I have given the following density function: $$ f_c(x) = \begin{cases} \frac{2c^2}{x^3} & \;\mbox{for}\; x\geq {c}\\ 0 &\;\mbox{else}\end{cases} $$ with $c > 0$
I need now to find a Maximum Likelihood Estimator for $c$.
My current approach is the following: $$ L(\vec{x}, c) = \prod_{i=1}^{n}{f_x(x_i, c)} = \prod_{i=1}^{n}{\frac{2c^2}{x^3}} = (2c^2)^n\prod_{i=1}^{n}{\frac{1}{x^3}} = 2^n*c^{2n} * \prod_{i=1}^{n}{\frac{1}{x^3}} $$
Now I calculate the first derivative and set it to zero: $$\frac{\delta L(\vec{x}, c)}{\delta c} = 2^n * 2n * c^{2n - 1} * \prod_{i=1}^{n}{\frac{1}{x^3}} = 0$$
But the only way to fulfill the equation is to set c to 0 which makes no sense to me.
I also tried the same approach with the Log-Likelihood Function: $$2^n*c^{2n} * \prod_{i=1}^{n}{\frac{1}{x_i^3}} = n * ln(2) + 2n * ln(c) + \sum\limits_{i=1}^{n} [-3 * ln(x_i)]$$
And first derivative again: $$\frac{\delta L(\vec{x}, c)}{\delta c} = \frac{2n}{c} = 0$$ But the same thing again only $c = 0$ works...
Does someone have an idea what I'm doing wrong?
Yes. As you said :
$$ \mathcal L(x;c) = \prod_{i=1}^n f_x(x_i;c)$$
So it gives :
$$ \mathcal L(x;c) = \prod_{i=1}^n \frac{2c^2}{x_i^3}\mathbb I_{\{x_i\geqslant c\}} $$
and you have to write the term $\mathbb I_{\{x_i\geqslant c\}}$
It leads to :
$$ \mathcal L(x;c) = (2c^2)^n \prod_{i=1}^n \frac{1}{x_i^3}\mathbb I_{\{x_i\geqslant c\}} = (2c^2)^n \mathbb I_{\{\min x_i\geqslant c\}}\prod_{i=1}^n \frac{1}{x_i^3}$$
Indeed, the terms $\mathbb I_{\{\min X_i \geqslant c \}}$ are all equal to $1$ if, and only if $\mathbb I_{\{\min x_i\geqslant c\}}$ is equal to $1$. Since you want to find the maximum of $\mathcal L$ and since $c \mapsto (2c^2)^n$ is increasing on $\mathbb R_+$ and since $c>0$ you have :
$$ \hat{c} = \min X_i$$