MHB Finding the Maximum Likelihood Estimator for a Density Function

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The discussion centers on calculating the maximum likelihood estimator (MLE) for the parameter \( c \) in the density function \( f_x(x) = \frac{2c^2}{x^3} \). The likelihood function is derived as \( L(c) = \prod_{i=1}^n f_{X_i}(x_i; c) \), and its logarithm is expressed as \( \ell(c) = n \ln 2 + 2n \ln c - 3 \sum_{i=1}^n \ln(x_i) \). The derivative \( \ell'(c) \) indicates that the log-likelihood function is increasing, suggesting that the maximum occurs at the boundary of the domain. However, participants highlight that the density function does not integrate to one, indicating it cannot be a valid probability density function, thus invalidating the likelihood analysis.
mathmari
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Hey! :o

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 $\ell (c)$ ?

(Wondering)
 
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mathmari said:
Hey! :o

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

Hey mathmari!

Shouldn't we have that the integral of the density function is $1$?
It seems to me that whatever we pick for $c$ this is not the case. (Worried)

mathmari said:
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 $\ell (c)$ ?

It means that the maximum is at the right boundary of the domain, doesn't it?
And we could already observe that in $L(c)$ since it's a linear function of $c^{2n}$.

However, it seems that the density function does not satisfy the requirements of a density function.
Consequently we cannot do a likelihood analysis. (Thinking)
 
Please go back and check this problem again! First, as I like Serena said, there is NO value of c that makes this a probability density function. Second, you are trying to take a product over integer values of n but there is no "n" in the problem! The only variable is "x" and it is a continuous variable.
 
Ah ok! Thank you! (Smile)
 
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