逻辑概率分布的推导过程

f(εiy
- εin ) = f(y) =
e -y (1+e-y )2
,
(7)
To get the distribution function:
Hence
∫t
e-y
t1
1
F(y<t) = -∞ (1+e-y )2 dy = −∞ 1 + e−y = 1+e-t
P(ε - ε < e - e ) = 1 + e 1 in iy
Set v = e-z (1 + e−y )
Note that 0 < v < ∞ because 0 < e-z < ∞ as -∞ < z < ∞ Hence, ln(v) = -z + ln(1+e-y )
and z = ln(1+e-y ) - ln(v)
Therefore, ∂z = - 1 and J = ∂z = 1
P(εin - εiy < e−di2y - e−di2n ) = P(εiy - εin > e−di2n - e−di2y )
(3)
In order to get the distribution of εiy - εin set up the joint density and then do a change
of variables (note that the distribution of εin - εiy will be the same as the distribution
Hale Waihona Puke Baidu
of εiy - εin ):
f (ε , ε ) = e e iy in
-(εiy + εin ) −(e− εiy + e-εin )
− y −2 2 −{(1+e− y )-1v[1+e− y ]}
1dv
=
+∞
e-y (1 + e−y )−2 v2e−v
1 dv
=
0
v
0
v
+∞
∫ e-y (1 + e−y )−2 ve−vdv = e-y (1 + e−y )−2 Γ(2) = e-y (1 + e−y )−2 (6) 0
Therefore,
− di2y
− d i2n
−
-(e
di2y
- e− di2n )
(8)
3
(4)
1
Set y = εiy - εin and z = εin
Hence εiy = y + z and εin = z
and the Jacobian is:
Hence
∂εiy ∂y J= ∂εin ∂y
∂εiy
∂z 1 1
=
=1
∂εin 0 1
∂z
f (y + z , z) = e e = e e e -(y + 2z) −(e−(y+z) + e-z )
-y −2z −[e− z (1+e− y )]
To get the distribution of y = εiy - εin integrate out z:
+∞
∫ e e e dz -y −2z −[e−z (1+e−y )]
(5)
-∞
This requires another change in variables:
∂v v
∂v v
Hence
2
+∞
1 -y −2z −[e− z (1+e− y )]
∫ e e e dz = e ∫ e e e vdv = -∞
+∞ -y
−2ln(1+e− y ) 2ln( v ) −{e−[ln(1+e− y )−ln( v )][1+e− y ]}
0
∫ ∫ +∞
e (1 + e ) v e -y
and
d
2 in
are the squared distances from the ith legislator to the Yea and Nay
choices and the ε are distributed as the logarithm of the inverse of an exponential variable
Logit_Probability.doc
Derivation of the Logit Probability
Utility function for Yea and Nay choices:
Uiy = e-di2y + εiy and Uin = e-di2n + εin
(1)
where
di2y
(Dhrymes, 1978, p. 342). Namely
f (ε) = e e-ε −e−ε , -∞ < ε < +∞
(2)
The probability that the legislator will choose the Yea alternative is:
P(Uiy > Uin ) = P(e−di2y + εiy > e−di2n + εin ) = P(e−di2y - e−di2n > εin - εiy ) =