6
$\begingroup$

Suppose the incomes of the employees in a firm follow a Pareto distribution as follows:$$f(x)=\dfrac{cA^c}{x^{c+1}}$$ where $x\geq A>0$.

Suppose you take a random sample of the incomes $(X_1,X_2,...,X_n)$ of $n$ employees. Find the MLE of the proportion of employees whose income falls in the interval $[I_1,I_2]$.

This is a question that came in a semester exam in my university. Here is my working:

The question actually means that we have to find the MLE of $P(I_1\leq X\leq I_2)$ where $X$ is a r.v. following the given Pareto distribution. It turns out that $$P(I_1\leq X\leq I_2)=A^c\left[\dfrac{1}{I_1^c}-\dfrac{1}{I_2^c}\right]$$assuming that $I_1\geq A$. If $I_2<A$ then $P(I_1\leq X\leq I_2)=0$. And if $I_1\leq A$ but $I_2\geq A$ then $$P(I_1\leq X\leq I_2)=P(A\leq X\leq I_2)=1-\dfrac{A^c}{I_2^c}$$

Now by Invariance property of MLE, if $\hat{\theta}$ is the MLE of $\theta$ and $\tau$ is any function then $\tau(\hat{\theta})$ is the MLE for $\tau(\theta)$.

Notice that $P(I_1\leq X\leq I_2)$ is a function of $A$ in each case. Hence, to find the MLE of the probability means to find the probability based on the MLE of $A$, which is $X_{(1)}$. Thus, our required MLE of the proportion of employees with incomes in $[I_1,I_2]$ turns out to be $$X_{(1)}^c\left[\dfrac{1}{I_1^c}-\dfrac{1}{I_2^c}\right]\space\space\space,I_1\geq X_{(1)},I_2\geq X_{(n)}$$$$1-\dfrac{X_{(1)}^c}{I_2^c}\space\space\space,I_1\leq X_{(1)}\leq I_2\leq X_{(n)}$$$$0\space\space\space,X_{(1)}\geq I_2\space\text{or}\space X_{(n)}\leq I_1$$$$1\space\space\space,I_1\leq X_{(1)}<X_{(n)}\leq I_2$$.

I believe I have heavily messed up the last part of the solution where I have to identify where the probability is what. Even if I haven't, I am not sure why I selected this way. Help is appreciated.

Improve this question
$\endgroup$
12
  • 1
    $\begingroup$ Having just spent some time looking at it closely, I don't believe your answer to be correct; I don't see how $X_{(n)}$ comes into the MLE. It seems to me you can work out the MLE of $c$ and $A$ (neither of which involve $X_{(n)}$) and substitute them into the relevant places to get the MLE of the probability.. $\endgroup$
    – Glen_b
    Commented Jul 12, 2015 at 12:46
  • $\begingroup$ Is there a reason you haven't written down the likelihood function for the observed data? $\endgroup$
    – probabilityislogic
    Commented Jul 12, 2015 at 15:00
  • $\begingroup$ As @probabilityislogic pointed out, my answer was strange. It didn't make sense, so I deleted it until I can fix it. (I really shouldn't have answered until I'd spent the time to think about it properly.) $\endgroup$
    – Glen_b
    Commented Jul 12, 2015 at 15:10
  • $\begingroup$ I did not get @probabilityislogic's comment. Could either of you two explain what he meant? $\endgroup$
    – Landon Carter
    Commented Jul 12, 2015 at 15:13
  • $\begingroup$ My point is that you are trying to find the maximum likelihood estimate of "something". The first step is to write down the function you will be maximising. You haven't done this yet. What is the probability of observing the data set you actually observed $(X_1, ...X_n) $? $\endgroup$
    – probabilityislogic
    Commented Jul 12, 2015 at 22:39

1 Answer 1

Reset to default
2
$\begingroup$

Unless I've made an error, you're very close to the right answer.

I don't see how $X_{(n)}$ comes into the MLE. It looks to me like you can work out the MLE of $c$ and $A$ (neither of which involve $X_{(n)}$) and substitute them into the relevant places to get the MLE of the probability. After removing reference to $X_{(n)}$, I believe the only relevant cases are the first three, and they can all be written in one reasonably simple expression for the MLE of the required probability:

$\qquad\min((\frac{X_{(1)}}{I_1})^\hat{c},1)-\min((\frac{X_{(1)}}{I_2})^\hat{c},1)$

where $\hat{c}$ is the usual MLE for $c$ (which I'll leave for you to deal with).

As you see, aside from minor details already mentioned, that's very close to what you had already.

Improve this answer
$\endgroup$
5
  • $\begingroup$ One thing I find strange about this answer is that it doesn't involve the observed proportion of employees with incomes in the range. If you write $ \theta=Pr (I_1 \leq X\leq I_2) $ then the likelihood function for $\theta $ of the data is $ p (X_1,..., X_n|\theta)\propto \theta^y(1-\theta)^{n-y} $ which has the "well known" maximum of $\hat {\theta}_{mle}=\frac {y}{n} $. Here $ y $ is the observed number of $ X_i $ falling in the interval $ [I_1, I_2] $. $\endgroup$
    – probabilityislogic
    Commented Jul 12, 2015 at 13:53
  • $\begingroup$ @Glen_b I think you need to take into account $X_{(n)}$ even though it is not in any MLE but because otherwise you can't carve out the regions appropriately. For example, your expression does not show what happens if $X_{(n)}<I_1$ (which also implies that the MLE of $P(I_1<X<I_2)=0$). Also, if say $I_1<X_{(1)}<X_{(n)}<I_2$ then we can safely conclude that given our sample, as all sample points lie in $[I_1,I_2]$, the MLE of the probability is $1$, which does not follow from what you have written. Afterall, if all points are in $[I_1,I_2]$ then we can trivially estimate $P(I_1<X<I_2)$ by $1$. $\endgroup$
    – Landon Carter
    Commented Jul 12, 2015 at 14:16
  • $\begingroup$ Also, considering $X_{(1)}<I_1<X_{(n)}<I_2$, we have MLE of $P(I_1<X<I_2)=$MLE of $P(X<I_2)-$MLE of $P(X<I_1)=\dfrac{X_{(1)}^c}{I_1^c}$ where $c$ is the mle of $c$ for simplicity. This is because MLE of $P(a<X<b)=X_{(1)}^c(\dfrac{1}{a^c}-\dfrac{1}{b^c})$ and $P(X<I_1)=P(X_{(1)}<X<I_1)$. At least this is what I feel. As I said, things are getting extremely messy. I am not happy with my work either: it lacks rigor. I am not happy in writing MLE of $P(a<X<b)$ as MLE of $P(X<b)-$MLE of $P(X<a)$. I am not happy in saying $P(X<I_1)=P(X_{(1)}<X<I_1)$ if $X_{(1)}<I_1$. $\endgroup$
    – Landon Carter
    Commented Jul 12, 2015 at 14:25
  • $\begingroup$ @probabilityislogic Yes, that's exceedingly strange. The answer doesn't make sense. $\endgroup$
    – Glen_b
    Commented Jul 12, 2015 at 15:08
  • $\begingroup$ @Glen_b: I do not understand why the answer does not make sense, this is standard likelihood theory. $\endgroup$
    – Xi'an
    Commented Nov 26, 2015 at 18:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.