Monotone likelihood ratio for logistic distribution $f(x;\theta) = e^{-x - \theta} (1 + e^{-x - \theta})^{-2}$

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Balkys

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Balkys Asks: Monotone likelihood ratio for logistic distribution $f(x;\theta) = e^{-x - \theta} (1 + e^{-x - \theta})^{-2}$
This is a question from Problems 9.4 of An Introduction to Probability and Statistics by Rohatgi.

Let $X$ have logistic distribution with the PDF $$f(x;\theta) = e^{-x - \theta} (1 + e^{-x - \theta})^{-2}, x \in \mathbb{R}.$$ Does ${f}$ have a monotone likelihood ratio?
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I have tried to look at this related question, but the theory was beyond what I have learned.

My attempt: Let $\theta_1, \theta_0 \in \mathbb{R}$. Let $\theta_1 > \theta_0.$

Consider the likelihood ratio, $T_{NP}(\mathbf{x}) = \dfrac{f_n(\mathbf{x};\theta_1)}{f_n(\mathbf{x};\theta_0)} = \dfrac{ e^{-n \bar{x} - n \theta_1} \prod (1+e^{-x_i - \theta_1})^{-2} }{ e^{ -n \bar{x} - n \theta_0} \prod (1+e^{-x_i - \theta_0})^{-2} }.$

After simplifying more, I get $T_{NP}(\mathbf{x}) = e^{n(\theta_0 - \theta_1)} \prod \left(\dfrac{1+e^{-x_i - \theta_0}}{1+e^{-x_i - \theta_1}}\right)^2.$

I found that $\dfrac{d}{dx} \dfrac{1+e^{-x - \theta_0}}{1+e^{-x - \theta_1}}$ was negative for all $x$.

I tried to relate this to what I had done so far. I used that $\log$ is monotone increasing. Taking the log of the likelihood ratio, I get that the log-likelihood ratio is decreasing in $\sum \log \left(\dfrac{1+e^{-x_i - \theta_0}}{1+e^{-x_i - \theta_1}}\right).$ However, I am not sure how to get the required test statistic that is independent of the parameter $\theta$.

I think that a relevant result is that for a distribution from the exponential family, $f(x;\theta) = c(\theta)h(x)\exp(\pi(\theta)T(x))$, the likelihood ratio is monotone in $T(x)$.

I tried to use this by writing $f(x;\theta) = \exp{(-x-\theta)} \exp{(-2\log(1+e^{-x-\theta}))}.$ But I cannot identify $T(x)$ from this form.

Could someone please help me? Thank you very much.

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