Maximum likelihood for a single sample of normal distribution

Question

Let $\theta \in [0,1]$ and $x$ sampled from $N(\theta, 1)$. what is $\hat \theta$?

So the solution suggests:

$$L = \frac{1}{2\pi} \exp \left( -\frac{(x-\theta)^2}{2} \right)$$

Then, $$l = -(x-\theta)^2 /2 + const$$

$$\frac{\partial l}{\partial \theta} = -(x-\theta) = 0 \implies \theta = x$$

And of course, we bound it in the range $[0,1]$.

I'm used to just taking the derivative of $L$.

I don't understand what $l$ is and I'd be glad for an explanation.

Answer 1

$$l = \log L$$

It is the log-likelihood funciton.

The constant is just $-\log 2\pi$.

Answer 2

It is the log-likelihood, $l=\log L$. We do this so that the likelihood is easier to differentiate and maximize.

In this case we have $$l=\log \left(\frac{1}{2\pi} \exp \left( -\frac{(x-\theta)^2}{2} \right)\right)=\log\left(\frac1{2\pi}\right)-\frac{(x-\theta)^2}{2}$$