# Unbiased estimator/MSE from a Gamma dist.

1. Mar 9, 2010

### daoshay

I have given some serious effort to working out and understanding the MLE of a distribution. From the distribution f(x;$$\theta$$)= $$x^{3}$$$$e^{-x/\theta}$$/(6$$\theta$$$$^{4}$$), I have gotten the MLE theta-hat = xbar/4

I have a lot of difficulty figuring out if it is an unbiased estimator or not. How do I determine the expected value of theta-hat?

2. Mar 9, 2010

### SW VandeCarr

You're saying MLE in the text and MSE in the title. For the Mean Squared Error, you can get a Bayesian minimum. (see Gamma Distribution in the Wiki). You make a Maximum Likelihood Estimate of a parameter, not a distribution. You can write a likelihood function for a distribution.

Last edited: Mar 9, 2010
3. Mar 10, 2010

### daoshay

I'm sorry, I was rushed while typing that up and I'm afraid I wasn't clear. I found the MLE for the parameter theta. I am supposed to test it for all theta for bias and then find the MSE of theta-hat.

Based on the gamma family, the mean of this distribution should be 4theta --> theta = mu/4
Bias is E(theta-hat)-theta, right?

Now, I'm supposed to find the MSE of theta-hat E[(theta-hat -theta)^2] right?
So, am I supposed to use the value of theta based on the distribution? I'll check my work on the expansion and check back later. Thanks for your patience. =)

4. Mar 10, 2010

### SW VandeCarr

Your equation has only one parameter so it's a simple exponential distribution (or gamma with k=1). $$\theta$$ is the reciprocal of the rate parameter which is often written as $$\lambda$$. So $$E|X|=\theta=\frac{1}{\lambda}$$. $$Var|X|=\frac{1}{\lambda^2}$$. For MSE use the Baysian minimum that I referred to earlier. Are you using any dummy (or real) data here?

EDIT: I'm using $$\theta$$ above as the mean of the distribution $$F(x;\theta)$$. Otherwise, this is not making any sense to me.

Last edited: Mar 11, 2010
5. Mar 12, 2010

### SW VandeCarr

Are you sure that shouldn't be $$f(x|\theta)$$? ie $$L(\theta|x)=f(x|\theta)$$.

EDIT: OK. I see that k=4. Now for some N you can estimate $$\theta$$ for the gamma distribution. As far as I know the ML estimate of $$\theta$$ is unbiased, assuming an unbiased sample. $$\hat{\theta} =\frac{1}{kN}\sum_{i=1}^{N}x_{i}$$.

Last edited: Mar 13, 2010
6. Mar 13, 2010

### SW VandeCarr

I looked further into the issue of estimator bias for the gamma distribution. Numerical analysis for moderate sized samples indicate a small "upward" or positive bias for ML estimates of the mean and variance.

http://web.uvic.ca/econ/research/papers/ewp0908.pdf

The bias of $$\hat{\theta}$$ is shown on page 8, equation (14).

Last edited: Mar 14, 2010