极大似然函数为
[L(alpha,eta) = (dfrac{exp(alpha+eta)}{1+exp(alpha+eta)})^a(dfrac{exp(alpha)}{1+exp(alpha)})^b(dfrac{\,1\,}{1+exp(alpha+eta)})^c(dfrac{\,1\,}{1+exp(alpha)})^dquad(1)
]
取自然对数为
[ln L(alpha,eta) = a(alpha+eta-ln(1+exp(alpha+eta)))+b(alpha-ln(1+exp(alpha)))-cln(1+exp(alpha+eta))-dln(1+exp(alpha))^dquad(2)
]
分别关于 (alpha,eta) 求偏导，得对数似然方程组
[egin{cases}
dfrac{ ext{d}ln L}{ ext{d}alpha} = a-adfrac{exp(alpha+eta)}{1+exp(alpha+eta)}+b-bdfrac{exp(alpha)}{1+exp(alpha)}-cdfrac{exp(alpha+eta)}{1+exp(alpha+eta)}-ddfrac{exp(alpha)}{1+exp(alpha)}=0 quad (3.1)\
dfrac{ ext{d}ln L}{ ext{d}eta} = a-adfrac{exp(alpha+eta)}{1+exp(alpha+eta)}-cdfrac{exp(alpha+eta)}{1+exp(alpha+eta)}=0quad (3.2)\
end{cases}]
由 ((3.2)) 可得$$dfrac{exp(alpha+eta)}{1+exp(alpha+eta)}=dfrac{a}{a+c},quad(4)$$
代入 ((3.1)) 可得
[a-adfrac{a}{a+c}+b-bdfrac{exp(alpha)}{1+exp(alpha)}-cdfrac{a}{a+c}-ddfrac{exp(alpha)}{1+exp(alpha)}=0
]
整理可得$$dfrac{exp(alpha)}{1+exp(alpha)} = dfrac{b}{b+d}quad(5)$$
即
[b+bexp(alpha) = bexp(alpha)+dexp(alpha),
]
亦即 $$exp(alpha) = dfrac{,b,}{d},quad(6)$$
故$$alpha = ln(dfrac{b}{d}).$$
由 ((4)) 可得$$aexp(alpha+eta)+cexp(alpha+eta) = a+a exp(alpha+eta),$$
整理得 $$exp(alpha+eta) = dfrac{a}{c},$$
从而$$eta = ln(dfrac{a}{c})-alpha =ln(dfrac{a}{c})- ln(dfrac{b}{d}) = ln(dfrac{ad}{bc}).$$