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I have a shifted double exponential distribution with density $$f(x;\theta)=\frac{1}{2}e^{-|x-\theta|}$$ Now I have a test statistic given by $$T(x_1,...,x_n;\theta_0) = \sum_{i=0}^n |x_i - \theta_0| - \sum_{i=0}^n |x_i - \text{ median}(x_1,...,x_n)|$$ Which is exactly the likelihood ratio test statistic. The test is given by $H_0 = \theta_0 \ \ \ \ \ \ H_1 \neq \theta_0$

Now I have to show that the distribution of this $T$ under $H_0$ does not depend on $\theta_0$. I was going to say that $Y_i = X_i - \theta_0$ to show that $Y_i \sim f(\cdot;\theta_0)$ and to show that $T(Y_1,...,Y_n;0) = T(X_1,...,X_n;\theta_0)$. How can I do this? How can I find that distribution?

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This is a location family of distributions. Under $H_0,$ $X_i\sim f(x;\theta_0).$ This implies $Y_i=X_i-\theta_0\sim f(y;0),$ i.e., $Y_i$ has pdf $f(y;0)=\dfrac{1}{2}e^{-|y|}.$ The distribution of $Y_i$ is hence free of $\theta_0$.

Median$(Y_1,\dots,Y_n)=$Median$(X_1,\dots,X_n)-\theta_0.$ As a result, $$X_i-\text{Median}(X_1,\dots,X_n)=Y_i+\theta_0-\text{Median}(Y_1,\dots,Y_n)-\theta_0.$$ So we can write $T(X_1,\dots,X_n;\theta_0)=\displaystyle\sum_{i=1}^n|Y_i|-\sum_{i=1}^n|Y_i-\text{Median}(Y_1,\dots,Y_n)|=T(Y_1,\dots,Y_n;0).$

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