这是一份umass麻省大学 STAT 310作业代写的成功案例
$$
L(\pi)=\left(\begin{array}{c}
N \
x
\end{array}\right) \pi^{x}(1-\pi)^{N-x}
$$
and the log likelihood is
$$
\ell(\pi)=\log \left(\begin{array}{c}
N \
x
\end{array}\right)+x \log \pi+(N-x) \log (1-\pi) .
$$
Thus the total score is
$$
U(\pi)=\frac{d \ell}{d \pi}=\frac{x}{\pi}-\frac{(N-x)}{1-\pi}=\frac{x-N \pi}{\pi(1-\pi)} .
$$
When set equal to zero, the solution yields the $M L E$ :
$$
\widehat{\pi}=x / N=p .
$$
The observed information is
$$
i(\pi)=\frac{-d^{2} \ell}{d \pi^{2}}=\frac{x}{\pi^{2}}+\frac{(N-x)}{(1-\pi)^{2}}
$$
Since $E(x)=N \pi$, the expected information is
$$
I(\pi)=E[i(\pi)]=\frac{E(x)}{\pi^{2}}+\frac{E(N-x)}{(1-\pi)^{2}}=\frac{N}{\pi(1-\pi)}
$$
STAT 310 COURSE NOTES ：
Now consider the score test for $H_{0}: \beta=0$ in the conditional logit model. It is readily shown that
$$
U(\beta){\mid \beta=0}=(f-g) / 2 $$ and that $$ I(\beta){\mid \beta=0}=\frac{E(M)}{4} \cong \frac{M}{4}
$$
Therefore, the efficient score test is
$$
X^{2}=\frac{(f-g)^{2}}{M}
$$
which is McNemar’s test.