1/2F1+F3
0 |
-1/2F2+ F4
0 | -1/2 | 3/4 | -1/4 | -1/4 | 0 | 1 | 0 |
0 |1/2 | -1/4 | 3/4 | -1/4 | 0 | 0 | 1 |
1 | 0 | 1/2 | 0 | 1/2 | -1/2 | 0 | 0 |
0 | 1 | -1/2 | 1/2 | -1/2 | 1 | 0 | 2F3
0 |
0 | 0 | 1/2 | 0 | -1/2 | 1/2 | 1 | 0 |
-1/2F3+ F1
0 | 0 | 0 | 1/2 | 0 |-1/2 | 0 | 1 |
1 | 0 | ½ | 0 | ½ | -½ | 0 | 1/2F3+F2
0 |
0 | 1 | -½ | ½ | -½ | 1 | 0 | 0 |
0 | 0 | 1 | 0 | -1 | 1 | 2 | 2F4
0 |
0 | 0 | 0 | ½ | 0 | -½ | 0 | 1 |
1 | 0 | 0 | 0 | 1 | -1 |-1 | -1/2F4+F2
0 |
0 | 1 | 0 | ½ | -1 | 3/2 | 1 | 0 |
0 | 0 | 1 | 0 | -1 | 1 | 2 | 0 |
0 | 0 | 0 | 1 | 0 | -1 | 0 | 2 |
1 | 0 | 0 | 0 | 1 | -1 | -1 | 0 |
0 | 1 | 0 | 0 | -1 | 2 | 1 | -1 |0 | 0 | 1 | 0 | -1 | 1 | 2 | 0 |
0 | 0 | 0 | 1 | 0 | -1 | 0 | 2 |
Por tanto se tiene:

XιX-1=1-1-10-121-1-101-12002

iv. Reemplazando datos en la ecuación (1)

γ=Xι X-1Xι Yγ=1-1-10-121-1-101-12002Y11+Y21+Y12+Y22Y11+ Y12Y21Y11

γ= Y11+Y21+Y12+Y22- Y11-Y12- Y21-Y11- Y21- Y12- Y22+2Y11+2Y12+Y21-Y11-Y11-Y21-Y12-Y22+Y11+Y12+2Y21-Y11-Y12+2Y11

γ= Y22Y12- Y22Y21-Y22Y11-Y12...