Likelihood function for only one trial

Question

I have a trial $\mathbb X = (X_1,X_2,....X_n)$. $X_i$ has specified distribution with unknow parameter $\theta$. I want to find an estimator of this parameter. So I can use methods like Likelihood function $$L(\theta; \mathbb X) = L(\theta; X_1,X_2,...X_n)$$. But I have only a trial which looks like: $\mathbb X = X_1$ (so I made only one experiment). Can I still use likelihood function to find the estimator? If I will use that it will be: $$L(\theta, \mathbb X) = L(\theta, X_1) = P_{\theta}(X_1 = x_1).$$ Is this still correct?

Answer 1

What you are talking about is in principe correct but be careful about the probability $P_\theta(X_1=x_1)$ which is $=0$ when your density is continuous. Actually you are looking for $$\hat{\theta}:=\arg\max_\theta f_\theta({\bf x})$$ where ${\bf x}:=x_1$ a scalar.