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- Apr 14, 2013

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We have the density function $f_x(x)=\frac{2c^2}{x^3}, x\geq 0, c\geq 0$.

I want to calculate the maximum Likelihood estimator for $c$.

We have the Likelihood Function $$L(c)=\prod_{i=1}^nf_{X_i}(x_i;c)=\prod_{i=1}^n\frac{2c^2}{x_i^3}$$

The logarithm of the Likelihood function is \begin{align*}\ell (c)&=\ln L(c)=\ln \left (\prod_{i=1}^n\frac{2c^2}{x_i^3}\right )=\sum_{i=1}^n\ln \frac{2c^2}{x_i^3}=\sum_{i=1}^n \left [\ln (2c^2)-\ln (x_i^3)\right ]\\ &=n\ln (2c^2)-\sum_{i=1}^n \ln (x_i^3) =n\left (\ln 2+\ln c^2\right )-3\sum_{i=1}^n \ln (x_i)\\ &=n\left (\ln 2+2\ln c\right )-3\sum_{i=1}^n \ln (x_i)=n\ln 2+2n\ln c-3\sum_{i=1}^n \ln (x_i)\end{align*}

The deivative is $$\ell'(c)=\frac{\partial}{\partial{c}}\left (n\ln 2+2n\ln c-3\sum_{i=1}^n \ln (x_i)\right )= \frac{2n}{c}>0$$

So $\ell$ is increasing.

How can we get from here the maximum?

Or is it easier to calculate the maximum of $L(c)$ instead of $\ell (c)$ ?