The Exact Distribution of the TSLS Estimator for a NonGaussian Just-Identified Linear Structural Equation by Giovanni Forchini

September 13, 2006

Supplementary results Expressing φw2 ( t1 ) as c2 Φ c ( t1 | w22 ) φm ( w22 ) 2

Notice that

φw ( t1 ) = c2 2 2

=

∫

0< q
∫

φ (q + w22 )q −1/ 2 dq × c2φm ( w22 ) 2 φm ( w2 )

∫

pdf ⎛⎜⎝ t1 | w22 ⎞⎟⎠ dq × c2φm ( w22 )

0< q
=

φ (q + w22 )q −1/ 2 dq

0< q
= Φ c ( t1 | w22 ) c2φm ( w22 ) .

Detailed proof of Proposition 1 The starting point is the density of βˆ when Π = 0 ,

( )

−1/ 2 pdf βˆ = c2 det ( Ω )

∫

ˆ −∞<Π<+∞

( )

( )

ˆ. ˆ φ ⎛⎜⎜ a βˆ Π ˆ 2 ⎞⎟⎟ d Π Π ⎝

( )

⎠

ˆ to Π ˆ = x / a βˆ . The Jacobian is 1/ a βˆ , so We transform Π

1

( )

−1/ 2 pdf βˆ = c2 det ( Ω )

∫

−∞< x <+∞

= c2 det ( Ω )

( )

−1/ 2

a βˆ

⎛

( )

( )

( )

2⎞

( )

⎜ ⎟ x / a βˆ φ ⎜⎜ a βˆ ⎡ x / a βˆ ⎤ ⎟⎟ / a βˆ dx ⎢ ⎥ ⎜ ⎣ ⎦ ⎟⎠ ⎝

−1

∫

−∞< x <+∞

x φ ⎛⎜⎝ x 2 ⎞⎟⎠ dx

Now we observe that the integrand is symmetric around zero so that

( )

( ) ∫

−1 / 2 pdf βˆ = 2c2 det ( Ω ) a βˆ

−1

0 < x <+∞

x φ ⎛⎜⎝ x 2 ⎞⎟⎠ dx ,

( )

and set x = t . The Jacobian of this transformation is 1/ 2 t , so

( )

( ) ∫ −1

−1 / 2 pdf βˆ = c2 det ( Ω ) a βˆ

φ ( t ) dx .

0 < t <+∞

Equation (13) follows from the fact that (10) integrates to one so that

∫

π c2

φ ( t ) dx = 1 .

0 < t <+∞

Detailed proof of Proposition 2 ˆ as follows We start from equation (12) In the general case, we transform Π

( )

ˆ = a βˆ Π

−1/ 2

(

( )( )

s + b βˆ a βˆ

−1/ 2

( )

) . The Jacobian is 1/

( ) ∫

−1/ 2 pdf βˆ = c2 det ( Ω ) a βˆ

−1

( )( )

s + b βˆ a βˆ

−∞< s <+∞

( )

a βˆ , and the density of βˆ is −1/ 2

(

( ) ( )) 2

φ s 2 + c − b βˆ /a βˆ ds .

Next we transform s using polar coordinates, s = ht 1/ 2 , where h 2 = 1 and t > 0 (the Jacobian is (1/ 2)t −1/ 2 ) we have:

( )

( ) ∫

−1/ 2 pdf βˆ = c2 det ( Ω ) a βˆ

(A.1) (1/ 2)

−1

0
∫

( )( )

ht1/ 2 + b βˆ a βˆ

h =1

(

( ) ( )) 2

φ t + c − b βˆ /a βˆ t −1/ 2 −1 / 2

( dh ) dt.

2

The integral over h 2 = 1 can be evaluated explicitly. Since h can take on only the values ±1 , the integral is in fact a sum:

2

( )( )

1 ht1/ 2 + b βˆ a βˆ ∫ 2 h2 =1

−1/ 2

( dh ) =

( )( )

1⎡ ( +1) t1/ 2 + b βˆ a βˆ ⎢ 2⎣

−1 / 2

( )( )

−1/ 2

( ) ( )

−1 / 2

+ ( −1) t1/ 2 + b βˆ a βˆ

( ) ( )

2 The term on the right-hand-side equals t 1/ 2 if t > b βˆ /a βˆ

and b βˆ a βˆ

⎤ ⎥⎦ .

if

( ) ( ) 2

0 < t < b βˆ /a βˆ . Inserting the last display into equation (A.1) we obtain

( )

( ) ( )

−1/ 2 pdf βˆ = c2 det ( Ω ) b βˆ a βˆ

−3 / 2

∫

(

( ) ( ) 2

0
(A.2) +c2 det ( Ω )

−1/ 2

( )

a βˆ

−1

∫

∫

0< t < b ( βˆ )2 /a ( βˆ )

(

2

(

( ) ( )) 2

φ t + c − b βˆ /a βˆ dt.

b ( βˆ ) 2 /a ( βˆ )
Notice that using (11) the integral in the first line is c2

( ) ( ))

φ t + c − b βˆ /a βˆ t −1/ 2 dt

( ) ( )) 2

φ t + c − b βˆ /a βˆ t −1/ 2 dt = φ

b( βˆ ) /a ( βˆ ) ) . ( ) ( )( 2

2

c −b βˆ /a βˆ

( ) ( )

2 To evaluate the integral in the second line of (A.2), let t = z − ⎡c − b βˆ /a βˆ ⎤ , (the ⎢⎣ ⎥⎦

Jacobian is 1) so that

π c2

∫

φ ( z ) dz = 1 − Φ ( c ) .

c < z <+∞

3