Source of the question: the Example 7.2.8 of Statistical Inference Book by Casella and Berger.
Let $X_1,...,X_n$ be iid $n(\theta,1)$, where it is known that $\theta$ must be nonnegative. With no restrictions on $\theta$, we saw that the MLE of $\theta$ is $\bar{X}$; however, if $\bar{X}$ is negative, it will be outside the range of the parameter.
I don't understand the below and I need help. How to get the MLE of $\theta$ in normal distribution when $\theta$ is known nonnegative?
If $\bar{x}$ is negative, it is easy to check that the likelihood function $L(\theta|x)$ is decreasing in $\theta$ for $\theta \geq 0$ and is maximized at $\hat{\theta}=0$. Hence, in this case, the MLE of $\theta$ is
$$\hat{\theta}= \bar{X} \text{ if } \bar{X} \geq 0 \text{ and > } \hat{\theta}=0 \text{ if } \bar{X} < 0.$$
Normally, you're trying to maximize the likelihood function $L(\theta, x)$.
Which is defined as: $$L(\theta, x) = \prod f_\theta(x_i)$$ Where $f_\theta$ is the density function of $\mathcal{N}(\theta, 1)$. However, most of the time, we prefer to maximize the logarithm of the likelihood function, as it is just so much more convenient. $$l(\theta, x) = \log L (\theta, x)=\sum \log f_\theta(x_i)$$ The big advantage about this log likelihood function is that has a close relationship with the likelihood function: a lower/higher log-likelihood implies a lower/higher likelihood and vice verse. So we can just focus on this log likelihood. Working out the math, we find (while leaving out some
$$l(\theta, x) = -n\log (\sqrt{2\pi}) + \sum -0.5 (x_i - \theta)^2$$
If we try to optimize it, we can just take the derivative of this function w.r.t. $\theta$ to find: $$l'(\theta, x) = - n\cdot \theta + \sum x_i$$
Which is zero if $\theta = \overline x$, therefore this is the optimizer. However, the big question is, how do we maximize the log-likelihood function under the constraint that $\theta \geq 0$?
Basically one way to see this is by defining that $l(\theta, x) = -\infty$ if $\theta < 0$: it cannot happen, so there is absolutely no likelihood attributed to it. So the next question will be: for which value for $\theta \geq 0$ does $l$ give the highest score?
As you can see, if $\overline x < 0$ and $\theta \geq 0$, the derivative $l'$ becomes smaller than 0. In other words, $l$ is strictly decreasing. So if $\overline x < 0$, then $\theta = 0$ will always have a higher log-likelihood than any $\theta > 0$.
Hence the conclusion you can draw is that the maximum likelihood estimator under the constraints posed is $\theta = \overline x$ if $\overline x > 0$, and $\theta = 0$ else.