Consider a random sample of size $n$ from a distribution with probability density function (pdf) is given by
$$f(x;\theta)=\left\{\begin{matrix} \sqrt{\frac{2}{\pi}}e^{-\frac{1}{2}(x-\theta)^2}&\text{if }\space x\geq \theta \\ 0&\text{elsewhere} \end{matrix}\right.$$
Find the maximum likelihood estimator of $\theta$.
My attempt,
$$L(\theta)=\prod_{i=1}^{n}\sqrt{\frac{2}{\pi}}e^{-1/2(x_i-\theta)^2}$$
$$=\exp(-\frac{1}{2}\sum_{i=1}^{n}(x_i-\theta)^2)\cdot constant$$
$$\log L=-\frac{1}{2}\sum_{i=1}^{n}(x_i-\theta)^2+constant$$
$$\frac{\delta}{\delta \theta}\log L=\sum_{i=1}^{n}(x_i-\theta)$$
$$=\sum_{i=1}^{n}x_i-n\theta$$
$$n\theta=\sum_{i=1}^{n}x_i$$
$$\hat{\theta}=\bar{x}$$
Am I correct?