Show the likelihood equation for Logistic$(\theta,1)$ distribution has a unique root. Find the asymptotic distribution of MLE $\hat\theta$. 
Let $X_1,\ldots, X_n$ i.i.d. from the Logistic$(\theta,1)$ distribution.
(a) Show that the likelihood equation has a unique root.
(b) Find the asymptotic distribution of MLE $\hat\theta$.

(a) the PDF of logistic dist. $f(x; \theta, 1) = \frac{e^{\theta-x}}{(1+e^{\theta-x})^2}$
So, $l(\theta) = \sum_{i=1}^{n} \ln f_\theta(X_i) = \sum_{i=1}^{n}(\theta-X_i)-2\sum_{i=1}^{n}\ln(1+e^{\theta-Xi})$
$l'(\theta) = n-2\sum_{i=1}^{n}(\frac{e^{\theta-X_i}}{1+e^{\theta-X_i}})=0 \iff \frac{n}{2}=\sum_{i=1}^{n}(\frac{e^{\theta-X_i}}{1+e^{\theta-X_i}})=\sum_{i=1}^{n}(\frac{1}{1+e^{X_i-\theta}})$
let $g(\theta)=\sum_{i=1}^{n}(\frac{1}{1+e^{X_i-\theta}})$
$\Rightarrow g(\theta)\rightarrow \sum_{i=1}^{n}(1) = n$ as $\theta \rightarrow \infty$
$\Rightarrow g(\theta)\rightarrow 0$ as $\theta \rightarrow -\infty$
$\Rightarrow$ solution $\hat{\theta}$ is unique.
(b)
since $\hat{\theta}$ is unique, so $\hat{\theta}$ is MLE
So, $\hat{\theta}$ is is asymptotic efficient by Thm 3.10.
$\sqrt{n}(\hat{\theta}-\theta) \xrightarrow{\mathcal{L}} N(0, \frac{1}{I(\theta)})$
Need to Calculate $I(\theta)$
since $f_\theta(x) = \frac{e^{\theta-x}}{(1+e^{\theta-x})^2}$, $l(x) = \ln f_\theta(x) = (\theta-x)-2\ln(1+e^{\theta-x})$
$$\frac{\partial{\ln f_\theta(x)}}{\partial\theta}=1-2\frac{e^{\theta-x}}{1+e^{\theta-x}}$$
$$I(\theta) = E\left[1-2\frac{e^{\theta-x}}{1+e^{\theta-x}}\right]^2=\int_{-\infty}^{\infty}\left(\frac{1-e^{\theta-x}}{1+e^{\theta-x}}\right)^2.\frac{e^{\theta-x}}{(1+e^{\theta-x})^2}dx$$
Let $t=e^{\theta-x}$, $t \in (0, \infty)$,
$\Rightarrow -\int_{0}^{\infty} \frac{(1-t)^2}{(1+t)^4}dt=-\int_{0}^{\infty} \frac{1-t}{(1+t)^3}dt =[-\frac{1-t}{2\left(1+t\right)^2}+\frac{1}{2\left(1+t\right)}]_{0}^{\infty} = 0$
Is there error in my calculation? Thank you!
 A: For finding the information function, it is perhaps easier to work with the alternative formula $$I(\theta)=-E_{\theta}\left[\frac{\partial^2}{\partial\theta^2}\ln f_{\theta}(X_1)\right]$$
For $\theta\in \mathbb R$, you have $$f_{\theta}(x)=\frac{e^{-(x-\theta)}}{(1+e^{-(x-\theta)})^2}\quad ,\,x\in \mathbb R$$
Then as you have found, $$\frac{\partial}{\partial\theta}\ln f_{\theta}(x)=1-2\left(\frac{e^{-(x-\theta)}}{1+e^{-(x-\theta)}}\right)$$
This can be rewritten as $$\frac{\partial}{\partial\theta}\ln f_{\theta}(x)=1-2\left(1-\frac1{1+e^{-(x-\theta)}}\right)=\frac 2{1+e^{-(x-\theta)}}-1$$
Therefore, $$\frac{\partial^2}{\partial\theta^2}\ln f_{\theta}(x)=-\frac{2e^{-(x-\theta)}}{(1+e^{-(x-\theta)})^2}$$
And so
\begin{align}
I(\theta)&=2 E_{\theta}\left[\frac{e^{-(X_1-\theta)}}{(1+e^{-(X_1-\theta)})^2}\right]
\\&=2\int_{-\infty}^{\infty} \left\{\frac{e^{-(x-\theta)}}{(1+e^{-(x-\theta)})^2}\right\}^2 dx
\\&=2\int_{-\infty}^{\infty} \left\{\frac{e^{-y}}{(1+e^{-y})^2}\right\}^2 dy
\end{align}
Now if you substitute $t=1/(1+e^{-y})$, this eventually reduces to
$$I(\theta)=2\int_0^1 t(1-t)\,dt=\frac13$$
