# Thread: Point Estimates

1. ## Point Estimates

Point Estimate Question

Hi,
May I confirm if my answers to the questions are correct?

For Q1) I used the following limits for my integral
$\displaystyle \int_{x_{0}}^{0} \alpha x_{0}^{\alpha}\ x^{-\alpha-1}\ dx = 1$

Q2) My final answer is:
$\displaystyle \alpha = \frac{\mu_{1}}{\mu_{1} - x_{0}}$

Q3) My answer after solving the derivative of the log likelihood
$\displaystyle \alpha = \frac{1}{\log \left( \frac{x}{x_{0}} \right)}$

Thank you in advance
Linda

2. ## Re: Point Estimates

the upper bound should be infinity not zero
I get a slightly differenet answer for part 2
and for the MLE, do we have a sample or just one observation?

3. ## Re: Point Estimates

Hi Matt,

Thank you for pointing out that the limit bound should be infinity!

May I ask what you got for part 2?
They way I arrived at my answer was I found the expectation of the distribution first:
$\displaystyle E[X] = \alpha x_{0}^{\alpha} \int_{x_{0}}^{\infty} x \times x^{-\alpha-1}\ dx$
$\displaystyle =\frac{\alpha x_{0}}{\alpha -1}$

I then substituted the $\displaystyle E[X]$ with the first moment $\displaystyle \mu_{1}$ and then rearranged the above term to obtain $\displaystyle \alpha$

w.r.t MLE, that was how the question was given, we had no further data to work with.

4. ## Re: Point Estimates

it's math eagle
the mean is correct, next you solve for alpha

$\displaystyle \frac{\alpha x_{0}}{\alpha -1}=\bar X$

$\displaystyle \frac{\alpha -1}{\alpha}= {x_0\over \bar X}$

$\displaystyle 1-\frac{1}{\alpha}= {x_0\over \bar X}$

ok, now you finish

5. ## Re: Point Estimates

Originally Posted by matheagle
it's math eagle
Oh no! I am so sorry about getting you're alias incorrect. Forgive me for my dumbness.
I know someone with the surname Heagle, and so originally read it as Mat Heagle. My apologies!!!

$\displaystyle 1-\frac{1}{\alpha}= {x_0\over \bar X}$

ok, now you finish
$\displaystyle \\ -\frac{1}{\alpha} = \frac{x_{0}}{\bar{X}} - 1 \\ \frac{1}{\alpha} = 1 - \frac{x_{0}}{\bar{X}} \\ \frac{1}{1 - \frac{x_{0}}{\bar{X}}} = \alpha \\ \alpha = \frac{\bar{X}}{\bar{X}-x_{0}}$

I see that I should have substituted $\displaystyle \bar{X}$ for $\displaystyle \mu_{1}$. Thank you!!

As a further I have updated my answer for the MLE to:
$\displaystyle \alpha = \frac{n}{\sum\log \left( \frac{x_{i}}{x_{0}} \right)}$

6. ## Re: Point Estimates

its ok, I've been called much worse, its just not matt
again with the MLE, usually you have a sample of i.i.d. rvs.
But sometimes it's just one observation.
It isn't clear, but most likely its the sample of n observations.