I have the following density function:
$f(x;\omega) = \omega*x^{(\omega-1)}*I_{(0,1)}(x)$ for $\omega > 0$
First I want to have the Likelihoodfuntion, which is
$\prod_{i=1}^n f(x_i;\omega)$
I used the log and got:
$ln L(\omega) = ln \sum_{i=1}^n\omega*x^{(\omega -1)} $
Im wondering if this is correct or not, since when I wanna get the ML estimator (derive the function and set it to zero) I cant get a correct result.
thanks for any help in advance
On
I assume $I_{(0,1)}$ is a characteristic function, and $*$ is multiplication. If you ignore $I$ when finding the maximum likelihood, make sure to take it into consideration later.
Your $\ln$ should be inside the $\sum$. This comes from the property:
$$\ln(xy) = \ln x + \ln y$$
for $x, y > 0$. If you wanted to convince yourself that it's true for any number of factors, you can use induction, but you probably believe it. Perhaps the summation notation confused you.
$$\ln(x_1x_2x_3 \ldots x_n) = \ln\left({\prod_{i = 1}^n x_i}\right)$$ $$= \ln x_1 + \ln x_2 + \ln x_3 + \ldots + \ln x_n = \sum_{i = 1}^ n \ln x_i$$
It's just better to write out everything step by step.
$$\begin{align*} \mathcal L(\omega \mid \boldsymbol x) &= \prod_{i=1}^n \omega x_i^{\omega - 1} \mathbb 1(0 < x_i < 1) \\ &= \omega^n \biggl( \prod_{i=1}^n x_i \biggr)^{\omega - 1} \mathbb 1 (0 < x_{(1)}) \; \mathbb 1 (x_{(n)} < 1) , \end{align*}$$ where $x_{(1)} = \min_i x_i$ is the first order statistic and $x_{(n)} = \max_i x_i$ is the last order statistic. Then the log-likelihood is $$\ell(\omega \mid \boldsymbol x) = n \log \omega + (\omega - 1) \sum_{i=1}^n \log x_i = n \log \omega + (\omega - 1) \overline{\log x},$$ if $0 < x_{(1)} \le x_{(n)} < 1$. Taking the derivative with respect to $\omega$ gives $$\frac{\partial \ell}{\partial \omega} = \frac{n}{\omega} + \overline{\log x}$$ and solving for the critical point gives the maximum likelihood $\hat \omega$.