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[SOLVED] Maximum Likelihood estimator

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,003
Hey!! :eek:

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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,684
Hey!! :eek:

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)

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)
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
368
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.
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,003
Ah ok! Thank you!! (Smile)