THE QUADRATIC UTILITY MODEL
With the quadratic distribution utility model, legislator i’s utility for the Yea outcome on
roll call j is just:
2
uijy = -dijy
= -  Xik  O jky 
s
2
(4.4)
k=1
The difference between the two deterministic utility functions is:
s
2
2
uijy - uijn = dijy
 dijn
 2 Xik (O jkn - O jky ) +
k 1
s
 (O
jkn
- O jky )(O jkn + O jky )
k=1
Now consider the geometry of O jn  O jy and O jn - O jy where
O jn
 O j1n + O j1y 
 O j1n - O j1y 
O + O 
O - O 
j2 n
j2y 
j2 n
j2y 

 O jy 
and O jn - O jy  
.
.




.
.




 O jsn + O jsy 
 O jsn - O jsy 
1
(4.5)
So that the vector Ojn - Ojy is simply a constant times the normal vector:
jNj = Ojn - Ojy
where
1
 s
2
j =   (O jkn - O jky ) 2  if OjnNj > OjyNj or
 k 1

1
 s
2
j = -   (O jkn - O jky ) 2  if OjnNj < OjyNj
 k 1

2
(4.6)
and the vector Ojn + Ojy divided by 2 is the midpoint of the Yea and Nay outcomes:
Zj =
O jn  O jy
2
The outcomes are only identified up to parallel tracks through the space:
uiy - uin = d iy2  din2 = -(C2 + A2) + [C2 + (A+B)2] = 2AB + B2
uiy - uin = d iy2  d in2 = -[(C+D)2 + A2] + [(C+D)2 + (A+B)2] = 2AB + B2
This allows equation (4.5) to be rewritten as the vector equation:
3
uijy - uijn = d iy2  d in2 = 2j(ZjNj - XiNj ) = 2j(cj – wi)
(4.7)
If you assume that the error process is normally distributed:
ijn - ijy ~ N(0, 1)
Then the distribution of the difference between the overall utilities is
Uijy - Uijn ~ N(uijy - uijn , 1)
Hence the probability that legislator i votes Yea on the jth roll call can be rewritten as:
Pijy = P(Uijy > Uijn) = P(ijn - ijy < uijy - uijn) = [uijy - uijn]
(4.8)
Therefore, the probability of the observed choice can be written as:
Φ[2j(cj – wi)]
(4.10)
where 2j(cj – wi) > 0 if the legislator is on the same side of the cutting plane as the policy
outcome she voted for (this is picked up by the signed distance term, j).
For the quadratic utility model the probability of the observed choice can also be written
in the more traditional form:
 2 (c -w ) 
 j j i 
i


(4.12)
where i2 is the legislator specific variance, that is:
ijn - ijy ~ N(0, i2 )
This model is explored in depth in Poole (2001).
The likelihood function is:
p
q
2
L =  Pijij
i=1
C
(4.15)
j=1  =1
where τ is the index for Yea and Nay, Pijτ is the probability of voting for choice τ, and C ijτ = 1 if
the legislator’s actual choice is τ and zero otherwise.
4
It is standard practice to work with the natural log of the likelihood function:
p
q
2
=  Cij ln Pij
(4.16)
i 1 j1 1
The Quadratic-Normal (QN) model is isomorphic with the standard IRT model.
Specifically, using equations (4.5) to (4.7) let
s
s
k 1
k 1
αj = (  O2jkn   O2jky ) = 2jZjNj and
 2(O j1n  O j1y ) 
 2(O  O ) 
j2n
j2y 
βj = 
= -2jNj




 2(O jsn  O jsy ) 
(4.30)
This transformation allows the difference between the latent utilities for Yea and Nay to be
written in the same form as the item response model; namely:
y*ij = Uijy - Uijn = αj + Xiβj + ij
(4.31)
where y*ij is the difference between the latent utilities and
ij = ijn - ijy ~ N(0, 1)
Let τ=Yea, then
 2 (c -w ) 
Pijτ = [uijy - uijn] =   j j i  =   j + Xi  j 
i


To set up the Bayesian Framework let the prior distribution for the legislator ideal points
be:
Xi ~ N(0, Is ) = ξ(Xi)
5
(4.34)
Where 0 is a s length vector of zeroes and Is is an s by s identity matrix. The prior distribution
for the roll call outcome parameters is:
 j 
   ~ N(b0 , B0) = ξ(αj , βj)
 j
(4.35)
where b0 is an s+1 length vector and B0 is a s+1 by s+1 variance-covariance matrix. Clinton,
Jackman, and Rivers (2004) set b0 to a vector of zeroes and B0 to ηIs+1 where η is a large positive
constant (typically 25). The posterior distribution is:
ξ(α, β, X | Y)  f(Y | α, β, X)ξ(X)ξ(α , β)  L(α, β, X | Y)ξ(X)ξ(α , β)
(4.36)
The conditional distributions for (Y* , , , X | Y) that implement equation (4.36) are:
 N[ 0, ) (  j  X'i j , 1) if y ij  Yea


*
1) g y*  y ij |  j ,  j , X i , y ij    N (  ,0] (  j  X'i j , 1) if y ij  Nay
ij

'
 N(  , ) (  j  X i j , 1) if y ij  Mis sin g
where the subscript on the normal distribution indicates the range. For Yea and Nay the normal
is truncated as indicated and the missing data is sampled over the entire real line.
2) g  ,   j ,  j | Yj* , X, Yj   N(  j ,  j )
where Yj* and Yj are the jth columns of Y* and Y, respectively, νj is an s+1 length vector of
means, and Ξj is the s+1 by s+1 variance-covariance matrix. Specifically,
νj = Ξj  X*'Yj*  B01b0 


and
Ξj =  X*'X*  B01 


1
where X* is the p by s+1 matrix of legislator ideal points bordered
by ones; that is, X* = [Jp|X] and Jp is a p length vector of ones.
6
3) g X  X i | , , Yi* , Yi   N(t i , Ti )
where Yi* and Yi are the ith rows of Y* and Y, respectively, ti is an s length vector of means, and
Ti is the s by s variance-covariance matrix. Specifically,
ti = Ti (   Yi* ) 
and
Ti =   Is 
1
Note that if a more general prior distribution is used for the legislator ideal points,
namely:
Xi ~ N(t0, T0)
Where t0 is a s length vector of means and T0 is a s by s variance-covariance matrix, then the
expressions for ti and Ti are:
ti = Ti (   Yi* )  T01t 0 
and
Ti =   T0 
1
At the hth iteration the MCMC sampling procedure outlined in equation set (4.29) for
these conditional distributions is:
1)
1)
1)
, (h
, X (h
, y ij  , i=1,…,p , j=1,…,q
1) Sample y*(ij h) from g y*  y*ij | (h
j
j
i
ij
(h)
2) Sample (h)
from g  ,   j ,  j | Yj*(h1) , X (h1) , Yj  , j=1,…,q
j , j
(4.37)
3) Sample X (h)
from g X  X i | (h1) , (h1) , Yi*(h1) , Yi  , i=1,…,p
i
Sampling from these conditional distributions is straightforward because they are all
normal distributions with known means and variances.
7
This boils down to three simple OLS type steps:
1) Given  j ,  j , Xi , y ij , sample y *ij (the latent Yea/Nay variable) from the appropriate
normal distribution:

g y* y*ij |  j ,  j , Xi , y ij
ij

 N[ 0 , ) ( j  X'i j ,1) if y ij  Yea


  N (  ,0 ] (  j  X'i j ,1) if y ij  Nay

'
 N(  , ) ( j  Xi j ,1) if y ij  Mis sin g
2) Given Xi , y ij , y*ij , sample  j ,  j from the normal distribution:


1
1


g  ,  j ,  j | Yj* , X, Yj  N   X*'X*  B01   X*'Yj*  B 01b 0  ,  X*'X*  B 01  








3) Given  j ,  j , y*ij , y ij sample Xi from the normal distribution:

g X  Xi | , , Yi* , Yi   N   I s  (  Yi* )  ,   I s 
1
8
1
