Let we have a exponential distribution, where $f(x,\theta)=c(\theta)e^{-(x-\theta)}$ where $x\geq 2\theta$. It is required to obtain the Maximum likelihood estimator and value of $c(\theta)$.
My approach
The value of $c(\theta)$ is turned out to be $e^{\theta}$ after solving the total probability law. Then, the range of theta is $(-\infty,\frac{min({x_1,x_2,....,x_n})}{2}]$. So, the Maximum likelihood estimator should be $\frac{X_{(1)}}{2}$. But I am confused between $\frac{X_{(1)}}{2}$ and $X_{(1)}$. Any help would be great. Thanks
Let the estimator be$$\hat\theta=E(\theta)=\arg\min_{\theta}p(\theta|x)$$also$$f(x;\theta)=e^{2\theta-x}$$for $x\ge 2\theta$ therefore $$\hat\theta=\arg\min_{\theta}p(\theta|x)\\=\arg\min_{\theta}\dfrac{p(\theta,x)}{p(x)}\\=\arg\min_{\theta}p(\theta,x)\\=\arg\min_{\theta}p(\theta)p(x|\theta)\\=\arg\min_{\theta}p(\theta)f(x;\theta)\\=\arg\min_{\theta}p(\theta)e^{2\theta-x}$$if we know nothing about $\theta$ i.e. our ambiguity is homogeneous for $\theta$ then $p(\theta)$ is constant and can be removed therefore $$\hat\theta=\arg\min_{\theta}e^{2\theta-x}=\arg\min_{\theta}2\theta-x=\dfrac{x}{2}$$since $2\theta$ is at most $x$ by assumption. Then we have$$\hat\theta=\dfrac{x}{2}$$