# Thread: integration and moment estimators

1. ## integration and moment estimators

Well, the question and my working out is shown below. I think that I know how to do the moment estimator maybe.. but I'm just having trouble with the integration part. I have a STRONG feeling that I'm doing it wrong. Can someone please help me?

Let X1 .. Xn be a random sample with pdf f(x) = e^-(x - theta), x> theta. Find the moment estimator and the maximum likelihood estimator of the parameter theta.

2. Originally Posted by gconfused
Well, the question and my working out is shown below. I think that I know how to do the moment estimator maybe.. but I'm just having trouble with the integration part. I have a STRONG feeling that I'm doing it wrong. Can someone please help me?

Let X1 .. Xn be a random sample with pdf f(x) = e^-(x - theta), x> theta. Find the moment estimator and the maximum likelihood estimator of the parameter theta.
Where have the 0 and 1 come from? Are you saying that the support of the distribution is [0, 1]? In which case you get the parameter by putting the integral of the pdf over [0, 1] equal to 1.

3. Well, I'm not sure where the 0 and 1 came from. I'm really lost, and just put that there because all the examples in my book use 0 and 1 so i just assumed.. well i guess Im wrong then SIGH

So um how do you know what the range is then? ><

4. Originally Posted by gconfused
Well, I'm not sure where the 0 and 1 came from. I'm really lost, and just put that there because all the examples in my book use 0 and 1 so i just assumed.. well i guess Im wrong then SIGH

So um how do you know what the range is then? ><
It looks like the support is $\displaystyle [\theta, + \infty)$.

5. yup, the support is theta to infinity.
I noticed that immediately.
It's a shift of an exponential with mean equal to 1.
Y=X-theta.

6. Originally Posted by matheagle
yup, the support is theta to infinity.
I noticed that immediately.
It's a shift of an exponential with mean equal to 1.
Y=X-theta.

Hi guys, also looking at this question.

I got theta = X(bar) -1

as my answer to the integral from theta to infinity.

Also, how do I get started on finding the likelihood function?

7. $\displaystyle f_X(x)=\exp^{(x-\theta)}u(x-\theta)$ where $\displaystyle u(.)$ is the unit step function.

The likelihood function is
$\displaystyle L(\theta) = \prod_{i=1}^{n}f_X(x_i)$

Hint: Find the log-likelihood function. There is no need to differentiate to find the MLE(this is usually the case when the parameter is the upper/lower bound of the support)

8. and in your integration you should use $\displaystyle e^{a+b}=e^ae^b$
so you can pull the $\displaystyle e^{\theta}$ out of the integral.

9. Thanks guys, I've solved this problem