MLE for the mean of a distribution?

Let $X_1, \ldots, X_n$ be a random sample of size $n$ from a distribution having the pdf $$f(x)=\frac{2x}{\theta^n}.$$ I need to find the MLE for the mean of the distribution but am not sure how.

1 Answer

The idea behind MLE is to pick the value of the parameter to maximize the chances that the currently observed sample would be generated. Assuming $\{X_k\}$ are iid and actually result in the values $\{x_k\}$, we have for $\epsilon \to 0$ $$\begin{split} p(\theta) &= \mathbb{P}\left[ X_1 \in [x_1 - \epsilon, x_1 + \epsilon], \ldots, X_n \in [x_n - \epsilon, x_n + \epsilon] \right] \\ &= \prod_{k=1}^n \mathbb{P}\left[ X_k \in [x_k - \epsilon, x_k + \epsilon] \right] \\ &= \prod_{k=1}^n f(x_k) = \prod_{k=1}^n \frac{2 x_k}{\theta^n} = \left(\frac{2}{\theta^n}\right)^n \prod_{k=1}^n x_k. \end{split}$$ It helps to maximize $$L(x) = \ln f(x) = n \ln \left(\frac{2}{\theta^n}\right) + \sum_{k=1}^n \ln(x_k)$$ with respect to $\theta$ getting the best value for $\theta$ in terms of the $x_k$'s...